NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry

FAQ on Chapter 5 Introduction to Euclid’s Geometry

What are the benefits of NCERT Solutions for Chapter 5 Introduction to Euclid’s Geometry Class 9 NCERT Solutions?

Chapter 5 Introduction to Euclid’s Geometry Class Maths NCERT Solutions is nice way through which one can make themselves doubt free and understand all the important points of the chapter. We have provided detailed NCERT Solutions of every questions that you can check here.

What is a theorem?

A mathematical statement whose truth has been logically established is called a theorem.

What is the Euclid’s second axiom?

If equals be added to equals, the wholes are equal.

How many dimensions does a surface has?

A surface has 2 dimensions.

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

January 17, 2021, 9:33 pm

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles| PDF Download

Here you will find Class 9 Maths Chapter 6 Lines and Angles NCERT Solutions that will help you in knowing the important concepts provided in the chapter and completing your homework on time. You can also Download PDF of Chapter 6 Lines and Angles NCERT Solutions Class 9 Maths to practice in a better manner. Through help of these Chapter 6 NCERT Solutions, you can raise the level of quality of your studies and at the same time it will also help in solving the difficulties that lie ahead with ease.

There are variety of concepts given in this chapter 6 Class 9 Maths textbook that will develop your necessary skill to solve more and more questions not only in this class 9 but upcoming board exams also. It will introduce to a news set of topics through which you can understand the geometry better. These NCERT Solutions will provide good experience and provide opportunities to learn new things.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Page No: 96

Exercise 6.1

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer
Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°

Page No: 97

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer

Given,
∠POY = 90° and a : b = 2 : 3
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°

3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer

Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° - ∠PQR --- (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° - ∠PRQ
⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Answer

Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Answer

Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS --- (i)
∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)
⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS
⇒ ∠QOS - ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.

6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Answer

Given,
∠XYZ = 64°
YQ bisects ∠ZYP

∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠ZYP = 2∠ZYQ
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58°
⇒ ∠XYQ = 122°
also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
⇒ ∠QYP = 302°

Page No: 103

Exercise 6.2

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

Answer

x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.

Page No: 104

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Answer

Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y + ∠O = 180° (Linear pair)
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°

3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer

Given,
AB || CD
EF ⊥ CD
∠GED = 126°
A/q,
∠FED = 90° (EF ⊥ CD)
Now,
∠AGE = ∠GED (Since, AB || CD and GE is transversal. Alternate interior angles.)
∴ ∠AGE = 126°
Also, ∠GEF = ∠GED - ∠FED
⇒ ∠GEF = 126° - 90°
⇒ ∠GEF = 36°
Now,
∠FGE +∠AGE = 180° (Linear pair)
⇒ ∠FGE = 180° - 126°
⇒ ∠FGE = 54°

4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]

Answer

Given,

PQ || ST, ∠PQR = 110° and ∠RST = 130°

Construction,

A line XY parallel to PQ and ST is drawn.

∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY + ∠QRS = 180°
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°

5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Answer

Given,
AB || CD, ∠APQ = 50° and ∠PRD = 127°
A/q,
x = 50° (Alternate interior angles.)
∠PRD + ∠RPB = 180° (Angles on the same side of transversal.)
⇒ 127° + ∠RPB = 180°
⇒ ∠RPB = 53°
Now,
y + 50° + ∠RPB = 180° (AB is a straight line.)
⇒ y + 50° + 53° = 180°
⇒ y + 103° = 180°
⇒ y = 77°

6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Answer

Let us draw BE ⟂ PQ and CF ⟂ RS.

As PQ || RS
So, BE || CF

By laws of reflection we know that,

Angle of incidence = Angle of reflection

Thus, ∠1 = ∠2 and ∠3 = ∠4 --- (i)

also, ∠2 = ∠3 (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C) --- (ii)

From (i) and (ii),

∠1 + ∠2 = ∠3 + ∠4

⇒ ∠ABC = ∠DCB

⇒ AB || CD (alternate interior angles are equal)

Page No: 107

Exercise 6.3

1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer

Given,
∠SPR = 135° and ∠PQT = 110°
A/q,
∠SPR +∠QPR = 180° (SQ is a straight line.)
⇒ 135° +∠QPR = 180°
⇒ ∠QPR = 45°
also,
∠PQT +∠PQR = 180° (TR is a straight line.)
⇒ 110° +∠PQR = 180°
⇒ ∠PQR = 70°
Now,
∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)
⇒ 70° + 45° + ∠PRQ = 180°
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 65°

2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Answer

Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle.)
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 64°
Now,
∠OZY = 1/2∠XZY (ZO is the bisector.)
⇒ ∠OZY = 32°
also,
∠OYZ = 1/2∠XYZ (YO is the bisector.)
⇒ ∠OYZ = 27°
Now,
∠OZY +∠OYZ + ∠O = 180° (Sum of the interior angles of the triangle.)
⇒ 32° + 27° + ∠O = 180°
⇒ 59° + ∠O = 180°
⇒ ∠O = 121°

3. In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer

Given,
AB || DE, ∠BAC = 35° and ∠ CDE = 53°
A/q,
∠BAC = ∠CED (Alternate interior angles.)
∴ ∠CED = 35°
Now,
∠DCE +∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle.)
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE + 88° = 180°
⇒ ∠DCE = 92°

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer

Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
A/q,
∠PRT +∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle.)
⇒ 40° + 95° + ∠PTR = 180°
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 45°
∠PTR = ∠STQ = 45° (Vertically opposite angles.)
Now,
∠TSQ +∠PTR + ∠SQT = 180° (Sum of the interior angles of the triangle.)
⇒ 75° + 45° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 60°

Page No: 108

5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Answer

Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°

6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.

Answer

Given,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS - ∠TQR --- (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR = 2∠TRS - 2∠TQR
⇒ 1/2∠QPR = ∠TRS - ∠TQR --- (ii)
Equating (i) and (ii)
∠QTR - ∠TQR = 1/2∠QPR
Hence proved.

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Chapter 6 Lines and Angles is a great chapter which must be in your strategy to improve your marks in geometry section. You will learn variety of new terms and definitions which are going to help in solving questions.

• Basic Terms and Definitions: A part of a line with two end points is called a line-segment and a part of a line with one end point is called a ray. If three or more points lie on the same line, they are called collinear points otherwise they are called non-collinear points. An angle is formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle.

An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. An angle greater than 90° but less than 180° is called an obtuse angle. A straight angle is equal to 180°. An angle which is greater than 180° but less than 360° is called a reflex angle. Two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles. Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm.

• Intersecting Lines and Non-intersecting Lines: If two lines intersect each other, then the vertically opposite angles are equal.

• Pairs of Angles: There is a very important theorem is given in this section. On the basis of which you have to solve questions given in the exercise 6.1 If two lines intersect each other, then the vertically opposite angles are equal.

• Parallel Lines and a Transversal: There are four theorems given in this section which is to going to help you in solving questions effectively.

(i) If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

(ii) If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

(iii) If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

(iv) If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

• Lines Parallel to the Same Line: Lines which are parallel to the same line are parallel to each other.

• Angle Sum Property of a Triangle: (i) The sum of the angles of a triangle is 180º. (ii) If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Three exercises are given in Chapter 6 Lines and Angles NCERT Solutions which will improve your knowledge of Geometry. Every students must try to solve each questions given in the exercise that is why we have also provided exercise wise solutions of every problem that you can find below.

Studyrankers experts at every step, they have tried to prepare these Class 9 Maths NCERT Solutions in such a way that you can easily understand even the most difficult problems. You can always clear related to this chapter just by visiting this page.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 6 Lines and Angles

How many exercises in Chapter 6 Lines and Angles?

There are only three exercise in Chapter 6 Lines and Angles NCERT Solutions which are also important for competitive exams and higher grades. We have detailed detailed every step through which one can always clear their doubts.

If the supplement of an angle is 4 times of its complement, find the angle.

Let the required angle be x

∴ (180°- x) = 4 (90° - x)

⇒ x = 60°

What is the measure of an angle whose measure is 32° less than its supplement?

Let the required angle be x

∴ x = (180°- x) - 32°

⇒ x = 74°

Angles ∠P and 100° form a linear pair. What is the measure of ∠P?

Since, the sum of the angles of a linear pair equal to 180°.

∴ ∠ P + 100° = 180°

⇒ ∠ P = 180° - 100 = 80°.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

January 17, 2021, 10:28 pm

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles| PDF Download

On this page you will get Chapter 7 Triangles Class 9 Maths NCERT Solutions that is going to help you in understanding the concepts in a better way and prepare for the examinations, You can download PDF of NCERT Solutions for Class 9 Maths Chapter 7 Triangles in order to learn at your ease. You can also complete your homework on time through the help of these NCERT Solutions and able to solve the difficult problems in a given in a exercise.

Class 9 Maths NCERT Solutions will help in developing your problem solving skills and be aware of the concepts. These solutions are prerequisites before solving exemplar problems and going for supplementary Maths Books.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Page No: 118

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Answer

Given,
AC = AD and AB bisects ∠A
To prove,
ΔABC ≅ ΔABD
Proof,
In ΔABC and ΔABD,
AB = AB (Common)
AC = AD (Given)
∠CAB = ∠DAB (AB is bisector)
Therefore, ΔABC ≅ ΔABD by SAS congruence condition.
BC and BD are of equal length.

Page No: 119

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Answer

Given,
AD = BC and ∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,
AB = BA (Common)
∠DAB = ∠CBA (Given)
AD = BC (Given)
Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ≅ ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ≅ ΔBAC
Therefore ∠ABD = ∠BAC by CPCT

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Answer

Given,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In ΔAOD and ΔBOC,
∠A = ∠B (Perpendicular)
∠AOD = ∠BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, ΔAOD ≅ ΔBOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

Answer

Given,
l || m and p || q
To prove,
ΔABC ≅ ΔCDA
Proof,
In ΔABC and ΔCDA,
∠BCA = ∠DAC (Alternate interior angles)
AC = CA (Common)
∠BAC = ∠DCA (Alternate interior angles)
Therefore, ΔABC ≅ ΔCDA by ASA congruence condition.

5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Answer

Given,
l is the bisector of an angle ∠A.
BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,
∠P = ∠Q (Right angles)
∠BAP = ∠BAQ (l is bisector)
AB = AB (Common)
Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.

Page No: 120

6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Answer

Given,
AC = AE, AB = AD and ∠BAD = ∠EAC
To show,
BC = DE
Proof,
∠BAD = ∠EAC (Adding ∠DAC both sides)
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔABC and ΔADE,
AC = AE (Given)
∠BAC = ∠EAD
AB = AD (Given)
Therefore, ΔABC ≅ ΔADE by SAS congruence condition.
BC = DE by CPCT.

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Answer

Given,
P is mid-point of AB.
∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In ΔDAP ≅ ΔEBP,
∠DPA = ∠EPB
AP = BP (P is mid-point of AB)
∠BAD = ∠ABE (Given)
Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.
(ii) AD = BE by CPCT.

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB

Answer

Given,
∠C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC (byy CPCT, already proved)
Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.

(iv) DC = AB (ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (M is mid-point)
⇒ DM + CM = AM + BM
⇒ CM + CM = AB
⇒ CM = 1/2AB

Page No: 123

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC (ii) AO bisects ∠A

Answer

Given,
AB = AC, the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,
∴ ∠B = ∠C
⇒ 1/2∠B = 1/2∠C
⇒ ∠OBC = ∠OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.
∠BAO = ∠CAO (by CPCT)
Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

Answer

Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
In ΔADB and ΔADC,
AD = AD (Common)
∠ADB = ∠ADC
BD = CD (AD is the perpendicular bisector)
Therefore, ΔADB ≅ ΔADC by SAS congruence condition.
AB = AC (by CPCT)

Page No: 124

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Answer

Given,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In ΔAEB and ΔAFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.
Thus, BE = CF by CPCT.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer

Given,
BE = CF

(i) In ΔABE and ΔACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

Answer

Given,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Answer

Given,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° - 2∠ACB --- (i)
Similarly in ΔADC,
∠CAD = 180° - 2∠ACD --- (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° - 2∠ACB + 180° - 2∠ACD
⇒ 180° = 360° - 2∠ACB - 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer

Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Answer

Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

Page No: 128

Exercise 7.3

1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.

Answer

Given,
ΔABC and ΔDBC are two isosceles triangles.

(i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ΔABC is isosceles)
BD = CD (ΔDBC is isosceles)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition.

(ii) In ΔABP and ΔACP,
AP = AP (Common)
∠PAB = ∠PAC (ΔABD ≅ ΔACD so by CPCT)
AB = AC (ΔABC is isosceles)
Therefore, ΔABP ≅ ΔACP by SAS congruence condition.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.
AP bisects ∠A. --- (i)
also,
In ΔBPD and ΔCPD,
PD = PD (Common)
BD = CD (ΔDBC is isosceles.)
BP = CP (ΔABP ≅ ΔACP so by CPCT.)
Therefore, ΔBPD ≅ ΔCPD by SSS congruence condition.
Thus, ∠BDP = ∠CDP by CPCT. --- (ii)
By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD ≅ ΔCPD)
and BP = CP --- (i)
also,
∠BPD + ∠CPD = 180° (BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° ---(ii)
From (i) and (ii),
AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠A.

Answer

Given,

AD is an altitude and AB = AC

(i) In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)
Therefore, ΔABD ≅ ΔACD by RHS congruence condition.
Now,
BD = CD (by CPCT)
Thus, AD bisects BC

(ii) ∠BAD = ∠CAD (by CPCT)
Thus, AD bisects ∠A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR

Answer

Given,
AB = PQ, BC = QR and AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)
also,
BC = QR
⇒ 1/2 BC = 1/2QR
⇒ BM = QN
In ΔABM and ΔPQN,
AM = PN (Given)
AB = PQ (Given)
BM = QN (Proved above)
Therefore, ΔABM ≅ ΔPQN by SSS congruence condition.

(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR (by CPCT)
BC = QR (Given)

Therefore, ΔABC ≅ ΔPQR by SAS congruence condition.

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer

Given,
BE and CF are two equal altitudes.
In ΔBEC and ΔCFB,
∠BEC = ∠CFB = 90° (Altitudes)
BC = CB (Common)
BE = CF (Common)
Therefore, ΔBEC ≅ ΔCFB by RHS congruence condition.
Now,
∠C = ∠B (by CPCT)
Thus, AB = AC as sides opposite to the equal angles are equal.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Answer

Given,
AB = AC
In ΔABP and ΔACP,
∠APB = ∠APC = 90° (AP is altitude)
AB = AC (Given)
AP = AP (Common)
Therefore, ΔABP ≅ ΔACP by RHS congruence condition.
Thus, ∠B = ∠C (by CPCT)

Page No: 132

Exercise 7.4

1. Show that in a right angled triangle, the hypotenuse is the longest side.

Answer

ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Answer

Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° - ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° - ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.

3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Answer

Given,
∠B < ∠A and ∠C < ∠D
Now,
AO < BO --- (i) (Side opposite to the smaller angle is smaller)
OD < OC ---(ii) (Side opposite to the smaller angle is smaller)
Adding (i) and (ii)
AO + OD < BO + OC
⇒ AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that ∠A > ∠C and ∠B > ∠D.

Answer

In ΔABD,
AB < AD < BD
∴ ∠ADB < ∠ABD --- (i) (Angle opposite to longer side is larger.)
Now,
In ΔBCD,
BC < DC < BD
∴ ∠BDC < ∠CBD --- (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠ADC < ∠ABC
⇒ ∠B > ∠D
Similarly,
In ΔABC,
∠ACB < ∠BAC --- (iii) (Angle opposite to longer side is larger.)
Now,
In ΔADC,
∠DCA < ∠DAC --- (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠BCD < ∠BAD
⇒ ∠A > ∠C

5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Answer

Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ --- (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS --- (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS --- (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS --- (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

Page No: 133

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer

Let l is a line segment and B is a point lying o it. We drew a line AB perpendicular to l. Let C be any other point on l.
To prove,
AB < AC
Proof,
In ΔABC,
∠B = 90°
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
∴ ∠C must be acute angle. or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is larger.)

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Chapter 7 Triangles NCERT Solutions which will be helpful in revising the important theorems and topics. A closed figure formed by three intersecting lines is called a triangle. We will be studying the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle.

• Congruence of Triangles: Two congruent figures have exactly the same shape and size. In congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.

• Criteria for Congruence of Triangles:

SAS congruence rule- Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

ASA congruence rule- Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. This is called the AAS Congruence Rule.

• Some Properties of a Triangle: Angles opposite to equal sides of an isosceles triangle are equaland the converse is the sides opposite to equal angles of a triangle are equal.

• Some More Criteria for Congruence of Triangles:

SSS congruence rule- If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

RHS congruence rule- If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

• Inequalities in a Triangle:

(i) If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).

(ii) In any triangle, the side opposite to the larger (greater) angle is longer.

(iii) The sum of any two sides of a triangle is greater than the third side.

There are total five exercises in which last one is optional. These Chapter 7 NCERT Solutions are will increase your understanding of Triangles and will increase concentration among students. Below we have provided exercisewise NCERT Solutions which you can check.

Exercise 7.1 Chapter 7 Class 9 Maths NCERT Solutions
Exercise 7.2 Chapter 7 Class 9 Maths NCERT Solutions
Exercise 7.3 Chapter 7 Class 9 Maths NCERT Solutions
Exercise 7.4 Chapter 7 Class 9 Maths NCERT Solutions
Exercise 7.5 Chapter 7 Class 9 Maths NCERT Solutions

Our subject matter experts have prepared these NCERT Solutions through which one can clear their doubts and understand them easily.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 7 Triangles

How many exercises in Chapter 7 Triangles?

Chapter 7 Triangles consists of total five exercises however one is optional not useful for the purpose of exams but will check your in depth knowledge. Here Studyrankers experts have provided accurate and detailed solutions of every question.

Each of the equal angles of an isosceles triangle is 38°, what is the measure of the third angle?

Let the third angle = x

∴ x + 38° + 38° = 180°

⇒ x = 180° - 38° - 38°

= 104º

Find the measure of each of acute angle in a right angle isosceles triangle.

Let the measure of each of the equal acute angle of the Δ be x

∴ We have: x + x + 90° = 180°

⇒ x + x = 180° - 90° = 90°

⇒ x= (90°/2)= 45°

If two angles are (30 ∠ a)º and (125 + 2a)º and they are supplement of each other. Find the value of ‘a’.

(30 - a)º and (125 + 2a)º are supplement to each other.

∴ (30 - a + 125 + 2a)º = 180º

⇒ a = 180º - 125º - 30º = 25º

⇒ Value of a = 25°

↧

MCQ Questions for Class 9 History: Chapter 6 किसान और काश्तकार

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MCQ Questions for Class 9 History: Chapter 6 किसान और काश्तकार

Chapter 6 किसान और काश्तकार Class 9 History MCQ Questions with answers is available on this page that will help you in covering more topics in less time and improving marks in the exams. MCQ Questions for Class 9 will help you in understanding latest pattern of the exams released by CBSE.

1. सोलहवीं सदी में, निम्न में से किस कारणवश इंग्लैंड में खुले खेतों और मुक्त और साझी जमीन की अर्थव्यवस्था बदलने लगी थी ?

(क) खाद्यान्न मूल्यों में बढ़ोतरी के कारण

(ख) विश्व बाजार में ऊन के दाम बढ़ जाने के कारण

(ग) सिंचाई सुविधा की कमी के कारण

(घ) आवासीय भूमि की कमी के कारण

► (ख) विश्व बाजार में ऊन के दाम बढ़ जाने के कारण

2. अठारहवीं सदी के अंत में, भूमि की बाड़ाबंदी करने का मुख्य उद्देश्य निम्न में से क्या था ?

(क) सिंचाई की सुविधाएं बढ़ाना।

(ख) अनाज उत्पादन को बढ़ाना।

(ग) दलहन उत्पादन को बढ़ाना।

(घ) भेड़ पालन का विकास करना

► (ख) अनाज उत्पादन को बढ़ाना।

3. इग्लैंड के मेहनतकशों द्वारा भेजे गए धमकी भरे खतों पर निम्न में से किसके हस्ताक्षर होते थे ?

(क) कैप्टन कुक

(ख) कैप्टन नील

(ग) कैप्टन स्विंग

(घ) कैप्टन फ्लाईंग

► (ग) कैप्टन स्विंग

4. सोलहवीं सदी में, इग्लैंड में जमीन की बाड़ाबंदी का मुख्य उद्देश्य क्या था?

(क) अनाज उत्पादन को बढ़ाना

(ख) मृदा के उपजाऊपन को बढ़ाना

(ग) दुग्ध उत्पादन को बढ़ाना

(घ) भेड़-पालन को बढ़ाना

► (घ) भेड़-पालन को बढ़ाना

5. अठारहवीं सदी के अंतिम वर्षों तथा उन्नीसवीं सदी के शुरुआती दौर में निम्न में से किस प्रकार की जमीन इंग्लैंड के गरीबों के लिए जिंदा रहने का बुनियादी साधन थी?

(क) साझा जमीन

(ख) चरागाह भूमि

(ग) कृषि भूमि

(घ) उपजाऊ भूमि

► (क) साझा जमीन

6. 1660 के दशक में, इंग्लैंड में शलजम और तिपतिया घास (क्लोवर) की खेती क्यों की जाती थी ?

(क) इन फसलों से जमीन की पैदावार बढ़ती थी।

(ख) इन फसलों को बोना आसान था|

(ग) इन्हें कम सिंचाई सुविधा की आवश्यकता होती थी।

(घ) इन्हें भिन्न-भिन्न उर्वरकों की आवश्यकता होती थी।

► (क) इन फसलों से जमीन की पैदावार बढ़ती थी।

7. सन् 1820-1850 के बीच मिसीसिपी की घाटी में निम्न में से किस प्रकार की कृषि की जाती थी?

(क) रोपण कृषि

(ख) गहन कृषि

(ग) कर्तक दहन कृषि

(घ) निर्वाह कृषि

► (ग) कर्तक दहन कृषि

8. इंग्लैंड में गरीब मध्यवर्ती क्षेत्रों से दक्षिणी प्रांतों की ओर अग्रसर क्यों होने लगें?

(क) गरीबों ने साझा भूमि को खो दिया था।

(ख) इन्हें साझा भूमि से बेदखल कर दिया गया।

(ग) कृषक मेहनतकशों की माँग पूर्ति के लिए।

(घ) रोजगार की तलाश में|

► (ग) कृषक मेहनतकशों की माँग पूर्ति के लिए।

9. यू०एस०ए० में प्रवासी पश्चिम की ओर अग्रसर क्यों हुए?

(क) अधिक फसल उत्पादन के लिए

(ख) नई भूमि की खोज तथा नई फसलें उगाने के लिए

(ग) अधिक लाभ प्राप्ति के लिए

(घ) विभिन्न फसलों के उत्पादन को बढ़ाने के लिए

► (ख) नई भूमि की खोज तथा नई फसलें उगाने के लिए

10. 1800 वीं सदी के अंत तक इंग्लैंड में मजदूरों की दशा के संदर्भ में, निम्न में से कौन-सा वक्तव्य अन्य तीन वक्तव्यों से भिन्न है?

(क) अनिश्चित काम

(ख) असुरक्षित रोजगार

(ग) अस्थिर आय

(घ) उच्च जीवन स्तर

► (घ) उच्च जीवन स्तर

11. निम्न में से किस कारणवश सन् 1910 से 1919 के मध्य गेहूँ उत्पादक क्षेत्रों में वृद्धि हुई थी?

(क) उर्वरकों का प्रयोग।

(ख) नई तकनीकी का प्रयोग

(ग) उच्च नस्ल के बीजों का प्रयोग

(घ) कुशल श्रमिक

► (ख) नई तकनीकी का प्रयोग

12. निम्न में से किसकी सहायता से खाद्यान्नों को गेहूँ उत्पादक क्षेत्रों से निर्यात के लिए पूर्वी तट पर | ले जाना आसान हो गया?

(क) रेलवे का प्रसार

(ख) मृदा का उपजाऊपन

(ग) उपयुक्त सिंचाई सुविधाएँ

(घ) सम जलवायु

► (क) रेलवे का प्रसार

13. किसानों ने गेहँ की कृषि के लिए विशाल मैदानी क्षेत्रों को किस प्रकार साफ किया?

(क) पेड़ों को जला कर।

(ख) लम्बी घास को जला कर।

(ग) सख्त जमीन को ट्रैक्टर और डिस्क हलों की मदद से कृषि योग्य बना कर।

(घ) पेड़ों को काटकर भूमि के एक बड़े भाग को साफ करके।

► (ग) सख्त जमीन को ट्रैक्टर और डिस्क हलों की मदद से कृषि योग्य बना कर।

14. निम्न में से कौन अंग्रेजी सत्ता के लिए सरकारी आय का एक बड़ा स्रोत था?

(क) कच्चे माल का निर्यात

(ख) चाय तथा कॉफी का उत्पादन

(ग) भू-राजस्व

(घ) कर संग्रह

► (ग) भू-राजस्व

15. 1930 के दशक में, अमेरिका के दक्षिणी भाग में आए तूफानों का मुख्य कारण इनमें से क्या था?

(क) वर्षा का अभाव

(ख) बाढ़ का आगमन

(ग) सूखे का आगमन

(घ) चक्रवातीय विक्षोभ

► (ग) सूखे का आगमन

16. उन्नीसवीं शताब्दी के आरंभिक वर्षों में निम्न में से कौन-सी व्यावसायिक फसल भारत में उगाई जाती थी?

(क) नील तथा अफीम

(ख) कपास तथा जूट

(ग) चाय तथा कॉफी

(घ) गन्ना तथा चाय

► (क) नील तथा अफीम

17. निम्न में से किस युद्ध के पश्चात्, भारत में धीरे-धीरे अंग्रेजी राज स्थापित हो गया ?

(क) भारत-चीन युद्ध

(ख) प्लासी का युद्ध

(ग) पानीपत का युद्ध

(घ) सिंध का युद्ध

► (ख) प्लासी का युद्ध

18. भारत में निम्न में से किस स्थान पर अफीम का उत्पादन किया जाता है?

(क) मद्रास

(ख) बंगाल

(ग) उड़ीसा

(घ) केरल

► (ख) बंगाल

19. सोलहवीं शताब्दी के प्रारंभ में, निम्न में से किसने चीन में अफीम के व्यापार की शुरुआत की थी?

(क) भारतीयों ने

(ख) पुर्तगालियों ने

(ग) ब्रिटिश लोगों ने

(घ) अफ्रीकी लोगों ने

► (ख) पुर्तगालियों ने

20. भारतीय कृषक अपने खेतों में अफीम उगाने के लिए तैयार क्यों नहीं थे ?

(क) उपजाऊ मृदा की कमी के कारण।

(ख) सिंचाई सुविधाओं की कमी के कारण

(ग) सरकार अफीम बोने के बदले किसान को बहुत कम दाम देती थी।

(घ) किसानों के पास आधुनिक उपकरणों तथा तकनीकों की कमी थी।

► (ग) सरकार अफीम बोने के बदले किसान को बहुत कम दाम देती थी।

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NIOS Classes 10 and 12 Admit Card Released

January 17, 2021, 11:32 pm

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NIOS Classes 10 and 12 Admit Card Released| Download at nios.ac.in

National Institute of Open Schooling (NIOS) has issued the admit card for appearing in 10th, 12th. Students can download the admit card from NIOS website. According to the notification, practical exams for Class 10 and 12 class are being conducted from January 12 to January 25, 2021 and theory examinations will commence from January 22 to February 15, 2021. Earlier, these exams were scheduled to be held in October-November 2020 but due to Covid-19 pandemic, the examination was postponed.

Those students who have registered for the examinations in January and February 2021 and also paid the examination fee can only they download their Admit Card online nios.ac.in.

How to Download NIOS Classes 10 and 12 Admit Card

• Go to the official website sdmis.nios.ac.in.

• On the homepage, click on the link ,"Public Examination Hall Ticket January / February 2021" that appear above. on the screen.

• Enter your credentials- ID, password, and login.

• NIOS Class 10 or Class 12 admit card 2021 will be displayed on the screen.

• Download the admit card and print it.

According to the datasheet of the 10 and 12 board exams, the exams of both classes will be between 2.30 pm and 5.30 pm in the afternoon. Theory and practical exam will be held in the same regional centers, where the candidate had registered during admission. NIOS conducts board exams twice every year. The first examination of NIOS is held in the month of March-April and the second exam is held in the month of September-October.

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MCQ Questions for Class 9 History: Chapter 7 इतिहास और खेल: क्रिकेट की कहानी

January 18, 2021, 3:10 am

≫ Next: MCQ Questions for Class 9 History: Chapter 8 पहनावे का सामाजिक इतिहास

≪ Previous: NIOS Classes 10 and 12 Admit Card Released

MCQ Questions for Class 9 History: Chapter 7 इतिहास और खेल: क्रिकेट की कहानी

Here you will find Chapter 7 इतिहास और खेल: क्रिकेट की कहानी Class 9 History MCQ Questions with answers which is very much helpful in understanding latest pattern of the exams released by CBSE and scoring good marks in the exams. MCQ Questions for Class 9 will be useful in understanding topics in less time.

1. विश्व का प्रथम क्रिकेट क्लब निम्न में से किस स्थान पर बना?

(a) पर्थ

(b) एडीलेड

(d) पोलैंड

► (c) हैम्बल्डन

2. क्रिकेट गेंद बनाने वाली सामग्री के संदर्भ में, निम्न में से कौन अन्य तीन वस्तुओं से भिन्न है ?

(a) रबड़

(b) काग (कार्क)

(d) चमड़ा

► (a) रबड़

3. निम्न में से किस क्रिकेट क्लब की स्थापना सन् 1787 में की गई ?

(a) मेरिलिबॉन क्रिकेट क्लब

(b) भारतीय क्रिकेट क्लब

(d) स्टॉकहोम क्रिकेट क्लब

► (a) मेरिलिबॉन क्रिकेट क्लब

4. पहला लेग बिफोर विकेट (पगबाधा) नियम निम्न में से किस वर्ष प्रकाशित हुआ?

(a) 1780

(b) 1784

(d) 1774

► (c) 1775

5. प्राचीन समय में, क्रिकेट मूलतः निम्न में से किस स्थान पर खेला जाता था?

(a) स्टेडियम में

(b) खेल के मैदान में

(d) बाड़-रहित सार्वजनिक स्थानों पर

► (d) बाड़-रहित सार्वजनिक स्थानों पर

6. मजे के लिए क्रिकेट खेलने वाले अमीर लोग निम्न में से किस नाम से जाने जाते थे?

(a) शौकिया

(b) चंदे

(d) क्रिकेट के खिलाड़ी

► (a) शौकिया

7. निम्न में से किस वर्ष क्रिकेट पैड का प्रचलन आरम्भ हुआ था?

(a) 1849

(b) 1848

(d) 1846

► (b) 1848

8. एल्युमीनियम के बल्ले से खेलने की कोशिश करने पर निम्न में से किस क्रिकेटर को अंपायरो ने अवैध करार दिया था?

(a) कपिल देव

(b) अनिल कुम्बले

(d) शेन वार्न

► (c) डेनिस लिली

9. पेशेवर खिलाड़ियों को दिया जाने वाला मेहनताना निम्न में से किस रूप में इकट्ठा किया जाता था?

(a) शौकिया

(b) चंदे

(d) पेशेवर

► (b) चंदे

10. इंग्लैंड में क्रिकेट के आयोजन पर निम्न में से किस समाज की छाप साफ दिखाई देती थी?

(a) यूरोपीय समाज

(b) अफ्रीकी समाज

(d) अंग्रेजी समाज

► (d) अंग्रेजी समाज

11. निम्न में से किस स्थान पर प्रथम गैर-गोरा क्रिकेट क्लब स्थापित किया गया था?

(a) अफ्रीका

(b) आस्ट्रेलिया

(d) मध्य अमेरिका

► (c) वेस्ट इंडीज

12. निम्न में से किस वर्ष कलकत्ता क्रिकेट क्लब की स्थापना हुई?

(a) 1792

(b) 1790

(d) 1910

► (a) 1792

13. निम्न में से कौन प्रथम भारतीय क्लब था?

(a) मुम्बई क्लब

(b) कलकत्ता क्लब

(d) बम्बई क्लब

► (b) कलकत्ता क्लब

14. निम्न में से किस स्थान पर ओरिएंटल क्रिकेट क्लब की स्थापना की गई थी?

(a) बम्बई

(b) कलकत्ता

(d) रायपुर

► (a) बम्बई

15. निम्न में से किस भारतीय समुदाय ने सर्वप्रथम क्रिकेट खेलना आरम्भ किया था?

(a) ईसाई

(b) हिन्दू

(d) सिख

► (c) ज़रतुश्तियों (पारसी)

16. निम्न में से कौन-सा भारत का प्रथम क्रिकेट क्लब है?

(a) कलकत्ता क्रिकेट क्लब

(b) मुम्बई क्रिकेट क्लब

(d) ओरिएंटल क्रिकेट क्लब

► (d) ओरिएंटल क्रिकेट क्लब

17. निम्न में से किस वर्ष भारत ने विश्व टेस्ट क्रिकेट में प्रवेश किया था?

(a) 1930

(b) 1929

(d) 1928

18. इम्पेरियल (Imperial) क्रिकेट कॉन्फ्रेन्स का नाम बदलकर क्या रखा गया ?

(a) इंडियन क्रिकेट कॉन्फ्रेन्स

(b) ओरिएंटल क्रिकेट कॉन्फ्रेन्स

(d) इन्डो-यूरोपियन क्रिकेट कॉन्फ्रेन्स

► (c) इंटरनैशनल क्रिकेट कॉन्फ्रेन्स

19. आधुनिक क्रिकेट में निम्न में से किसका वर्चस्व है ?

(a) टेस्ट और एकदिवसीय अन्तर्राष्ट्रीय

(b) अनेक टेस्ट

(d) पंचकोणीय

► (a) टेस्ट और एकदिवसीय अन्तर्राष्ट्रीय

20. वि-औपनिवेशीकरण के कारण ब्रिटिश साम्राज्य पर निम्न में से क्या प्रभाव पड़ा?

(a) भारतीयों को क्रिकेट खेलने का मौका मिल गया।

(b) व्यापार, वाणिज्य, सैन्य क्षेत्र, खेल क्षेत्र आदि पर ब्रिटिश प्रभाव कम हो गया।

(d) इंग्लैंड के खिलाफ टेस्ट मैचों की शुरूआत हुई।

► (b) व्यापार, वाणिज्य, सैन्य क्षेत्र, खेल क्षेत्र आदि पर ब्रिटिश प्रभाव कम हो गया।

↧

MCQ Questions for Class 9 History: Chapter 8 पहनावे का सामाजिक इतिहास

January 18, 2021, 4:38 am

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MCQ Questions for Class 9 History: Chapter 8 पहनावे का सामाजिक इतिहास

On this page you will find Chapter 8 पहनावे का सामाजिक इतिहास MCQ Questions for Class 9 History with answers that will be useful in covering all the essential topics in less time and getting good marks in the exams. Class 9 MCQ Questions will help you in improving knowledge and concentration among students.

1. पहनावे की सादगी से निम्न में से कौन-सा भाव प्रकट होता है?

(a) समानता का भाव

(b) श्रेष्ठता का भाव

(d) धार्मिक समानता का भाव विक्टोरियन समाज में

► (a) समानता का भाव

2. विक्टोरियन इंग्लैंड में लड़कियाँ बड़े होने पर निम्न में से क्या पहनती थी?

(a) साड़ी

(b) जींस

(d) स्टेज

► (b) जींस

3. स्वतंत्रता की लाल टोपी तथा लंबी-पतलून के लिए निम्न में से किस शब्दावली का प्रयोग किया जाता है ?

(a) इर्माइन

(b) सौं कुलॉत

(d) घुटने वाले

► (c) कॉकेड

4. निम्न में से किसने इग्लैंड में पोशाक सुधार के लिए मुहिम चलाई?

(a) राष्ट्रीय महिला मताधिकार सभा तथा अमेरिकन महिला मताधिकार सभा

(b) अन्तर्राष्ट्रीय मताधिकार सभा

(d) विक्टोरियन मताधिकार आंदोलन

► (a) राष्ट्रीय महिला मताधिकार सभा तथा अमेरिकन महिला मताधिकार सभा

5. विक्टोरियन समाज में निम्न में से किस प्रकार की नारी को आदर्श माना जाता था?

(a) जो बहुत सुन्दर होती थी।

(b) जो दुख-दर्द सह सके।

(d) जो गंभीर और बलवान होती थी।

► (b) जो दुख-दर्द सह सके।

6. सफ्रेज (Suffrage) आंदोलन से अभिप्राय है।

(a) समानता का अधिकार

(b) गाने का अधिकार

(d) मताधिकार (वोट डालने का अधिकार)

► (d) मताधिकार (वोट डालने का अधिकार)

7. महिलाओं की पोशाक सुधार के संदर्भ में, निम्न में से कौन-सा वक्तव्य अन्य तीन वक्तव्यो से भिन्न है ?

(a) कपड़ों को सरल बनाओं

(b) स्कर्ट छोटी करो

(d) कॉर्सेट का त्याग करो

► (c) लम्बे गाऊन

8. निम्न में से किसने इग्लैड में पोशाक सुधार के लिए मुहिम चलाई ?

(क) राष्ट्रीय महिला मताधिकार सभा तथा अमेरिकन महिला मताधिकार सभा

(ख) अन्तर्राष्ट्रीय मताधिकार सभा

(ग) मताधिकार आंदोलन

(घ) विक्टोरियन मताधिकार आंदोलन

► (क) राष्ट्रीय महिला मताधिकार सभा तथा अमेरिकन महिला मताधिकार सभा

9. बीसवीं सदी की शुरूआत में, निम्न में से किस रेशे के बने कपड़े सस्ते तथा साफ करने आसान हो

(a) कृत्रिम रेशा

(b) सिल्क

(d) कपास

► (a) कृत्रिम रेशा

10. महिलाओं के वस्त्रों में मूलभूत परिवर्तन निम्न में से किस कारणवश हुआ?

(a) जातीय आंदोलन

(b) प्रथम विश्व युद्ध

(d) मताधिकार आंदोलन

► (b) प्रथम विश्व युद्ध

11. सन् 1600 के बाद भारत के साथ व्यापार में निम्न में से कौन-सी वस्तु इंग्लैड ले जाई गई?

(a) भारतीय सूती वस्त्र

(b) भारतीय ऊन

(d) भारतीय सिल्क

► (c) भारतीय छींट

12. निम्न में से किसने सत्र न्यायधीश की, अदालत में जूते उतारने से इन्कार कर दिया?

(a) मनोकजी कोवासजी एन्टी

(b) लेडी बच्चूबाई

(d) सर एम. विश्वेश्वरैया

► (a) मनोकजी कोवासजी एन्टी

13. निम्न में से किस नियम को लार्ड डलहौजी ने भारतीयों के लिए सख्त कर दिया था?

(a) समाज के वरिष्ठ लोगों के सामने पगड़ी उतारना

(b) दफ्तर में भारतीय पोशाक ग्रहण करे।

(d) सरकारी संस्थाओं में जूते उतार कर प्रवेश करना।

► (d) सरकारी संस्थाओं में जूते उतार कर प्रवेश करना।

14. निम्न में से कौन आई. सी. एस (ICS) के पहले भारतीय सदस्य बने?

(a) सरला

(b) बच्चूबाई

(d) कस्तूरबा गाँधी

► (c) सत्येन्द्रनाथ टैगोर

15. निम्न में से किसे पुरुषों की सर्वोत्तम राष्ट्रीय पोशाक माना गया?

(a) कमर तक लंबा कोट

(b) चपकन (बटनदार लंबा कोट)

(d) ढ़ीली-ढ़ाली पतलून

► (b) चपकन (बटनदार लंबा कोट)

16. लार्ड कर्जन ने बंगाल को विभाजित करने का फैसला क्यों किया?

(a) ब्रिटिश राज के खिलाफ बढ़ते विरोध को नियन्त्रित करने के लिए।

(b) कपास तथा नील के निर्यात को बढ़ाना।

(d) कपास तथा नील जैसे कच्चे माल को बढ़ाने के लिए।

► (a) ब्रिटिश राज के खिलाफ बढ़ते विरोध को नियन्त्रित करने के लिए।

17. निम्न में से किस प्रकार की साड़ी को ब्रह्मा समाजी महिलाओं ने अपनाया?

(a) बनारसी साड़ी

(b) दक्षिण भारतीय साड़ी

(d) ब्रह्मिका साड़ी

► (d) ब्रह्मिका साड़ी

18. सत्रहवीं सदी में पूरे विश्व के उत्पाद का निम्न में से कितना भाग भारत में बनता था?

(क) दो-तिहाई

(ख) एक-चौथाई

(ग) तीन-चौथाई

(घ) एक का पाँचवा भाग

► (ख) एक-चौथाई

19. यूरोप में मौजूदा ड्रेस कोड खत्म हुए-

(a) अमेरिकी क्रांति

(b) फ्रांसीसी क्रांति

(d) प्रथम विश्व युद्ध

► फ्रांसीसी क्रांति

20. कुछ भारतीयों के लिए पश्चिमी कपड़े एक संकेत थे

(a) प्रगति

(b) आधुनिकता

(d) (a) और (b) दोनों

► (d) (a) और (b) दोनों

↧

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

January 18, 2021, 4:47 am

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals| PDF Download

Here Chapter 8 Quadrilaterals Class 9 Maths NCERT Solutions are available that will be helpful in understanding the key concepts in an easy way. Here you can download PDF of NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals that will serve as beneficial tool that can be used to recall various questions any time. You can also complete your homework on time through the help of these Chapter 8 NCERT Solutions and able to solve the difficult problems in a given in a exercise.

For application of concepts, an individual should first need to focus on the Quadrilaterals NCERT Solutions as it will tell you about the difficulty of questions. These NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Page No: 146

Exercise 8.1

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer

Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360°
Now,
3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.
∠A = ∠B (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus ABCD is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer

Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
AC = BD, AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
∠ABC = ∠BAD = 90°
AC = AD (Given)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer

Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by CPCT. --- (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT. --- (ii)
also,
AD = BC and AD = CD
⇒ AD = BC = CD = AB --- (ii)
also, ∠ADC = ∠BCD by CPCT.
and ∠ADC + ∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒ ∠ADC = 90° --- (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Answer

(i)
In ΔADC and ΔCBA,
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer

Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.

Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.

8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.

Answer

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.

(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
∠CBD = ∠ADB
⇒ ∠CDB = ∠ADB
Thus, BD bisects ∠D

Page No: 147

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram

Answer

(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

Answer

(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

Answer

(i) AB = DE and AB || DE (Given)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

Thus, AD = CF.

Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)

Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ΔABC ≅ ΔDEF by SSS congruence condition.

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C

(iii) In ΔABC and ΔBAD,
AB = AB (Common)
∠DBA = ∠CBA
AD = BC (Given)
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Page No: 150

Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that :
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Answer

(i) In ΔDAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SR || AC and SR = 1/2 AC

(ii) In ΔBAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQ || AC and PQ = 1/2 AC
also, SR = 1/2 AC
Thus, PQ = SR

(iii) SR || AC - from (i)
and, PQ || AC - from (ii)
⇒ SR || PQ - from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer

Given,
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
AC and BD are joined.
Proof,
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by CPCT --- (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.
RQ = SP by CPCT --- (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
⇒ QR || BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS || BD
⇒ QR || PS
Thus, PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Thus, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Answer

Given,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In ΔABC
P and Q are the mid-points of AB and BC respectively
Thus, PQ || AC and PQ = 1/2 AC (Mid point theorem) --- (i)
In ΔADC,
SR || AC and SR = 1/2 AC (Mid point theorem) --- (ii)
So, PQ || SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PS || QR and PS = QR (Opposite sides of parallelogram) --- (iii)
Now,
In ΔBCD,
Q and R are mid points of side BC and CD respectively.
Thus, QR || BD and QR = 1/2 BD (Mid point theorem) --- (iv)
AC = BD (Diagonals of a rectangle are equal) --- (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.

Answer

Given,
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
In ΔBAD,
E is the mid point of AD and also EG || AB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In ΔBDC,
G is the mid point of BD and also GF || AB || DC.
Thus, F is the mid point of BC (Converse of mid point theorem)

Page No: 151

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Answer

Given,
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,

ABCD is a parallelogram
Therefor, AB || CD

also, AE || FC
Now,
AB = CD (Opposite sides of parallelogram ABCD)
⇒ 1/2 AB = 1/2 CD
⇒ AE = FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AF || EC (Opposite sides of a parallelogram)
Now,
In ΔDQC,
F is mid point of side DC and FP || CQ (as AF || EC).
P is the mid-point of DQ (Converse of mid-point theorem)
⇒ DP = PQ --- (i)
Similarly,
In APB,
E is mid point of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
⇒ PQ = QB --- (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now,
In ΔACD,
R and S are the mid points of CD and DA respectively.
Thus, SR || AC.
Similarly we can show that,
PQ || AC
PS || BD
QR || BD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = 1/2 AB

Answer

(i) In ΔACB,
M is the mid point of AB and MD || BC
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90°
Thus, ∠ADM = 90° and MD ⊥ AC

(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common)
Thus, ΔAMD ≅ ΔCMD by SAS congruence condition.
AM = CM by CPCT
also, AM = 1/2 AB (M is mid point of AB)
Hence, CM = MA = 1/2 AB

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Chapter 8 Quadrilaterals NCERT Solutions is very important for the preparation of exams. A figure formed by joining four points in an order is called a quadrilateral. A quadrilateral has four sides, four angles and four vertices. In this chapter, we will be discussing different types of quadrilaterals, their properties and about parallelograms.

• Angle Sum Property of a Quadrilateral: The sum of the angles of a quadrilateral is 360º. This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

• Types of Quadrilaterals:

A square is a rectangle and also a rhombus.

A parallelogram is a trapezium.

A kite is not a parallelogram.

A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram).

A rectangle or a rhombus is not a square.

• Properties of a Parallelogram:

1. A diagonal of a parallelogram divides it into two congruent triangles.

2. In a parallelogram, opposite sides are equal.

3. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

4. In a parallelogram, opposite angles are equal.

5. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

6. The diagonals of a parallelogram bisect each other.

7. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

• Another Condition for a Quadrilateral to be a Parallelogram: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

• The Mid-point Theorem:

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

There are only two exercises in Chapter 8 Quadrilaterals NCERT Solutions which are provided below which can be helpful in completing your homework on time.

Studyrankers experts have taken every care while preparing these Class 9 Maths NCERT Solutions so you can easily clear your doubts. These NCERT Solutions are updated as per the latest marking scheme released by CBSE.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 8 Quadrilaterals

What are the benefits of NCERT Solutions for Chapter 8 Quadrilaterals Class 9 NCERT Solutions?

Through the help of NCERT Solutions of Chapter 8 Quadrilaterals you will be able to solve difficult questions easily and revising the chapter properly. It will improve the learning behaviour of the students and learning diverse topics.

What is a Rhombus?

A parallelogram having all sides equal is called a rhombus.

What is mid-point theorem?

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

In a quadrilateral, ∠ A : ∠ B : ∠ C : ∠ D = 1 : 2 : 3 : 4, then find the measure of each angle of the quadrilateral.

Since ∠ A : ∠ B : ∠ C : ∠ D = 1 : 2 : 3 : 4

∴ If ∠ A = x, then ∠ B = 2x, ∠ C = 3x and ∠ D = 4x. ∴ ∠ A + ∠ B + ∠ C + ∠ D = 360°

⇒ x + 2x + 3x + 4x = 360° ⇒ 10x = 36°

⇒ x= (360°/10)= 36°

∴ ∠ A = x = 36° ∠ B = 2x = 2 x 36° = 72° ∠ C = 3x = 3 x 36° = 108° ∠ D = 4x = 4 x 36° = 144°.

↧

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

January 18, 2021, 10:55 am

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles| PDF Download

Chapter 9 Areas of Parallelograms and Triangles Class 9 Maths NCERT Solutions is available on this page that will help you in completing your homework on time and obtain maximum marks in the exams. You can also Download PDF of Chapter 9 Areas of Parallelograms and Triangles NCERT Solutions Class 9 Maths to practice in a better manner. There are variety of concepts given in this chapter 6 Class 9 Maths textbook that will develop your necessary skill to solve more and more questions.

These NCERT Solutions for Class 9 will prove useful guide in making a student confident. These solutions are updated according to the latest NCERT Class 9 Maths textbook. Chapter 9 NCERT Solutions will develop you understanding of the chapter and make a student confident.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

Page No: 155

Exercise 9.1

1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Answer

(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ.
(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ.

Page No: 159

Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer

Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).

Answer

Given,
E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH --- (i)
also, area of GHF = 1/2 area of HFCD --- (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)

3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).

Answer

ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) --- (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) --- (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer

(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) --- (i)
Thus,
AD || BC ⇒ AG || BH --- (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) --- (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) --- (i)
Thus,
AB || CD ⇒ AE || DF --- (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) --- (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)

Answer

(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.

∴ ar(PQRS) = ar(ABRS) --- (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) --- (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)

Page No: 106

6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Answer

The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS --- (i)
Area of ΔPAQ = 1/2 area of PQRS --- (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS --- (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

Page No: 162

Exercise 9.3

1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).

Answer

Given,

AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.

∴ ar(ABD) = ar(ACD) --- (i)

also,

ED is the median of ΔABC.

∴ ar(EBD) = ar(ECD) --- (ii)

Subtracting (ii) from (i),

ar(ABD) - ar(EBD) = ar(ACD) - ar(ECD)

⇒ ar(ABE) = ar(ACE)

2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).

Answer

ar(BED) = (1/2) × BD × DE
As E is the mid-point of AD,

Thus, AE = DE
As AD is the median on side BC of triangle ABC,
Thus, BD = DC
Therefore,

DE = (1/2)AD --- (i)
BD = (1/2)BC --- (ii)

From (i) and (ii),

ar(BED) = (1/2) × (1/2) BC × (1/2)AD
⇒ ar(BED) = (1/2) × (1/2) ar(ABC)

⇒ ar(BED) = 1/4 ar(ABC)

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer

O is the mid point of AC and BD. (diagonals of bisect each other)

In ΔABC, BO is the median.

∴ ar(AOB) = ar(BOC) --- (i)

also,

In ΔBCD, CO is the median.

∴ ar(BOC) = ar(COD) --- (ii)

In ΔACD, OD is the median.

∴ ar(AOD) = ar(COD) --- (iii)

In ΔABD, AO is the median.

∴ ar(AOD) = ar(AOB) --- (iv)

From equations (i), (ii), (iii) and (iv),

ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)

So, the diagonals of a parallelogram divide it into four triangles of equal area.

4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that:

ar(ABC) = ar(ABD).

Answer

In ΔABC,

AO is the median. (CD is bisected by AB at O)

∴ ar(AOC) = ar(AOD) --- (i)

also,

In ΔBCD,

BO is the median. (CD is bisected by AB at O)

∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)

Page No: 163

5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
Show that
(i) BDEF is a parallelogram. (ii) ar(DEF) = 1/4 ar(ABC)
(iii) ar (BDEF) = 1/2 ar(ABC)

Answer

(i) In ΔABC,
EF || BC and EF = 1/2 BC (by mid point theorem)
also,
BD = 1/2 BC (D is the mid point)
So, BD = EF
also,
BF and DE will also parallel and equal to each other.
Thus, the pair opposite sides are equal in length and parallel to each other.
∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),
BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔBFD) = ar(ΔDEF) (For parallelogram BDEF) --- (i)
also,
ar(ΔAFE) = ar(ΔDEF) (For parallelogram DCEF) --- (ii)
ar(ΔCDE) = ar(ΔDEF) (For parallelogram AFDE) --- (iii)
From (i), (ii) and (iii)
ar(ΔBFD) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)
⇒ ar(ΔBFD) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = arar(ΔABC)
⇒ 4 ar(ΔDEF) = ar(ΔABC)
⇒ ar(DEF) = 1/4 ar(ABC)

(iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDE)
⇒ ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)
⇒ ar(parallelogram BDEF) = 2× ar(ΔDEF) ⇒ ar(parallelogram BDEF) = 2× 1/4 ar(ΔABC) ⇒ ar(parallelogram BDEF) = 1/2 ar(ΔABC)

6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]

Answer

Given,
OB = OD and AB = CD
Construction,
DE ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i) In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
Therefore, ΔDOE ≅ ΔBOF by AAS congruence condition.
Thus, DE = BF (By CPCT) --- (i)
also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) --- (ii)
Now,
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From i)
Therefore,ΔDEC ≅ ΔBFA by RHS congruence condition.
Thus, ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) --- (iii)
Adding (ii) and (iii),
ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)
⇒ ar (DOC) = ar (AOB)

(ii) ar(ΔDOC) = ar(ΔAOB)
⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB) (Adding ar(ΔOCB) to both sides) ⇒ ar(ΔDCB) = ar(ΔACB) (iii) ar(ΔDCB) = ar(ΔACB) If two triangles are having same base and equal areas, these will be between same parallels DA || BC --- (iv) For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel. Therefore, ABCD is parallelogram.

7. D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that DE || BC.

Answer

ΔDBC and ΔEBC are on the same base BC and also having equal areas. Therefore, they will lie between the same parallel lines. Thus, DE || BC.

8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar(ΔABE) = ar(ΔAC)

Answer

Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) --- (i)
also,
BE∥ CY (BE || AC) --- (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) --- (iii)
Also,
△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = 1/2ar(BEYC) --- (iv)
Similarly,
△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)
From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)

9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

Answer

AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) - ar(△ABQ) = ar(△APQ) - ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) --- (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) --- (ii)
ar(QBP) = 1/2 ar(PBQR) --- (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)

10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Answer

△DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
∴ ar(△DAC) = ar(△DBC)
⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
⇒ ar(△AOD) = ar(△BOC)

11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)

Answer

(i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.

∴ ar(△ACB) = ar(△ ACF)

(ii) ar(△ACB) = ar(△ACF)

⇒ ar(△ACB) + ar(△ACDE) = ar(△ACF) + ar(△ACDE)

⇒ ar(ABCDE) = ar(△AEDF)

Page No: 164

12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Answer

Let ABCD be the plot of the land of the shape of a quadrilateral.

Construction,

Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.
To prove:
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.
ar(△DEB) = ar(△DAB)
⇒ ar(△DEB) - ar△DOB) = ar(△DAB) - ar(△DOB)

⇒ ar(△DEO) = ar(△AOB)

13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

[Hint : Join CX.]

Answer

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

CX is joined.

To Prove,

ar(ADX) = ar(ACY)

Proof:

ar(△ADX) = ar(△AXC) --- (i) (On the same base AX and between the same parallels AB and CD)
also,

ar(△ AXC)=ar(△ ACY) --- (ii) (On the same base AC and between the same parallels XY and AC.)

From (i) and (ii),

ar(△ADX)=ar(△ACY)

14. In Fig.9.28, AP || BQ || CR. Prove that
ar(AQC) = ar(PBR).

Answer

Given,

AP || BQ || CR

To Prove,

ar(AQC) = ar(PBR)

Proof:

ar(△AQB) = ar(△PBQ) --- (i) (On the same base BQ and between the same parallels AP and BQ.)

also,

ar(△BQC) = ar(△BQR) --- (ii) (On the same base BQ and between the same parallels BQ and CR.)

Adding (i) and (ii),
ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)

⇒ ar(△ AQC) = ar(△ PBR)

15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.

Answer

Given,
ar(△AOD) = ar(△BOC)

To Prove,

ABCD is a trapezium.

Proof:

ar(△AOD) = ar(△BOC)

⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)

⇒ ar(△ADB) = ar(△ACB)

Areas of △ADB and △ACB are equal. Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD

Therefore, ABCD is a trapezium.

16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer

Given,

ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(△BDP) = ar(△ARC)

⇒ ar(△BDP) - ar(△DPC) = ar(△DRC)

⇒ ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC). Therefore, they must lying between the same parallel lines.

Thus, AB ∥ CD

Therefore, ABCD is a trapezium.

also,

ar(DRC) = ar(DPC). Therefore, they must lying between the same parallel lines.

Thus, DC ∥ PR

Therefore, DCPR is a trapezium.

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

In the Chapter 9 Areas of Parallelograms and Triangles, we are focused on the relationship between the areas of these geometric figures under the condition when they lie on the same base and between the same parallels.

• Figures on the Same Base and Between the Same Parallels: Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.

• Parallelograms on the same Base and Between the same Parallels: Parallelograms on the same base and between the same parallels are equal in area.

• Triangles on the same Base and between the same Parallels: Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

There are total 4 exercises in the Chapter 9 Class 9 Maths NCERT Solutions which are very important for examinations purpose. We have prepared exercisewise NCERT Solutions of every questions which can find from the given links.

Exercise 9.1 Chapter Class 9 Maths NCERT Solutions
Exercise 9.2 Chapter Class 9 Maths NCERT Solutions
Exercise 9.3 Chapter Class 9 Maths NCERT Solutions
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What is Area of a parallelogram?

Area of a parallelogram = base × height.

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NCERT Solutions for Class 9 Maths Chapter 10 Circles

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NCERT Solutions for Class 9 Maths Chapter 10 Circles| PDF Download

Here you will get Chapter 10 Circles NCERT Solutions for Class 9 Maths that are updated as per the latest marking scheme and syllabus prescribed by CBSE. Also you can Download PDF of Chapter 10 Circles NCERT Solutions Class 9 Maths to practice in a better way. There are variety of concepts given in this Chapter 10 Class 9 Maths textbook that will develop your problem solving skills and make aware of the difficulty of questions.

You can also complete your homework on time through the help of these Chapter 10 NCERT Solutions and able to solve the difficult problems in a given in a exercise. These NCERT Solutions will serve as beneficial tool that can be used to recall various questions any time. You can be able to solve more questions related to chapter 10 Circles after practising more questions that results in fetching more marks.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Page No: 171

Exercise 10.1

1. Fill in the blanks:
(i) The centre of a circle lies in ____________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a _____________ of the circle.
(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.

Answer

(i) The centre of a circle lies in interior of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

2. Write True or False: Give reasons for your answers.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.

Answer

(i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can't say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.

Exercise 10.2

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer

A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.

Given,
AB = CD (Equal chords)
To prove,
∠AOB = ∠COD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
∴ ΔAOB ≅ ΔCOD (SSS congruence condition)
Thus, ∠AOB = ∠COD by CPCT.
Equal chords of congruent circles subtend equal angles at their centres.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer

Given,
∠AOB = ∠COD (Equal angles)
To prove,
AB = CD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
∠AOB = ∠COD (Given)
OB = OD (Radii)
∴ ΔAOB ≅ ΔCOD (SAS congruence condition)
Thus, AB = CD by CPCT.
If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Page No: 176

Exercise 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Answer

No point is common.

One point P is common.

Two points P and Q are common.

No point is common.

2. Suppose you are given a circle. Give a construction to find its centre.

Answer

Steps of construction:
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.

3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer

Given,
Two circles which intersect each other at P and Q.
To prove,
OO' is perpendicular bisector of PQ.
Proof,
In ΔPOO' and ΔQOO',
OP = OQ (Radii)
OO' = OO' (Common)
O'P = OQ (Radii)
∴ ΔPOO' ≅ ΔQOO' (SSS congruence condition)
Thus,
∠POO' = ∠QOO' --- (i)
In ΔPOR and ΔQOR,
OP = OQ (Radii)
∠POR = ∠QOR (from i)
OR = OR (Common)
∴ ΔPOR ≅ ΔQOR (SAS congruence condition)
Thus,
∠PRO = ∠QRO
also,
∠PRO + ∠QRO = 180°
⇒ ∠PRO = ∠QRO = 180°/2 = 90°
Hence,
OO' is perpendicular bisector of PQ.

Page No: 179

Exercise 10.4

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer

OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.

In ΔPOR,
OP²= OR²+PR²
⇒ 5²= (4-x)²+PR²
⇒ 25 = 16 + x²- 8x +PR²
⇒ PR² = 9 - x² + 8x --- (i)

In ΔPRS,
PS²= PR²+RS²
⇒ 3²= PR² +x²
⇒ PR² = 9 - x² --- (ii)
Equating (i) and (ii),
9 - x² + 8x = 9 - x²
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR² = 9 - 0²
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND --- (i)
and MB = CN --- (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT --- (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii) and (iii) we get,
MB - ME = CN - EN
⇒ EB = CE

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ

4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer

OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD --- (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC --- (ii)
From (i) and (ii),
AM - BM = MD - MC
⇒ AB = CD

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Answer

Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA²= OM²+AM²
⇒ 5²= x²+y²--- (i)Applying Pythagoras theorem in ΔAMB,
AB²= BM²+AM²
⇒ 6²= (5-x)²+y² --- (ii)Subtracting (i) from (ii), we get
36 - 25 = (5-x)²-x² -
⇒ 11 = 25 - 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y²+ 49/25 = 25
⇒ y² = 25 - 49/25
⇒ y² = (625 - 49)/25
⇒ y² = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.

6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB²= BD²+AD²
⇒ AD^²= AB^²- BD^²
⇒ AD²= a²-(a/2)²
⇒ AD²= 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.

Page No: 184

Exercise 10.5

1. In Fig. 10.36, A,B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Answer

Here,
∠AOC = ∠AOB + ∠BOC
⇒ ∠AOC = 60° + 30°
⇒ ∠AOC = 90°
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
∠ADC = 1/2∠AOC = 1/2 × 90° = 45°

Page No: 185

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer

Given,
AB is equal to the radius of the circle.
In ΔOAB,
OA = OB = AB = radius of the circle.
Thus, ΔOAB is an equilateral triangle.
∠AOC = 60°
also,
∠ACB = 1/2 ∠AOB = 1/2 × 60° = 30°
ACBD is a cyclic quadrilateral,
∠ACB + ∠ADB = 180° (Opposite angles of cyclic quadrilateral)
⇒ ∠ADB = 180° - 30° = 150°
Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.

3. In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer

Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°
∴ ∠POR = 360° - 200° = 160°
In ΔOPR,
OP = OR (radii of the circle)
∠OPR = ∠ORP
Now,
∠OPR + ∠ORP +∠POR = 180° (Sum of the angles in a triangle)
⇒ ∠OPR + ∠OPR + 160° = 180°
⇒ 2∠OPR = 180° - 160°
⇒ ∠OPR = 10°

4. In Fig. 10.38, ∠ABC = 69°, ∠ ACB = 31°, find ∠BDC.

Answer

∠BAC = ∠BDC (Angles in the segment of the circle)
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of the angles in a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° - 100°
⇒ ∠BAC = 80°
Thus, ∠BDC = 80°

5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠ BAC.

Answer

∠BAC = ∠CDE (Angles in the segment of the circle)
In ΔCDE,
∠CEB = ∠CDE + ∠DCE (Exterior angles of the triangle)
⇒ 130° = ∠CDE + 20°
⇒ ∠CDE = 110°
Thus, ∠BAC = 110°

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Answer

For chord CD,
∠CBD = ∠CAD (Angles in same segment)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠BCD + 100° = 180°
⇒ ∠BCD = 80°
In ΔABC
AB = BC (given)
∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
∠BCA = 30°
also, ∠BCD = 80°
∠BCA + ∠ACD = 80°
⇒ 30° + ∠ACD = 80°
∠ACD = 50°
∠ECD = 50°

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer

Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90° (Angles in the semi-circle)
Thus, ABCD is a rectangle as each internal angle is 90°.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer

Given,
ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction,
DM and CN are perpendicular drawn on AB from D and C respectively.
To prove,
ABCD is cyclic trapezium.

Proof:
In ΔDAM and ΔCBN,
AD = BC (Given)
∠AMD = ∠BNC (Right angles)
DM = CN (Distance between the parallel lines)
ΔDAM ≅ ΔCBN by RHS congruence condition.
Now,
∠A = ∠B by CPCT
also, ∠B +∠C = 180° (sum of the co-interior angles)
⇒ ∠A + ∠C = 180°
Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.

Page No: 186

9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.

Answer

Chords AP and DQ are joined.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) --- (i)
For chord DQ,
∠DBQ = ∠QCD (Angles in same segment) --- (ii)
ABD and PBQ are line segments intersecting at B.
∠PBA = ∠DBQ (Vertically opposite angles) --- (iii)
By the equations (i), (ii) and (iii),
∠ACP = ∠QCD

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer

Given,

Two circles are drawn on the sides AB and AC of the triangle ΔABC as diameters. The circles intersected at D.

Construction,

AD is joined.

To prove,

D lies on BC. We have to prove that BDC is a straight line.

Proof:
∠ADB = ∠ADC = 90° (Angle in the semi circle)
Now,
∠ADB + ∠ADC = 180°
⇒ ∠BDC is straight line.
Thus, D lies on the BC.

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Answer

Given,
AC is the common hypotenuse. ∠B = ∠D = 90°.
To prove,
∠CAD = ∠CBD

Proof:
Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi circle and AC is the diameter of the circle.
⇒ Points A, B, C and D are concyclic.
Thus, CD is the chord.
⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

12. Prove that a cyclic parallelogram is a rectangle.

Answer

Given,
ABCD is a cyclic parallelogram.
To prove,
ABCD is rectangle.

Proof:
∠1 + ∠2 = 180° (Opposite angles of a cyclic parallelogram)
also, Opposite angles of a cyclic parallelogram are equal.
Thus,
∠1 = ∠2
⇒ ∠1 + ∠1 = 180°
⇒ ∠1 = 90°
One of the interior angle of the paralleogram is right angled. Thus, ABCD is a rectangle.

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Chapter 10 Circles NCERT Solutions will help you in understanding about circles, other related terms and some properties of a circle. NCERT Textbooks for Class 9 are prepared in such a way that students can understand topics in an interesting way.

• Circles and Its Related Terms: A Review: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. A piece of a circle between two points is called an arc. The length of the complete circle is called its circumference. The region between a chord and either of its arcs is called a segment of the circle. When two arcs are equal, that is, each is a semicircle, then both segments and both sectors become the same and each is known as a semicircular region.

• Angle Subtended by a Chord at a Point: Equal chords of a circle subtend equal angles at the centre. If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

• Perpendicular from the Centre to a Chord: The perpendicular from the centre of a circle to a chord bisects the chord. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

• Circle through Three Points: There is one and only one circle passing through three given non-collinear points.

• Equal Chords and Their Distances from the Centre: The length of the perpendicular from a point to a line is the distance of the line from the point.

(i) Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

(ii) Chords equidistant from the centre of a circle are equal in length.

• Angle Subtended by an Arc of a Circle: If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.

(i) The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

(ii) Angles in the same segment of a circle are equal.

(iii) If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

• Cyclic Quadrilaterals: A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

There are total six exercises in the Chapter 10 Class 9 NCERT Solutions which can be used to complete homework and understand the concepts behind the questions. You can find exercisewise Class 9 Maths NCERT Solutions from the links given below.

Exercise 10.1 Chapter Class 10 Maths NCERT Solutions
Exercise 10.2 Chapter Class 10 Maths NCERT Solutions
Exercise 10.3 Chapter Class 10 Maths NCERT Solutions
Exercise 10.4 Chapter Class 10 Maths NCERT Solutions
Exercise 10.5 Chapter Class 10 Maths NCERT Solutions
Exercise 10.6 Chapter Class 10 Maths NCERT Solutions

These NCERT Solutions prepared by Studyrankers experts have taken every care so students can easily understand and clear their doubts. These solutions are prerequisites before going for supplementary Maths Books.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 10 Circles

How many exercises in Chapter 10 Circles?

Chapter 10 Circles Class 9 NCERT Solutions for Maths contains total six exercises in which the last one if optional. You will get detailed and accurate NCERT Solutions of every question so you can always find them whenever any difficulty occur.

The diameter of circle is 3.8 cm. Find the length of its radius.

Since, the diameter of circle is double its radius.

∴ Diameter = 2 x Radius

⇒ (1/2)x Diameter = Radius

⇒ Radius = (1/2) x 3.8 cm

= 1/2 x 38/10 cm

= 19/10 = 1.9 cm.

What is an arc?

The length of the complete circle is called its circumference, whereas a piece of a circle between two points is called an arc. If the length of an arc is less than the semicircle, then it is a minor arc, otherwise, it is a major arc.

What is a diameter?

A chord, passing through the centre is called a diameter of the circle.

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Board exams 2021, JEE, NEET will be based on reduced Syllabus: Education Minister

January 18, 2021, 9:16 pm

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Board exams 2021, JEE, NEET will be based on reduced Syllabus: Education Minister

Union Education Minister Ramesh Pokhriyal 'Nishank' has come live on social media platform Twitter on 12.20 pm on Monday. He clarified that all the questions asked in the JEE and NEET examinations will be asked from the part of the syllabus taught to them. Union Minister Nishank was interacting with students of Kendriya Vidyalayas across the country through an education dialogue program. Earlier, due to COVID-19, CBSE has reduced their syllabus by 30 percent.

Nishank told the students that it is not possible to call all the students in the classroom even after the opening of schools in the situation. In such a situation, online education of students will continue. On a question related to teaching Artificial Intelligence (AI) in schools under the new National Education Policy, the Union Education Minister said that at present it will be taught in the schools from Class 9.

On the webinar, he encouraged all the students and also appreciated the cooperation of teachers, principals and parents. The Union Education Minister said, "We have introduced a 'Manodarpan' platform to help students in improving their mental health in this tough time. This was done to improve the mental state of the students. Then Prime Minister Narendra Modi launched Fit India for physical efficiency."

During the webinar, he suggested that students should write their experiences of the Corona crisis period in their diary. Also, Students can share their any experience of Corona crisis by writing directly to Education Minister.

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NCERT Solutions for Class 9 Maths Chapter 11 Constructions

January 19, 2021, 3:43 am

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NCERT Solutions for Class 9 Maths Chapter 11 Constructions| PDF Download

Chapter 11 Constructions Class 9 Maths NCERT Solutions will be help you in knowing variety of concepts and developing problem solving skills. You can download PDF of Chapter 11 Constructions NCERT Solutions for Class 9 Maths from this page through which one can study at their ease. These Chapter 11 NCERT Solutions will help you in completing homework in no time and make aware of the difficulty of questions.

Class 9 Maths NCERT Solutions will be useful in building your own answers and getting good marks in the examination. These solutions are updated according to the latest NCERT Class 9 Maths textbook. After understanding the important concepts of the chapter, a student can minimise their errors.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Page No: 191

Exercise 11.1

1. Construct an angle of 90° at the initial point of a given ray and justify the construction.

Answer

Steps of construction:

Step 1: A ray YZ is drawn.
Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.
Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.
Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray XY making an angle 90° with YZ is formed.

Justification for construction:
We constructed ∠BYZ = 60° and also ∠AYB = 60°.
Thus, ∠AYZ = 120°.
Also, bisector of ∠AYB is constructed such that:
∠AYB = ∠XYA + ∠XYB
⇒ ∠XYB = 1/2∠AYB
⇒ ∠XYB = 1/2×60°
⇒ ∠XYB = 30°
Now,
∠XYZ = ∠BYZ + ∠XYB = 60° + 30° = 90°

2. Construct an angle of 45° at the initial point of a given ray and justify the construction.

Answer

Steps of construction:

Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.
Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.
Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray making an angle 90° with YZ is formed.
Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of ∠XOY is drawn.

Justification for construction:
By construction,
∠XOY = 90°
We constructed the bisector of ∠XOY as DOY.
Thus,
∠DOY = 1/2 ∠XOY
∠DOY = 1/2×90° = 45°

3. Construct the angles of the following measurements:
(i) 30° (ii) 22.5° (iii) 15°

Answer

(i) 30°

Steps of constructions:
Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.
Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.
Thus, ∠XOY is the required angle making 30° with OY.

(ii) 22.5°

Steps of constructions:
Step 1: An angle ∠XOY = 90° is drawn.
Step 2: Bisector of ∠XOY is drawn such that ∠BOY = 45° is constructed.
Step 3: Again, ∠BOY is bisected such that ∠AOY is formed.
Thus, ∠AOY is the required angle making 22.5° with OY.

(iii) 15°

Steps of constructions:
Step 1: An angle ∠AOY = 60° is drawn.
Step 2: Bisector of ∠AOY is drawn such that ∠BOY = 30° is constructed.
Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of ∠BOY is drawn.
Thus, ∠XOY is the required angle making 15° with OY.

4. Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°

Answer

(i) 75°

Steps of constructions:

Step 1: A ray OY is drawn.
Step 2: An arc BAE is drawn with O as a centre.
Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.
Step 4: With A and B as centres, arcs are made to intersect at X and ∠XOY = 90° is made.
Step 5: With A and C as centres, arcs are made to intersect at D
Step 6: OD is joined and and ∠DOY = 75° is constructed.
Thus, ∠DOY is the required angle making 75° with OY.

(ii) 105°

Steps of constructions:

Step 1: A ray OY is drawn.
Step 2: An arc ABC is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With B and C as centres, arcs are made to intersect at X
Step 6: OX is joined and and ∠XOY = 105° is constructed.
Thus, ∠XOY is the required angle making 105° with OY.

(iii) 135°

Steps of constructions:Step 1: A ray DY is drawn.
Step 2: An arc ACD is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With F and D as centres, arcs are made to intersect at X or bisector of ∠EOD is constructed.
Step 6: OX is joined and and ∠XOY = 135° is constructed.
Thus, ∠XOY is the required angle making 135° with DY.

5. Construct an equilateral triangle, given its side and justify the construction.

Answer

Steps of constructions:
Step 1: A line segment AB=4 cm is drawn.
Step 2: With A and B as centres, two arcs are made.
Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60° each.
Step 5: Lines from A and B are extended to meet each other at C.
Thus, ABC is the required triangle formed.

Justification:
By construction,
AB = 4 cm, ∠A = 60° and ∠B = 60°
We know that,
∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle)
⇒ 60° + 60° + ∠C = 180°
⇒ 120° + ∠C = 180°
⇒ ∠C = 60°
BC = CA = 4 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 4 cm
∠A = ∠B = ∠C = 60°

Page No: 195

Exercise 11.2

1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Answer

Steps of Construction:
Step 1: A line segment BC of 7 cm is drawn.
Step 2: At point B, an angle ∠XBC is constructed such that it is equal to 75°.
Step 3: A line segment BD = 13 cm is cut on BX (which is equal to AB+AC).
Step 3: DC is joined and ∠DCY = ∠BDC is made.
Step 4: Let CY intersect BX at A.
Thus, ΔABC is the required triangle.

2. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB - AC = 3.5 cm.

Answer

Steps of Construction:
Step 1: A line segment BC = 8 cm is drawn and at point B, make an angle of 45° i.e. ∠XBC.
Step 2: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.
Step 3: Join DC and draw the perpendicular bisector PQ of DC.
Step 4: Let it intersect BX at point A. Join AC.
Thus, ΔABC is the required triangle.

3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Answer

Steps of Construction:
Step 1: A ray QX is drawn and cut off a line segment QR = 6 cm from it.
Step 2:. A ray QY is constructed making an angle of 60º with QR and YQ is produced to form a line YQY'
Step 3: Cut off a line segment QS = 2cm from QY'. RS is joined.
Step 5: Draw perpendicular bisector of RS intersecting QY at a point P. PR is joined.
Thus, ΔPQR is the required triangle.

4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Answer

Steps of Construction:
Step 1: A line segment PQ = 11 cm is drawn. (XY + YZ + ZX = 11 cm)
Step 2: An angle, ∠RPQ = 30° is constructed at point A and an angle ∠SQP = 90° at point B.
Step 3: ∠RPQ and ∠SQP are bisected . The bisectors of these angles intersect each other at point X.
Step 4: Perpendicular bisectors TU of PX and WV of QX are constructed.
Step V: Let TU intersect PQ at Y and WV intersect PQ at Z. XY and XZ are joined.
Thus, ΔXYZ is the required triangle.

5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Answer

Steps of Construction:

Step 1: A ray BX is drawn and a cut off a line segment BC = 12 cm is made on it.

Step 2: ∠XBY = 90° is constructed.
Step 3: Cut off a line segment BD = 18 cm is made on BY. CD is joined.
Step 4: Perpendicular bisector of CD is constructed intersecting BD at A. AC is joined.
Thus, ΔABC is the required triangle.

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NCERT Solutions for Class 9 Maths Chapter 11 Constructions

NCERT Solutions of Chapter 11 Constructions provided here are detailed and accurate that will be given in depth study of concepts. Through this chapter you will learn some basic constructions and then after construct certain kinds of triangles.

• Basic Constructions: We will learn:

1. To construct the bisector of a given angle.

2. To construct the perpendicular bisector of a given line segment.

3. To construct an angle of 60° at the initial point of a given ray.

• Some Constructions of Triangles: We will learn:

1. To construct a triangle, given its base, a base angle and sum of other two sides.

2. To construct a triangle given its base, a base angle and the difference of the other two sides.

3. To construct a triangle, given its perimeter and its two base angles.

There are only two exercises in the whole chapter which will let you learn variety of constructions with given measures. You can always find exercisewise NCERT Solutions for Chapter 11 just by clicking on the links given below.

While preparing these NCERT Solutions for Class 9 Maths, Studyrankers experts have taken every care and provided a detailed step by step solutions of every question which will help you in learning the concepts embedded in the question.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 11 Constructions

Why we should solve NCERT Solutions for Chapter 11 Constructions Class 9?

These NCERT Solutions will make you equipped with variety of concepts which can be helpful in solving questions in the examinations. Also, our subject matter experts have prepared these Chapter 11 NCERT Solutions as per the latest marking scheme which let you score more marks in tests.

What are the instruments required for drawing geometrical figures?

A protractor, a pair of compasses, a pair of set squares, a pair of dividers, a graduated scale are required instruments for drawing geometrical figures.

How many exercises in Chapter 11 Constructions?

Chapter 11 Constructions consists of two exercises. In the first one, you will learn to draw basic constructions such as constructing a bisector while in the second one you will have to draw triangles with given measures.

How can I understand the topics given in Chapter 11 Class 9 Maths?

It is very necessary to understand the topics present in Chapter 11 Class 9 Maths and NCERT Solutions is one of the best way of doing. By practicing questions, you will get to know the concepts behind them and will make you ready for supplementary Maths Books of Class 9.

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NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula

January 19, 2021, 5:01 am

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NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula| PDF Download

On this page you will get Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths that will useful in the preparation of examinations and make you aware of concepts of the chapter. If you want to study offline then you can also Download PDF of NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula. You can also complete your homework on time through the help of these Chapter 12 NCERT Solutions and able to solve the difficult problems in a given in a exercise.

NCERT Solutions for Class 9 Maths is good way through which one can build their basic knowledge and getting good marks in the examination. These NCERT Solutions are updated according to the latest NCERT Class 9 Maths textbook and syllabus.

NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula

Page No: 202

Exercise 12.1

1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Answer

Length of the side of equilateral triangle = a
Perimeter of the signal board = 3a = 180 cm
∴ 3a = 180 cm ⇒ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
Using heron's formula,
Area of the signal board = √s (s-a) (s-b) (s-c)
                                       = √(3a/2) (3a/2 - a) (3a/2 - a) (3a/2 - a)
                                       = √3a/2 × a/2 × a/2× a/2
                                       = √3a⁴/16
                                       = √3a²/4
                                       = √3/4 × 60 × 60 = 900√3 cm²

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer

The sides of the triangle are 122 m, 22 m and 120 m.
Perimeter of the triangle is 122 + 22 + 120 = 264m
Semi perimeter of triangle (s) = 264/2 = 132 m
Using heron's formula,
Area of the advertisement = √s (s-a) (s-b) (s-c)
                                       = √132(132 - 122) (132 - 22) (132 - 120) m²
                                       = √132 × 10 × 110 × 12 m²
                                       = 1320 m²
Rent of advertising per year = ₹ 5000 per m²
Rent of one wall for 3 months = ₹ (1320 × 5000 × 3)/12 = ₹ 1650000

Page No: 203

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer

Sides of the triangular wall are 15 m, 11 m and 6 m.
Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m
Using heron's formula,
Area of the message = √s (s-a) (s-b) (s-c)
                                       = √16(16 - 15) (16 - 11) (16 - 6) m²
                                       = √16 × 1 × 5 × 10 m²= √800 m²
                                       = 20√2 m²

4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer

Two sides of the triangle = 18cm and 10cm
Perimeter of the triangle = 42cm
Third side of triangle = 42 - (18+10) cm = 14cm
Semi perimeter of triangle = 42/2 = 21cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √21(21 - 18) (21 - 10) (21 - 14) cm²
                                       = √21 × 3 × 11 × 7 m²
                                       = 21√11 cm²

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Answer

Ratio of the sides of the triangle = 12 : 17 : 25
Let the common ratio be x then sides are 12x, 17x and 25x
Perimeter of the triangle = 540cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
⇒ x = 10
Sides of triangle are,
12x = 12 × 10 = 120cm
17x = 17 × 10 = 170cm
25x = 25 × 10 = 250cm
Semi perimeter of triangle(s) = 540/2 = 270cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √270(270 - 120) (270 - 170) (270 - 250)cm²
                                       = √270 × 150 × 100 × 20 cm²
                                       = 9000 cm²

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer

Length of the equal sides = 12cm
Perimeter of the triangle = 30cm
Length of the third side = 30 - (12+12) cm = 6cm
Semi perimeter of the triangle(s) = 30/2 cm = 15cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 12) (15 - 12) (15 - 6)cm²
                                       = √15 × 3 × 3 × 9 cm²
                                       = 9√15 cm²

Page No: 206

Exercise 12.2

1. A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer

∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
BD is joined.

In ΔBCD,
By applying Pythagoras theorem,
BD² = BC²+ CD^²
⇒ BD² = 12²+ 5^²
⇒ BD² = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m²
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 13) (15 - 9) (15 - 8) m²
                                       = √15 × 2 × 6 × 7 m²
                                       = 6√35 m²= 35.5 m² (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m²+35.5m² = 65.5m²

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Answer

AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm

In ΔABC,
By applying Pythagoras theorem,
AC² = AB²+ BC^²
⇒ 5² = 3²+ 4^²
⇒ 25 = 25
Thus, ΔABC is a right angled at B.
Area of ΔBCD = 1/2 × 3 × 4 = 6 cm²
Now,
Semi perimeter of ΔACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
                                       = √7(7 - 5) (7 - 5) (7 - 4) cm²
                                       = √7 × 2 × 2 × 3 cm²
                                       = 2√21 cm²= 9.17 cm² (approx)
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm²+9.17 cm² = 15.17 cm²

3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

Answer

Length of the sides of the triangle section I = 5cm, 1cm and 5cm
Perimeter of the triangle = 5 + 5 + 1 = 11cm
Semi perimeter = 11/2 cm = 5.5cm
Using heron's formula,
Area of section I = √s (s-a) (s-b) (s-c)
                                       = √5.5(5.5 - 5) (5.5 - 5) (5.5 - 1) cm²
                                       = √5.5 × 0.5 × 0.5 × 4.5 cm²
                                       = 0.75√11 cm²= 0.75 × 3.317cm² = 2.488cm² (approx)
Length of the sides of the rectangle of section I = 6.5cm and 1cm
Area of section II = 6.5 × 1 cm² =6.5 cm²
Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.
Area of the trapezium = 3 × √3/4 × 1²cm²= 1.3 cm²(approx)
Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm
Area of region IV and V = 2 × 1/2 × 6 × 1.5cm²= 9cm²
Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm²=19.3 cm²

4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Answer

Given,
Area of the parallelogram and triangle are equal.

Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.
Perimeter of the triangle = 26 + 28 + 30 = 84 cm
Semi perimeter of the triangle = 84/2 cm = 42 cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √42(42 - 26) (46 - 28) (46 - 30) cm²
                                       = √46 × 16 × 14 × 16 cm²
                                       = 336 cm²Let height of parallelogram be h.
Area of parallelogram = Area of triangle
28cm × h = 336 cm²
⇒h = 336/28 cm
⇒h = 12 cm
The height of the parallelogram is 12 cm.

Page No: 207

5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Answer

Diagonal AC divides the rhombus ABCD into two congruent triangles of equal area.

Semi perimeter of ΔABC = (30 + 30 + 48)/2 m = 54 m
Using heron's formula,
Area of the ΔABC = √s (s-a) (s-b) (s-c)
                                       = √54(54 - 30) (54 - 30) (54 - 48) m²
                                       = √54 × 24 × 24 × 6 cm²
                                       = 432 m²
Area of field = 2 × area of the ΔABC = (2 × 432)m²= 864 m²
Thus,
Area of grass field which each cow will be getting = 864/18m²=48 m²

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer

Semi perimeter of each triangular piece = (50 + 50 + 20)/2 cm = 120/2 cm = 60cm
Using heron's formula,
Area of the triangular piece = √s (s-a) (s-b) (s-c)
                                       = √60(60 - 50) (60 - 50) (60 - 20) cm²
                                       = √60 × 10 × 10 × 40 cm²
                                       = 200√6 cm²
Area of triangular piece = 5 × 200√6 cm²= 1000√6cm²

7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

Answer

We know that,
As the diagonals of a square bisect each other at right angle.
Area of given kite = 1/2 (diagonal)²
                      = 1/2 × 32 × 32 = 512 cm²
Area of shade I = Area of shade II
⇒ 512/2 cm²= 256cm²
So, area of paper required in each shade = 256 cm²
For the III section,
Length of the sides of triangle = 6cm, 6cm and 8cm
Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm
Using heron's formula,
Area of the III triangular piece = √s (s-a) (s-b) (s-c)
                                       = √10(10 - 6) (10 - 6) (10 - 8) cm²
                                       = √10 × 4 × 4 × 2 cm²
                                       = 8√6 cm²

8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm² .

Answer

Semi perimeter of the each triangular shape = (28 + 9 + 35)/2 cm = 36 cm
Using heron's formula,
Area of the each triangular shape = √s (s-a) (s-b) (s-c)
                                       = √36(36 - 28) (36 - 9) (36 - 35) cm²
                                       = √36 × 8 × 27 × 1 cm²
                                       = 36√6 cm²= 88.2 cm²
Total area of 16 tiles = 16 × 88.2 cm²= 1411.2 cm²Cost of polishing tiles = 50p per cm²
Total cost of polishing the tiles = Rs. (1411.2 × 0.5) = Rs. 705.6

9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer

Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.
CM ⊥ AB and CE || AD.
In ΔBCE,
BC = 14m, CE = AD = 13 m and
BE = AB - AE = 25 - 10 = 15m
Semi perimeter of the ΔBCE = (15 + 13 + 14)/2 m = 21 m
Using heron's formula,
Area of the ΔBCE = √s (s-a) (s-b) (s-c)
                                       = √21(21 - 14) (21 - 13) (21 - 15) m²
                                       = √21 × 7 × 8 × 6 m²
                                       = 84 m²
also, area of the ΔBCE = 1/2 × BE × CM = 84 m²
⇒ 1/2 × 15 × CM = 84 m²
⇒ CM = 168/15 m²
⇒ CM = 56/5 m²
Area of the parallelogram AECD = Base × Altitude = AE × CM = 10 × 84/5 = 112 m²
Area of the trapezium ABCD = Area of AECD + Area of ΔBCE
                                                = (112+ 84) m²= 196 m²

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NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula

Chapter 12 Heron's Formula NCERT Solutions is a great way through which you can find area of triangles. You know area of a triangles is 1/2 × base × height. When it is not possible to find the height of the triangle easily and measures of all the three sides are known then we use Heron’s formula.

• Area of a Triangle — by Heron’s Formula:

where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = a+b+c/2

• Application of Heron’s Formula in Finding Areas of Quadrilaterals: We will extend the use of Heron's Formula to find the area of quadrilaterals. We can divide the quadrilateral in triangular parts and then use the formula for area of the triangle.

There are only two exercises in chapter 12 Heron's Formula where you will learn to grasp key concepts of the chapter properly. Below you can find exercisewise NCERT Solutions of Heron's Formula just by clicking on the links.

Studyrankers experts have prepared these Chapter 12 NCERT Solutions that are detailed and accurate through which a student can clear their doubts easily. By solving the questions from Class 9 NCERT textbook, a student will gain confidence that is going to help them in exams.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 12 Heron's Formula

What are the benefits of NCERT Solutions for Chapter 12 Heron's Formula Class 9 NCERT Solutions?

NCERT Solutions are very important in improving problem skills and understand all the important points of the chapter. Here we have detailed every solutions step by step so you can easily know the concepts.

How many exercises in Chapter 9 Areas of Parallelograms and Triangles?

Only two exercises are there in the chapter 12 Heron's Formula. In the first exercise, you need to find the area of triangles by using heron's formula. In the second exercise, you have to find the area of quadrilaterals. For finding area of a quadrilateral we divide it into various triangles. After we need to use Heron’s formula to find the area of the triangles.

Find the length of a diagonal of a square whose side is 2 cm.

The diagonal of a square = (√2) a cm

∴ Length of the diagonal = (√2) x 2 cm = 2√2cm.

Find the height of an equilateral triangle whose side is 2 cm.

Since height of an equilateral triangle is given by

Height = (√3/2) x side

⇒ height =(√3/2) x 2 cm = 3 cm.

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MCQ Questions for Class 9 Geography: Chapter 1 भारत - आकार और स्थिति

January 19, 2021, 10:40 pm

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MCQ Questions for Class 9 Geography: Chapter 1 भारत - आकार और स्थिति

In this page you will find Chapter 1 भारत - आकार और स्थिति Class 9 Geography MCQ Questions with answers through which you can know the important points given in the chapter. MCQ Questions for Class 9 will help you a lot in preparing for the exams and understand the latest pattern introduced by CBSE.

1 निम्न में से भारत का अक्षांशीय विस्तार कितना है ?

(a) 8° 4' उत्तर से 37° 6' उत्तर

(b) 8°2' उत्तर से 30°6' उत्तर

(d) 10° उत्तर से 30° 9' उत्तर

► (a) 8° 4' उत्तर से 37° 6' उत्तर

2. भारत निम्न में से किस गोलार्द्ध में स्थित है?

(a) उत्तर-पश्चिमी गोलार्द्ध

(b) पूर्वी-दक्षिणी गोलार्द्ध

(d) दक्षिणी गोलार्द्ध

► (c) उत्तर-पूर्वी गोलार्द्ध

3. निम्न में से कौन-सा द्वीप समूह अरब सागर में स्थित है?

(a) लक्षद्वीप

(b) अंडमान और निकोबार द्वीप समूह

(d) मेडागास्कर

► (a) लक्षद्वीप

4. भारत का देशान्तरीय विस्तार कितना है ?

(a) 66°7' पूर्व से 97°25' पूर्व

(b) 68°7' पूर्व से 97°25' पूर्व

(d) 65°7' पूर्व से 97°10' पश्चिम

► (b) 68°7' पूर्व से 97°25' पूर्व

5. भारत का अंडमान और निकोबार द्वीप समूह निम्न में से किस स्थान पर स्थित है?

(a) अरब सागर

(b) बंगाल की खाड़ी

(d) कोरल सागर

► (b) बंगाल की खाड़ी

6. भारत का क्षेत्रफल विश्व के कुल भौगोलिक क्षेत्रफल का कितना प्रतिशत है?

(a) 2.0%

(b) 2.5%

(d) 2.9%

► (c) 2.4%

7. निम्न में से कौन-सी अक्षांश रेखा भारत को लगभग दो बराबर भागों में बाँटती है?

(a) मकर रेखा

(b) आर्कटिक रेखा

(d) कर्क रेखा

► (d) कर्क रेखा

8. निम्न में से कौन सा महासागर भारतीय उप-महाद्वीप के सुदूर दक्षिण में स्थित है?

(a) हिन्द महासागर

(b) अन्ध महासागर

(d) आर्कटिक महासागर

► (a) हिन्द महासागर

9. निम्न में से किस नहर के खुलने से भारत और यूरोप के बीच की दूरी कम हो गई ?

(a) स्वेज़ नहर

(b) इंदिरा गाँधी नहर

(d) ऑरेंज नहर

► (a) स्वेज़ नहर

10. निम्न में से कौन-सा राज्य भारत की मानक याम्योत्तर पर स्थित है?

(a) उत्तर प्रदेश

(b) गुजरात

(d) आन्ध्र प्रदेश

► (b) गुजरात

11. निम्न में से किसे भारत की मानक याम्योत्तर के रूप में चुना गया है ?

(a) 82°15' पूर्व

(b) 82°30' पश्चिम

(d) 82°15' दक्षिण

► (c) 82°30' पूर्व

12. निम्न में से कौन-सा अक्षांश भारत की पश्चिमी सीमा निर्धारित करता है ?

(a) 8°4' उत्तर

(b) 68°7' पश्चिमी

(d) 97°25' पूर्व

► (b) 68°7' पश्चिमी

13. भारत की मानक याम्योत्तर निम्न में से किस स्थान से होकर गुजरती है?

(a) कानपुर (उत्तर प्रदेश)

(b) मिर्जापुर (उत्तर प्रदेश)

(d) गुड़गांव (हरियाणा)

► (b) मिर्जापुर (उत्तर प्रदेश)

14. पाक जलसंधि निम्न में से किस देश को भारत से अलग करती है?

(क) श्रीलंका

(ख) मालदीव

(ग) भूटान

(घ) नेपाल

► (क) श्रीलंका

15. भारत में निम्न में से कितने राज्य हैं?

(a) 26

(b) 27

(d) 29

► (d) 29

16. पूर्व में अरूणाचल प्रदेश तथा पश्चिम में गुजरात का समयान्तराल निम्न में गजरात का समयान्तराल निम्न में से कितना है?

(a) 2 घण्टे

(b) 3 घण्टे

(d) 1.45 घण्टे

► (a) 2 घण्टे

17. निम्न में से कौन-से देश की सीमा गजरात, राजस्थान, पंजाब, जम्मू और कश्मीर से मिलती है?

(a) श्रीलंका

(b) पाकिस्तान

(d) भूटान

► (b) पाकिस्तान

18. निम्न में से कौन-सा जलमार्ग पश्चिम में यूरोपीय देशों और पूर्व में एशियाई देशों को मिलाते हुए भारत को केंद्रीय स्थिति प्रदान करता है ?

(a) अन्ध महासागरीय मार्ग

(b) आन्तरिक कैनेडियन रेलवे मार्ग

(d) प्रशान्त महासागरीय मार्ग

► (c) हिन्द महासागरीय मार्ग

19. निम्न में से किस कारणवश भारतीय संघ राज्य का सबसे दक्षिणत्तम बिंदु (इंदिरा बिंदु) जलमग्न हा गया?

(a) भूकम्प के कारण

(b) ज्वालामुखी के कारण

(d) चक्रवातों के कारण

► (c) सुनामी के कारण

20. विश्व में भारत के आकार के सम्बन्ध में निम्न में से कौन सा कथन सत्य है?

(क) पाँचवाँ बड़ा देश

(ख) छठा बड़ा देश

(ग) नौवां बड़ा देश

(घ) साँतवाँ बड़ा देश

► (घ) साँतवाँ बड़ा देश

↧

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

January 20, 2021, 8:10 am

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes| PDF Download

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

Page No: 213

Exercise 13.1

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m² costs Rs 20.

Answer

Length of plastic box (l) = 1.5 m
Width of plastic box (b) = 1.25 m
Depth of plastic box (h) = 0.65 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×0.65] + (1.5 × 1.25) m²
= (3.575 + 1.875) m²
= 5.45m²
The sheet required required to make the box is 5.45 m²

(ii) Cost of 1 m²of sheet= Rs 20
∴ Cost of 5.45m²of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m².

Answer

length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
= 2(5+4)×3 + (5×4) m²
= (54 + 20) m²
= 74m²Cost of white washing = ₹7.50 per m²
Total cost = ₹(74×7.50) = ₹ 555

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m² is ₹15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ₹15000
Rate per m²=₹10
Area of four walls = 2(l + b) h m²= (250×h) m²
A/q,
(250×h)×10 = ₹15000
⇒ 2500×h = ₹15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Answer

Volume of paint = 9.375 m²=93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm²
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2×468.75 cm² = 937.5cm²
Number of bricks can be painted = 93750/937.5 = 100

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?

Answer

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm²
= 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5) cm²
= 2(125+80+100) cm²
= (2×305) cm²= 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?

Answer

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm²
= 2(750 + 625 + 750) cm²
= 4250 cm²

(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm² , find the cost of cardboard required for supplying 250 boxes of each kind.

Answer

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm²
= 2(500 + 100 + 125) cm²
= 1450 cm²

Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm²
= 2(180 + 60 + 75) cm²
= 630 cm²

Total surface area of 250 boxes of each type = 250(1450 + 630) cm²= 250×2080 cm²= 520000cm²
Extra area required = 5/100(1450 + 630) × 250 cm²= 26000cm²

Total Cardboard required = 520000 + 26000 cm² = 546000cm²
Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3m?

Answer

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb = [2(4+3)×2.5 + 4×3] m²
= (35+12) m²
= 47 m²
Page No: 216

Exercise 13.2

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Answer

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m²
= (2 × 22 × 0.1 × 1.7) m²
=7.48 m²

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Answer

Let R be external radius and r be the internal radius h be the length of the pipe.

R = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²
= 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh + 2πRh + 2π(R^{2 -} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 - 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm²= 2038.08 cm²

Page No: 217

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Answer

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m² = 3.168 m²
Area of the playground = (500 × 3.168) m²= 1584 m²

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m².

Answer

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = ₹12.50 per m²
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m²
=5.5 m²
Total cost of painting = (5.5 × 12.5) = ₹68.75

6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Answer

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹40 per m².

Answer

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = ₹40 per m²
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m²
= 110 m²

(ii) Cost of plastering = ₹(110 × 40) = ₹4400

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m² = 4.4 m²

9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

Answer

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
= (2 × 22/7 × 2.1 × 4.5) m²
= 59.4 m²

(ii) Total surface area of the tank = 2πr(r + h) m²
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
= 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 - 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35)cm²
= 2200 cm²

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr²
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
= (198 + 198/7) cm²
= 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
= 7920 cm²

Page No: 221

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Answer

Radius (r) = 10.5/2 cm = 5.25 cm
Slant height (l) = 10 cm
Curved surface area of the cone = (πrl) cm²
= (22/7 × 5.25 × 10) cm²
=165 cm²

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m²
= 22/7 × 12 × (21 + 12) m²
= (22/7 × 12 × 33) m²
= 1244.57 m²

3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.

Answer

(i) Curved surface of a cone = 308 cm²
Slant height (l) = 14cm
Let r be the radius of the base
∴ πrℓ = 308
⇒ 22/7 × r × 14 = 308
⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm²
= 22/7 × 7 ×(14 + 7) cm²
= (22 × 21) cm²
= 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70.

Answer

Radius of the base (r) = 24 m
Height of the conical tent (h) = 10 m
Let l be the slant height of the cone.
∴ l²= h²+ r²
⇒ l = √h²+ r²
⇒ l = √10²+ 24²
⇒ l = √100+ 576
⇒ l = 26 m
(ii) Canvas required to make the conical tent = Curved surface of the cone
Cost of 1 m² canvas = ₹70
∴ πrl = (22/7 × 24 × 26) m² = 13728/7 m²
∴ Cost of canvas = ₹ 13728/7 × 70 = ₹137280

5. What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π = 3.14).

Answer

Radius of the base (r) = 6 m
Height of the conical tent (h) = 8 m
Let l be the slant height of the cone.
∴ l = √h²+ r²
⇒ l = √8²+ 6²
⇒ l = √100
⇒ l = 10 m
CSA of conical tent = πrl
= (3.14 × 6 × 10) m² = 188.4 m²
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be x.
20 cm will be wasted in cutting.
So, the length will be (x - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(x - 0.2 m) × 3] m = 188.4 m²
⇒ x - 0.2 m = 62.8 m
⇒ x = 63 m

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹210 per 100 m².

Answer

Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = πrl m²
=(227×25×7) m²
=550 m²
Rate of white- washing = ₹210 per 100 m²
Total cost of white-washing the tomb = ₹(550 × 210/100) = ₹1155

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer

Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = √h²+ r²
⇒ l = √24²+ 7²
⇒ l = √625
⇒ l = 25 m
Sheet required for one cap = Curved surface of the cone
= πrl cm²
= (22/7 × 7 × 25) cm²
= 550 cm²
Sheet required for 10 caps = 550 × 10 cm²= 5500 cm²

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m² , what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)

Answer

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l be the slant height of a cone.
∴ l = √h²+ r²
⇒ l = √1²+ 0.2²
⇒ l = √1.04
⇒ l = 1.02 m
Rate of painting = ₹12 per m²

Curved surface of 1 cone = πrl m²
= (3.14 × 0.2 × 1.02) m²
= 0.64056 m²
Curved surface of such 50 cones = (50 × 0.64056) m² = 32.028 m²
Cost of painting all these cones = ₹(32.028 × 12) = ₹384.34

Page No: 225

Exercise 13.4

1. Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm

Answer

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
= (4 × 22/7 × 5.6 × 5.6) cm²
= 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
= (4 × 22/7 × 14 × 14) cm²
= 2464 cm²

2. Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m

Answer

(i) r = 14/2 cm = 7cm
Surface area = 4πr²
= (4 × 22/7 × 7 × 7)cm²
=616cm²

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr²
= (4 × 22/7 × 1.75 × 1.75) m²
= 38.5 m²

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer

r = 10 cm
Total surface area of hemisphere = 3πr²
= (3 × 3.14 × 10 ×10) cm²
= 942 cm²

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
= r²/R²
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Answer

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²
Rate of tin - plating is = ₹16 per 100 cm²
Therefor, cost of 1 cm²=₹16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72

6. Find the radius of a sphere whose surface area is 154 cm².

Answer

Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr²= 154
⇒ 4 × 22/7 × r²= 154
⇒ r²= 154/(4 × 22/7)
⇒ r²= 49/4
⇒ r = 7/2 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
= (5 + 0.25) cm = 5.25 cm
Outer curved surface = 2πR²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
= 2π × r × 2r
= 4πr²
(iii) Ratio of the areas = 4πr²:4πr² = 1:1

Page No: 228

Exercise 13.5

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

Answer

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
= (4 × 2.5 × 1.5) cm³= 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15) cm³= 180 cm³

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m³ = 1000 l)

Answer

Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m³
=(6×5×4.5)m3=135 m³
Therefore , the tank can hold = 135 × 1000 litres  [Since 1m³=1000litres]
= 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth)
= 380/(10×8) m
= 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³.

Answer

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
= (8×6×3) m³
= 144 m³
Rate of digging = ₹30 per m³
Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Answer

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m³ = 1000 litres
∴ 50000 litres = 50000/1000 m³ = 50 m³
Breadth = Volume of cuboid/(Length×Depth)
= 50/(2.5×10) m
= 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

Answer

Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m³
= (20×15×6) m³
= 1800 m³
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
= (4000×150)/1000
= 600 m³
Number of days the water will last = Capacity of tank Total water required per day
=(1800/600) = 3
The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

Answer

Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m³ = 10000 m³
Dimension of crates = 1.5m × 1.25m × 0.5m
Volume of 1 crates = (1.5 × 1.25 × 0.5) m³ = 0.9375 m³
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer

Edge of the cube = 12 cm.
Volume of the cube = (edge)³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm³ = 216 cm³
Side of the smaller cube = a
a³= 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)²
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
= 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer

Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute
Volume of water flowing into the sea in a minute = lbh m³
= (100/3 × 40 × 3) m³
= 4000 m³

Page No: 230

Exercise 13.6

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)

Answer

Let the base radius of the cylindrical vessel be ‘r’ cm.
∴ Circumference = 2πr
⇒ 2πr = 132   [∵ Circumference = 132 cm]
⇒ 2× 22/7 × r = 132 cm
R=(132×7)/(2×22) cm = 21 cm
∵ Height of the vessel = 25 cm
∴ Volume = πr² x h   [∵ Volume of a cylinder = πr²h]
= (22/7) (21)² × 25 cm³
= (22/7) × 21 × 21 × 25 cm³
= 22 × 3 × 21 × 25 cm³
= 34650 cm³
∵ Capacity of the vessel = Volume of the vessel
∴ Capacity of cylindrical vessel = 34650 cm³
Since 1000 cm³= 1 litre
⇒ 34650 cm³ = (34650/1000) litres = 34.65 l

2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer

Here, Inner diameter of the cylindrical pipe = 24 cm
⇒ Inner radius of the pipe (r) = (24/2) cm = 12 cm
Outer diameter of the pipe = 28 cm
⇒ Outer radius of the pipe (R) = (28/2)cm = 14 cm
Length of the pipe (h) = 35 cm
∵ Inner volume of the pipe = πr²h
Outer volume of the pipe = πr²h
∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume
= πR²h - πr²h
= πh(R²-r²)
= πh(R+r)(R-r) [ ∵ a² - b² = (a+b)(a-b)]
= 22/7 x 35 x (14+12) x (14-12) cm³
= 22 x 5 x 26 x 2 cm³

Mass of the wood in the pipe = [Mass of wood in 1 m³ of wood] x [Volume of wood in the pipe]
= [0.6g] x [22 x 5 x 26 x 2] cm³
= (6/10)x 22 x 10 x 26 g = 6 x 22 x 26 g
= 3432 g = (3432/1000)= 3.432 kg [∵ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.

3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer

For rectangular pack: Length (l) = 5 cm
Breadth (b) = 4 cm Height (h) = 15 cm
∴ Volume = l x b x h = 5 x 4 x 15 cm³
= 300 cm³
⇒ Capacity of the rectangular pack = 300 cm³...(1)
For cylindrical pack: Base diameter = 7 cm
⇒ Radius of the base (r) = (7/2)cm
Height (h) = 10 cm
∴ Volume = πr²h = (22/7) x (7/2)² x 10 cm³
= (22/7) x (7/2) x (7/2) x 10 cm³
= 11 x 7 x 5 cm³ = 385 cm³
⇒ Volume of the cylindrical pack = 385 cm³...(2)
From (1) and (2),
we have 385 cm³ – 300 cm³ = 85 cm³
⇒ The cylindrical pack has the greater capacity by 85 cm³.

4. If the lateral surface of a cylinder is 94.2 cm²and its height is 5 cm, then find: (i) radius of its base (ii) its volume. (Use π = 3.14)

Answer

(i) Since lateral surface of the cylinder = 2 πrh
But lateral surface of the cylinder = 94.2 cm²
2πrh = 94.2
2×3.14×r×5 =942/10
⇒{(10×314)/100} × r = 942/10
⇒ r=(942/10)×{100/(10×314)}cm
⇒ r =471/157 cm
Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr²h
⇒ Volume of the given cylinder = 3.14 x (3)² x 5 cm³
=314×100×3×3×5cm³
=(157×3×3)/10
= 1413/10 = 141.3 cm³

Thus, the required volume = 141.3 cm³

5. It costs ₹ 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m²; find: (i) inner curved surface of the vessel (ii) radius of the base (iii) capacity of the vessel.

Answer

(i) To find inner curved surface
Total cost of painting = ₹ 2200
Rate of painting = ₹ 20 per m²
∴ Area =cost/rate = 2200/20 = 110 m²
⇒ Inner curved surface of the vessel = 110 m²

(ii) To find radius of the base Let the base radius of the cylindrical vessel.
∵ Curved surface of a cylinder = 2 πrh
∴ 2πrh = 110
⇒ 2× 22/7 ×r×10 = 110 [∵Height=10 m ]
⇒ r= (110×7)/(2×22×10)m = 7/4 m
= 1.75 m
⇒ The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel
Since, volume of a cylinder = πr²h
∴ Volume (capacity) of the vessel =22/7 × (7/2)² × 10 m³
= 22/7 × 7/4 × 7/4 × 10 m³
= (11×7×5)/4 m³ = 385/4 m3 = 96.25 m³

Since, 1 m³ = 1000000 cm³ = 1000 l = 1 kl
∴ 96.5 m³ = 96.5 kl
Thus, the required volume = 96.25 kl

6. The capacity of closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Answer

Capacity of the cylindrical vessel = 15.4 l
= 15.4×1000 cm³

⇒ Volume of the vessel = (15.4/1000)m³
Height of the vessel = 1m Let ‘r’ metres be the radius of the base of the vessel
∴ Volume = πr²h

⇒ πr²h= 15.4/1000

Thus, the required sheet = 0.4708 m²

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer

Since, 10 mm = 1cm
∴ 1 mm = (1/10) cm
For graphite cylinder

Thus, the required volume of the graphite = 0.11 cm³
For the pencil Diameter of the pencil = 7 mm = (7/10)cm
∴ Radius of the pencil (R) = (7/20) cm

Height of the pencil (h) = 14 cm
Volume of the pencil = πr²h

Volume of the wood Volume of the wood = [Volume of the pencil] – [Volume of the graphite]
= 5.39 cm³ – 0.11 cm³ = 5.28 cm³
Thus, the required volume of the wood is 5.28 cm³.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer

The bowl is cylindrical.
Diameter of the base = 7 cm
⇒ Radius of the base (r) = (7/3) cm
Height (h) = 4 cm
Volume of soup = πr²h

= 38500 / 100 liters

Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

Page No: 233

Exercise 13.7

1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm    (ii) radius 3.5 cm, height 12 cm

Answer

(i) Radius (r) = 6 cm
Height (h) = 7 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 6 × 6 × 7) cm³
                                = 264 cm³
(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 3.5 × 3.5 × 12) cm³
                                = 154 cm³

2. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm    (ii) height 12 cm, slant height 13 cm

Answer

(i) Radius (r) = 7 cm
Slant height (l) = 25 cm
Let h be the height of the conical vessel.
∴ h = √l²- r²
⇒ h = √25²- 7²
⇒ h = √576
⇒ h = 24 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 7 × 7 × 24) cm³
                                = 1232 cm³
Capacity of the vessel = (1232/1000) l = 1.232 l

(i) Height (h) = 12 cm
Slant height (l) = 13 cm
Let r be the radius of the conical vessel.
∴ r = √l²- ^h²
⇒ r = √13²- 12²
⇒ r = √25
⇒ r = 5 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 5 × 5 × 12) cm³
                                = (2200/7) cm³
Capacity of the vessel = (2200/7000) l = 11/35 l

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Answer

Height (h) = 15 cm
Volume = 1570 cm³
Let the radius of the base of cone be r cm
∴ Volume = 1570 cm³
⇒ 1/3 πr²h = 1570
⇒ 13 × 3.14 × r²× 15 = 1570
⇒ r²= 1570/(3.14×5) = 100
⇒ r = 10

4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Answer

Height (h) = 9 cm
Volume = 48π cm³
Let the radius of the base of the cone be r cm
∴ Volume = 48π cm³
⇒ 1/3 πr²h = 48π
⇒ 13 × r²× 9 = 48
⇒ 3r²= 48
⇒ r²= 48/3 = 16
⇒ r = 4

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer

Diameter of the top of the conical pit = 3.5 m
Radius (r) = (3.5/2) m = 1.75 m
Depth of the pit (h) = 12 m
Volume = 1/3 πr²h
             = (13 × 22/7 × 1.75 × 1.75 × 12) m³
             = 38.5 m³
1 m³= 1 kilolitre
Capacity of pit = 38.5 kilolitres.

6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone      (ii) slant height of the cone
(iii) curved surface area of the cone

Answer

(i) Diameter of the base of the cone = 28 cm
Radius (r) = 28/2 cm = 14 cm
Let the height of the cone be h cm
Volume of the cone = 13πr²h = 9856 cm³
⇒ 1/3 πr²h = 9856
⇒ 1/3 × 22/7 × 14 × 14 × h = 9856
⇒ h = (9856×3)/(22/7 × 14 × 14)
⇒ h = 48 cm

(ii) Radius (r) = 14 m
Height (h) = 48 cm
Let l be the slant height of the cone
l² = h²+ r²
⇒ l² = 48²+ 14²
⇒ l² = 2304+196
⇒ l²= 2500
⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m
Slant height (l) = 50 cm
Curved surface area = πrl
                                 = (22/7 × 14 × 50) cm²
                                 = 2200 cm²

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer

On revolving the ⃤ ABC along the side 12 cm, a right circular cone of height(h) 12 cm, radius(r) 5 cm and slant height(l) 13 cm will be formed.
Volume of solid so obtained = 1/3 πr²h
                                             = (1/3 × π × 5 × 5 × 12) cm³
                                             = 100π cm³

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Answer

On revolving the ⃤ ABC along the side 12 cm, a cone of radius(r) 12 cm, height(h) 5 cm, and slant height(l) 13 cm will be formed.
Volume of solid so obtained =1/3 πr²h
                                              = (1/3 × π × 12 × 12 × 5) cm³
                                            = 240π cm³
Ratio of the volumes = 100π/240π = 5/12 = 5:12

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer

Diameter of the base of the cone = 10.5 m
Radius (r) = 10.5/2 m = 5.25 m
Height of the cone = 3 m
Volume of the heap = 1/3 πr²h
                                = (1/3 × 22/7 × 5.25 × 5.25 × 3) m³
                                = 86.625 m³
Also,
l² = h²+ r²
⇒ l² = 3²+ (5.25)²
⇒ l² = 9 + 27.5625
⇒ l²= 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
                         = πrl = (22/7 × 5.25 × 6.05) m²
                         = 99.825 m²(approx)

Page No: 236

Exercise 13.8

1. Find the volume of a sphere whose radius is
(i) 7 cm        (ii) 0.63 m

Answer

(i) Radius of the sphere(r) = 7 cm
Therefore, Volume of the sphere = 4/3 πr³
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm³
                                                     = 4312/3 cm³

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr³
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³
                                   = 1.05 m³

2. Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm                     (ii) 0.21 m

Answer

(i) Diameter of the spherical ball = 28 cm
Radius = 28/2 cm = 14 cm
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                        = (4/3 × 22/7 × 14 × 14 × 14) cm³
                              = 34496/3 cm³

(ii) Diameter of the spherical ball = 0.21 m
Radius (r) = 0.21/2 m = 0.105 m
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (43×227×0.105×0.105×0.105) m³
                          = 0.004851 m³

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr³
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm³
                               = 38.808 cm³
Density of the metal is 8.9g per cm3
Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon?

Answer

Let the diameter of the moon be r
Radius of the moon = r/2
A/q,
Diameter of the earth = 4r
Radius(r) = 4r/2 = 2r
Volume of the moon = v = 4/3 π(r/2)³
                                  = 4/3 πr³× 1/8
⇒ 8v = 4/3 πr³--- (i)
Volume of the earth = r³ = 4/3 π(2r)³
                            = 4/3 πr³× 8
⇒ V/8 = 4/3 πr³ --- (ii)
From (i) and (ii), we have
8v = V/8
⇒ v = 1/64 V
Thus, the volume of the moon is 1/64 of the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Answer

Diameter of a hemispherical bowl = 10.5 cm
Radius(r) = (10.5/2) cm = 5.25cm
Volume of the bowl = 2/3 πr³
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm³
                                = 303.1875 cm³
Litres of milk bowl can hold = (303.1875/1000) litres
                                 = 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer

Internal radius = r = 1m
External radius = R = (1 + 0.1) cm = 1.01 cm
Volume of iron used = External volume – Internal volume
                                  = 2/3 πR³- 2/3 πr³
                                  = 2/3 π(R³- r³)
                            = 2/3 × 22/7 × [(1.01)³−(1)³] m³
                                  = 44/21 × (1.030301 - 1) m³
                                  = (44/21 × 0.030301) m³
                                  = 0.06348 m³(approx)

7. Find the volume of a sphere whose surface area is 154 cm².

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm²
⇒ 4πr²= 154
⇒ 4 × 22/7 × r²= 154
⇒ r²= (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr³
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm³
             = 539/3 cm³

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,      (ii) volume of the air inside the dome.

Answer

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
                                                        = (498.96/2.00) m²= 249.48 m²

(ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × 22/7 × r²= 249.48
⇒ r²= (249.48×7)/(2×22) = 39.69
⇒ r²= 39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr³
= (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m³
                                                       = 523.9 m³(approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,    (ii) ratio of S and S′.

Answer

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr³ --- (i)
Volume of the new sphere of radius r′ = 4/3 πr'³ --- (ii)
A/q,
4/3 πr'³= 27 × 4/3 πr³
⇒ r'³= 27r³
⇒ r'^³= (3r)³
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr²/4πr′²= r²/(3r)²
                              = r²/9r² = 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Answer

Diameter of the spherical capsule = 3.5 mm
Radius(r) = 3.52mm
              = 1.75mm
Medicine needed for its filling = Volume of spherical capsule
                                                  = 4/3 πr³
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm³
                                                  = 22.46 mm³(approx.)

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

Chapter 13 NCERT Solutions available here is useful in preparing for the exams. We will learn to find the surface areas and volumes of cuboids and cylinders in details and extend this study to some other solids such as cones and spheres.

• Surface Area of a Cuboid and a Cube: Surface Area of a Cuboid = 2(lb + bh + hl) and Surface Area of a Cube = 6a²

• Surface Area of a Right Circular Cylinder: Curved Surface Area of a Cylinder = 2πrh and Total Surface Area of a Cylinder = 2πr(r + h).

• Surface Area of a Right Circular Cone: Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl and Total Surface Area of a Cone = πrl + πr² = πr(l + r).

• Surface Area of a Sphere: A sphere is like the surface of a ball. The word solid sphere is used for the

solid whose surface is a sphere. Surface Area of a Sphere = 4 πr², Curved Surface Area of a Hemisphere = 2πr² and Total Surface Area of a Hemisphere = 3πr².

• Volume of a Cuboid: The measure of this occupied space is called the Volume of the object. Volume of a Cuboid = base area × height = length × breadth × height and Volume of a Cube = edge × edge × edge = a³.

• Volume of a Cylinder: The volume of a cylinder can be obtained as : base area × height = area of circular base × height = πr²h So, Volume of a Cylinder = πr²h.

• Volume of a Right Circular Cone: Volume of a Cone = 1/3 πr²h where r is the base radius and h is the height of the cone.

• Volume of a Sphere: Volume of a Sphere = 4/3πr³ where r is the radius of the sphere.

• Volume of a Hemisphere = 2/3πr³ where r is the radius of the hemisphere.

Surface Areas and Volumes contains total nine exercises in which the last one is optional not important for examinations. Below, we have provided exercisewise Chapter 13 NCERT Solutions which you can check.

Exercise 13.1 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.2 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.3 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.4 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.5 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.6 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.7 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.8 Chapter 13 Class 9 Maths NCERT Solutions
Exercise 13.9 Chapter 13 Class 9 Maths NCERT Solutions

Studyrankers experts have prepared these NCERT Solutions for Class 9 Maths in which every question's answers are detailed step by step that will be give in depth study of concepts.

NCERT Solutions for Class 9 Maths Chapters:

FAQ on Chapter 13 Surface Areas and Volumes

Why we should solve NCERT Solutions for Chapter 13 Surface Areas and Volumes Class 9?

These Chapter 13 NCERT Solutions provided here will make you equipped with variety of concepts and help you in learning the concepts embedded in the question. Through these NCERT Solutions, one can easily complete their homework.

What is total surface area of a cone?

Total Surface Area of a Cone = πrl + r² = πr(l + r).

What do you mean by Solids?

The bodies occupying space are called solids, such as a cuboid, a cube, a cylinder, a cone, a sphere, etc. These solids have plane or curved surfaces.

The floor area of a room is 100 m² and its height is 8 m. Find its volume.

∵ Volume of a cuboid = [Base area] × Height

∴ Volume of the room = [Area of the floor] × height = 100 m² × 8 m = 800 m³

Thus, the volume of the room = 800 m³.

↧

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NCERT Solutions for Class 9 Maths Chapter 15 Probability| PDF Download

Here you will get Chapter 15 Probability NCERT Solutions for Class 9 Maths which will develop you understanding of the chapter and obtain maximum marks in the exams. These NCERT Solutions which are accurate and detailed and as per the latest syllabus released by CBSE. You can download PDF of Class 9 Maths Chapter 15 Probability NCERT Solutions make you able to solve the difficult problems in a exercise.

Studyrankers experts have taken every while preparing these Chapter 15 Probability NCERT Solutions that will develop your necessary skill to solve more and more questions. These NCERT Solutions will provide good experience and provide opportunities to learn new things. Chapter 15 NCERT Solutions help in solving the difficulties that lie ahead with ease.

NCERT Solutions for Class 9 Maths Chapter 15 Probability

Page No: 283

Exercise 15.1

1. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Answer

Total numbers of balls = 30
Numbers of boundary = 6
Numbers of time she didn't hit boundary = 30 - 6 = 24
Probability she did not hit a boundary = 24/30 = 4/5

2. 1500 families with 2 children were selected randomly, and the following data were recorded:

Number of girls in a family	2	1	0
Number of families	475	814	211

Compute the probability of a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.

Answer

Total numbers of families = 1500

(i) Numbers of families having 2 girls = 475
Probability = Numbers of families having 2 girls/Total numbers of families
= 475/1500 = 19/60
(ii) Numbers of families having 1 girls = 814
Probability = Numbers of families having 1 girls/Total numbers of families
= 814/1500 = 407/750
(iii) Numbers of families having 2 girls = 211
Probability = Numbers of families having 0 girls/Total numbers of families
= 211/1500
Sum of the probability = 19/60 + 407/750 + 211/1500
= (475 + 814 + 211)/1500 = 1500/1500 = 1
Yes, the sum of these probabilities is 1.

3. Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.

Answer

Total numbers of students = 40
Numbers of students = 6
Required probability = 6/40 = 3/20

4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome	3 heads	2 heads	1 head	No head
Frequency	23	72	77	28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Answer

Number of times 2 heads come up = 72
Total number of times the coins were tossed = 200
Required probability = 72/200 = 9/25

5. An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income (in ₹)	Vehicles per family
Monthly income (in ₹)	0	1	2	Above 2
Less than 7000	10	160	25	0
7000-10000	0	305	27	2
10000-13000	1	535	29	1
13000-16000	2	469	59	25
16000 or more	1	579	82	88

Suppose a family is chosen. Find the probability that the family chosen is

(i) earning ₹10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹7000 per month and does not own any vehicle.
(iv) earning ₹13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.

Answer

Total numbers of families = 2400

(i) Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29
Required probability = 29/2400

(ii) Number of families earning ₹16000 or more per month and owning exactly 1 vehicle = 579
Required probability = 579/2400

(iii) Number of families earning less than ₹7000 per month and does not own any vehicle = 10 Required probability = 10/2400 = 1/240

(iv) Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25
Required probability = 25/2400 = 1/96

(v) Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579

= 2062

Required probability = 2062/2400 = 1031/1200

Page No: 284

6. Refer to Table 14.7, Chapter 14.
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.

Marks	Number of students
0 - 20	7
20 - 30	10
30 - 40	10
40 - 50	20
50 - 60	20
60 - 70	15
70 - above	8
Total	90

Answer

Total numbers of students = 90

(i) Numbers of students obtained less than 20% in the mathematics test = 7
Required probability = 7/90

(ii) Numbers of student obtained marks 60 or above = 15+8 = 23
Required probability = 23/90

7. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

Opinion	Number of students
like	135
dislike	65

Find the probability that a student chosen at random
(i) likes statistics, (ii) does not like it.

Answer

Total numbers of students = 135 + 65 = 200

(i) Numbers of students who like statistics = 135
Required probability = 135/200 = 27/40

(ii) Numbers of students who does not like statistics = 65
Required probability = 65/200 = 13/40

8. Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within 1/2 km from her place of work?

Answer

The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5     3 10 20 25 11 13 7 12 31     19     10 12 17 18    11 3    2    17      16 2     7 9     7     8    3 5 12 15 18 3     12     14 2 9 6 15 15 7 6 12

Total numbers of engineers = 40
(i) Numbers of engineers living less than 7 km from her place of work = 9
Required probability = 9/40

(ii) Numbers of engineers living less than 7 km from her place of work = 40 - 9 = 31
Required probability = 31/40

(iii) Numbers of engineers living less than 7 km from her place of work = 0
Required probability = 0/40 = 0

Page No: 285

11. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
4.97      5.05      5.08     5.03     5.00     5.06     5.08    4.98       5.04       5.07       5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Answer

Total numbers of bags = 11
Numbers of bags containing more than 5 kg of flour = 7
Required probability = 7/11

12. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.
The data obtained for 30 days is as follows:
0.03      0.08      0.08      0.09      0.04      0.17      0.16      0.05      0.02      0.06      0.18      0.20    0.11    0.08      0.12    0.13      0.22      0.07      0.08      0.01      0.10      0.06      0.09      0.18      0.11      0.07      0.05      0.07      0.01      0.04

Answer

Total numbers of days data recorded = 30 days
Numbers of days in which sulphur dioxide in the interval 0.12-0.16 = 2
Required probability = 2/30 = 1/15

13. In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Answer

Total numbers of students = 30
Numbers of students having blood group AB = 3
Required probability = 3/30 = 1/10

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NCERT Solutions for Class 9 Maths Chapter 15 Probability

Chapter 15 Probability Class 9 Maths NCERT Solutions is good way through which one can build their basic knowledge. The numerical measure of uncertainty of an activity is called probability. The words ‘doubt’, ‘chances’, ‘most probably’ etc., show uncertainty or probability of happening of an action or activity.

• Probability – an Experimental Approach: A trial is an action which results in one or several outcomes. An event for an experiment is the collection of some outcomes of the experiment. Let n be the total number of trials. The empirical probability P(E) of an event E happening, is given by

P(E) = Number of trials in which the event happened/The total number of trials.

Only one exercise is present in the chapter 15 NCERT Solutions. We have also provided exercise wise solutions of every problem that you can find below.

Exercise 15.1 Chapter 15 Class 9 Maths NCERT Solutions

You will get accurate and detailed NCERT Solutions of Class 9 Maths prepared by Studyrankers experts that can help in easily understand even the most difficult problems. You will be able to solve the difficult problems given in a exercise.

NCERT Solutions for Class 9 Maths Chapters:

Chapter 13 Surface Areas and Volumes

Chapter 2 Linear Equations in One Variable

FAQ on Chapter 15 Probability

What are the benefits of NCERT Solutions for Chapter 15 Probability Class 9 NCERT Solutions?

Chapter 15 Probability NCERT Solutions that are very helpful in building your basic concepts embedded in the chapter. It is also very helpful in developing your problem solving skills.

What do you mean by random experiment?

An experiment in which all possible outcomes are known and the exact outcome cannot be predicted in advance, is called a random experiment.

What is uncertainty in Probability?

A doubtful situation of an action or activity that may or may not take place, is called ‘uncertainty’.

A coin is thrown once. Find the probability of getting a head.

Number of possible outcomes = 2

Number of favourable outcome = 1

∴ The required probability = (1/2).

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NCERT Solutions for Class 8 Maths Ch 1 Rational Numbers

January 20, 2021, 10:26 pm

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NCERT Solutions for Class 8 Maths Ch 1 Rational Numbers| PDF Download

Here, we are providing Chapter 1 Rational Number Class 8 Maths NCERT Solutions which are effective in the preparation of examinations. These NCERT Solutions for Class 8 are very much helpful in increasing concertation among students. It will make students able to solve the difficult problems in a given in exercise. The numbers of the form p/q where p and q are integers and (q + 0), are called rational numbers. These are very helpful in getting good marks in the examinations.

These solutions are as per the latest NCERT textbooks. These Chapter 1 NCERT solutions are prepared by faculty of Studyrankers who have provided step by step answers that will help you scoring more marks in the examinations. These can be used for completing your homework or understanding the basic concepts to solve problems.

NCERT Solutions for Class 8 Maths Ch 1 Rational Numbers

Study Materials for Class 8 Maths Chapter 1 Rational Numbers
CBSE Notes for Class 8 Maths Chapter 1 Rational Numbers MCQs Questions for Class 8 Maths Chapter 1 Rational Numbers

Study Materials for Class 8 Maths Chapter 1 Rational Numbers

Page No: 14

Exercise 1.1

1. Using appropriate properties find.
(i) -2/3 × 3/5 + 5/2 - 3/5 × 1/6   (ii) 2/5 × (-3/7) - 1/6 × 3/2 + 1/14 × 2/5

Answer

(i) -2/3 × 3/5 + 5/2 - 3/5 × 1/6
= -2/3 × 3/5 - 3/5 × 1/6 + 5/2    (by commutativity)
= 3/5(-2/3 - 1/6) + 5/2
= 3/5{(-4 - 1)/6} + 5/2
= 3/5(-5/6) + 5/2    (by distributivity)
= -15/30 + 5/2
= -1/2 + 5/2
= 4/2 = 2

(ii) 2/5 × (-3/7) - 1/6 × 3/2 + 1/14 × 2/5
= 2/5 × (-3/7) + 1/14 × 2/5 - (1/6 × 3/2)    (by commutativity)
= 2/5(-3/7 + 1/14) - 1/4
= 2/5{(-6 + 1)/14} - 1/4    (by distributivity)
= 2/5(-5/14) - 1/4
= -1/7 - 1/4
= (-4-7)/28
= -11/28

2. Write the additive inverse of each of the following.
(i) 2/8   (ii) -5/9   (iii) -6/-5   (iv) 2/-9   (v) 19/-6

Answer

(i) 2/8
Additive inverse = -2/8
(ii) -5/9
Additive inverse = 5/9
(iii) -6/-5 = 6/5
Additive inverse = -6/5
(iv) 2/-9 = -2/9
Additive inverse = 2/9
(v) 19/-6 = -19/6
Additive inverse = 19/6

3. Verify that : -(-x) = x for.
(i) x = 11/15 (ii) x = -13/17

Answer

(i) x = 11/15
The additive inverse of x = 11/15 is -x = -11/15 as 11/15 + (-11/15) = 0
The same equality 11/15 + (-11/15) = 0 , shows that the additive inverse of -11/15 is 11/15 or
-(-11/15) = 11/15 i.e. -(-x) = x

(ii) x = -13/17
The additive inverse of x = -13/17 is -x = 13/17 as (-13/17) + 13/17 = 0
The same equality 13/17 + (-13/17) = 0 , shows that the additive inverse of 13/17 is -13/17 or
-(13/17) = -13/17 i.e. -(-x) = x

4. Find the multiplicative inverse of the following.
(i) -13    (ii) -13/19    (iii) 1/5    (iv) -5/8 × -3/7    (v) -1 × -2/5    (vi) -1

Answer

The multiplicative inverse of a number is the reciprocal of that number.

(i) -13
Multiplicative inverse = -1/13
(ii) -13/19
Multiplicative inverse = -19/13
(iii) 1/5
Multiplicative inverse = 5
(iv) -5/8 × -3/7 = 15/56
Multiplicative inverse = 56/15
(v) -1 × -2/5 = 2/5
Multiplicative inverse = 5/2
(vi) -1
Multiplicative inverse = -1

5. Name the property under multiplication used in each of the following.
(i) -4/5 × 1 = 1 × -4/5 = -4/5
(ii) -13/17 × -2/7 = -2/7 × -13/17
(iii) -19/29 × 29/-19 = 1

Answer

(i) -4/5 × 1 = 1 × -4/5 = -4/5
Here 1 is the multiplicative identity.
(ii) -13/17 × -2/7 = -2/7 × -13/17
Commutavity
(iii) -19/29 × 29/-19 = 1
Multiplicative inverse

6. Multiply 6/13 by the reciprocal of -7/16.

Answer

Reciprocal of -7/16 = 16/-7
A/q,
6/13 × (Reciprocal of -7/16)
= 6/13 × 16/-7 = 96/-91 = -96/91

7. Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3.

Answer

By the property of associativity.
8. Is 8/9 the multiplicative inverse of

? Why or why not?

Answer

If it will be the multiplicative inverse then their product will be 1.

= -7/8
A/q,
8/9 × -7/8 = -7/9 ≠ 1
Hence, 8/9 is not the multiplicative inverse.

9. Is 0.3 the multiplicative inverse of

? Why or why not?

Answer

If it will be the multiplicative inverse then their product will be 1.

= 10/3

also, 0.3 = 3/10

A/q,
3/10 × 10/3 = 1
Hence, 0.3 is the multiplicative inverse.

Page No: 15

10. Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.

Answer

(i) 0 is the rational number that does not have a reciprocal.

(ii) 1 and -1 are the rational numbers that are equal to their reciprocals.

(iii) 0 is the rational number that is equal to its negative.

11. Fill in the blanks.
(i) Zero has ________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x ≠ 0 is ________.
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.

Answer

(i) Zero has no reciprocal.

(ii) The numbers 1 and -1 are their own reciprocals
(iii) The reciprocal of -5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational numbers.
(vi) The reciprocal of a positive rational number is positive rational numbers.

Page No: 20

Exercise 1.2

1. Represent these numbers on the number line. (i) 7/4 (ii) -5/6

Answer

(i) 7/4 on the number line.

Divide line between two natural number in 4 parts. Thus, the rational number 7/4 lies at a distance of 7 points from 0 towards positive number line.

(ii) -5/6 on the number line.

Divide line between two natural number in 6 parts. Thus, the rational number -5/6 lies at a distance of 5 points from 0 towards negative number line.

2. Represent -2/11, -5/11, -9/11 on the number line.

Answer

-2/11, -5/11, -9/11 on the number line.

Divide line between two natural number in 11 parts. Thus, the rational number -2/11, -5/11, -9/11 lie at a distance of 2, 5, 9 points from 0 towards negative number line respectively.

3. Write five rational numbers which are smaller than 2.

Answer

2 can be written as 10/5.

Thus, 5 natural numbers smaller than 2 are:

9/5, 8/5, 7/5, 6/5 and 5/5

4. Find ten rational numbers between -2/5 and 1/2.

Answer

The numbers -2/5 and 1/2 can be written as -8/20 and 10/20

Thus, ten rational numbers between -2/5 and 1/2 are:

-7/20, -6/20, -5/20, -4/20, -3/20, -2/20, -1/20, 0, 1/20 and 2/20

5. Find five rational numbers between.

(i) 2/3 and 4/5 (ii) -3/2 and 5/3 (iii) 1/4 and 1/2

Answer

(i) Five rational numbers between 2/3 and 4/5

The numbers 2/3 and 4/5 can be written as 30/45 and 36/45

Thus, five rational numbers are:

31/45, 32/45, 33/45, 34/45 and 35/45

(ii) Five rational numbers between -3/2 and 5/3

The numbers -3/2 and 5/3 can be written as -9/6 and 10/6

Thus, five rational numbers are:

-8/6, -5/6, -2/6, 0 and 2/6

(iii) Five rational numbers between 1/4 and 1/2

The numbers 1/4 and 1/2 can be written as 7/28 and 14/28

Thus, five rational numbers are:

8/28, 9/28, 10/28, 11/28 and 12/28

6. Write five rational numbers greater than -2.

Answer

-2 can be written as -16/8.

Five rational numbers greater than -2 are:

-15/8, -14/8, -13/8, -12/8 and -11/8

7. Find ten rational numbers between 3/5 and 3/4.

Answer

The numbers 3/5 and 3/4 can be written as 48/80 and 60/80

Thus, ten rational numbers between 3/5 and 3/4 are:

49/80, 50/80, 51/80, 52/80, 53/80, 54/80, 55/80, 56/80, 57/80 and 58/80.

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NCERT Solutions for Class 8 Maths Ch 1 Rational Numbers

You can find accurate detailed NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers which are prepared by Studyrankers experts who have in depth knowledge of the topics. These NCERT Solutions will improve the learning behaviour of the students.

• Properties of Rational Numbers: We are going to study various properties of rational numbers such as closure, commutativity, associativity and also about reciprocal of rational numbers and distributivity of multiplication over addition for rational numbers.

• Representation of Rational Numbers on the Number Line: We will try to represent rational numbers on number line.

• Rational Numbers between Two Rational Numbers: Between any two given rational numbers there are countless rational numbers. We will find rational numbers between two rational numbers.

On this page, you will also get exercisewise Chapter 1 Rational Numbers NCERT Solutions through which you can be able to solve difficult questions given in a exercise.

Studyrankers experts have prepared these Class 8 Maths NCERT Solutions by focusing the needs of the students and detailed every questions that will solve your doubts quickly and easily.

NCERT Solutions for Class 8 Maths Chapters:

Chapter 3 Understanding Quadrilaterals

Chapter 4 Practical Geometry

Chapter 5 Data Handling

Chapter 6 Square and Square Roots

Chapter 7 Cube and Cube Roots

Chapter 8 Comparing Quantities

Chapter 9 Algebraic Expressions and Identities

Chapter 10 Visualizing Solid Shapes

Chapter 11 Mensuration

Chapter 12 Exponents and Powers

Chapter 13 Direct and Inverse Proportions

Chapter 14 Factorization

Chapter 15 Introduction to Graphs

Chapter 16 Playing with Numbers

FAQ on Chapter 1 Rational Numbers

How many exercises in Chapter 1 Rational Numbers

There are total 2 exercise in the Chapter 1 Rational Numbers which are very much helpful in understand the basic concepts of the chapter and knowing properties of rational numbers.

What do you mean by Natural Numbers?

The counting numbers 1, 2, 3, 4, 5, … are called ‘natural numbers’. The smallest natural number is 1, but there is no last (or the greatest) natural number.

What is Commutative Property of Rational Numbers?

When two rational numbers are swapped between one operator and still their result does not change then we say that the rational numbers follow the commutative property for that operation.

What is Associative Property of Rational Numbers?

When rational numbers are rearranged among two or more same operations and still their result does not change then we say that the rational numbers follow the associative property for that operation.

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NCERT Solutions for Class 9 Maths Chapter 14 Statistics

January 20, 2021, 9:34 am

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NCERT Solutions for Class 9 Maths Chapter 14 Statistics| PDF Download

In this page, you will find NCERT Solutions for Chapter 14 Statistics Class 9 Maths which are accurate and detailed are prepared by Studyrankers experts. Here, you can download PDF of Class 9 Maths NCERT Solutions of Chapter 14 Statistics which will make you aware of the difficulty of questions. You can also complete your homework on time through the help of these NCERT Solutions and increase your concentration.

Class 9 Maths NCERT Solutions presented here are very helpful in developing problem solving skills and prepare for the examinations. These NCERT solutions are prerequisites before solving exemplar problems and going for supplementary Maths Books.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

Page No: 239

Exercise 14.1

1. Give five examples of data that you can collect from your day-to-day life.

Answer

Five examples from day-to-day life:
(i) Daily expenditures of household.
(ii) Amount of rainfall.
(iii) Bill of electricity.
(iv) Poll or survey results.
(v) Marks obtained by students.

2. Classify the data in Q.1 above as primary or secondary data.

Answer

Primary Data: (i) (iii) and (v)
Secondary Data: (ii) and (iv)

Page No: 245

Exercise 14.2

1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Answer

The frequency means the number of students having same blood group. We will represent the data in table:

Blood Group	Number of Students (Frequency)
A	9
B	6
O	12
AB	3
Total	30

Most common Blood Group (Highest frequency): O
Rarest Blood Group (Lowest frequency): AB

2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?

Answer

The given data is very large. So, we construct a group frequency of class size 5. Therefore, class interval will be 0-5, 5-10, 10-15, 15-20 and so on. The data is represented in the table as:

The classes in the table are not overlapping. Also, 36 out of 40 engineers have their house below 20 km of distance.

3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?

Answer

(i) The given data is very large. So, we construct a group frequency of class size 2. Therefore, class interval will be 84-86, 86-88, 88-90, 90-92 and so on. The data is represented in the table as:

Relative humidity (in %)	Frequency
84-86	1
86-88	1
88-90	2
90-92	2
92-94	7
94-96	6
96-98	7
98-100	4
Total	30

(ii) The humidity is very high in the data which is observed during rainy season. So, it must be rainy season.
(iii) Range of data = Maximum value of data - Minimum = 99.2 − 84.9 = 14.3

4. The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?

Answer

(i) The data with class interval 160-165, 165-170 and so on is represented in the table as:

Height (in cm)	No. of Students (Frequency)
150-155	12
155-160	9
160-165	14
165-170	10
170-175	5
Total	50

(ii) From the given data, it can be concluded that 35 students i.e. more than 50% are shorter than 165 cm.

5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Answer

(i) The data with class interval 0.00 - 0.04, 0.04 - 0.08 and so on is represented in the table as:

Concentration of sulphur dioxide in air (in ppm)	Frequency
0.00 − 0.04	4
0.04 − 0.08	9
0.08 − 0.12	9
0.12 − 0.16	2
0.16 − 0.20	4
0.20 − 0.24	2
Total	30

(ii) 2 + 4 + 2 = 8 days have the concentration of sulphur dioxide more than 0.11 parts per million.

Page No. 246

6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.

Answer

The frequency distribution table for the data given above can be prepared as follow:

Number of Heads	Frequency
0	6
1	10
2	9
3	5
Total	30

7. The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?

Answer

(i)The frequency is given as follow:

Digits	Frequency
0	2
1	5
2	5
3	8
4	4
5	5
6	4
7	4
8	5
9	8
Total	30

(ii) The digit having the least frequency occurs the least and the digit with highest frequency occurs the most. 0 has frequency 2 and thus occurs least frequently while 3 and 9 have frequency 8 and thus occur most frequently.

8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?

Answer

(i) The distribution table for the given data, taking class width 5 and one of the class intervals as 5-10 is as follows:

Number of Hours	Frequency
0-5	10
5-10	13
10-15	5
15-20	2
Total	30

(ii) We observed from the given table that 2 children television for 15 or more hours a week.

9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5.

Answer

A grouped frequency distribution table using class intervals of size 0.5 starting from the interval 2 - 2.5 is constructed.

Lives of batteries (in years)	No. of batteries (Frequency)
2-2.5	2
2.5-3	6
3-3.5	14
3.5-4	11
4-4.5	4
4.5-5	3
Total	40

Page No. 258

Exercise 14.3

1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):

S.No.	Causes	Female fatality rate (%)
1.	Reproductive health conditions	31.8
2.	Neuropsychiatric conditions	25.4
3.	Injuries	12.4
4.	Cardiovascular conditions	4.3
5.	Respiratory conditions	4.1
6.	Other causes	22.0

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Answer

(i) The data is represented below graphically.

(ii) From the above graphical data, we observe that reproductive health conditions is the major cause of women’s ill health and death worldwide.

(iii) Two factors responsible for cause in (ii)

• Lack of proper care and understanding.

• Lack of medical facilities.

2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

S.No.	Section	Number of girls per thousand boys
1.	Scheduled Caste (SC)	940
2.	Scheduled Tribe (ST)	970
3.	Non SC/ST	920
4.	Backward districts	950
5.	Non-backward districts	920
6.	Rural	930
7.	Urban	910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Answer

(i)

(ii) It can be observed from the above graph that the maximum number of girls per thousand boys is in ST. Also, the backward districts and rural areas have more number of girls per thousand boys than non-backward districts and urban areas.

Page No. 59

3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political party	A	B	C	D	E	F
Seats won	75	55	37	29	10	37

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?

Answer

(i)

(ii) The party named A has won the maximum number of seat.

4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

S.No.	Length (in mm)	Number of leaves
1.	118 - 126	3
2.	127 - 135	5
3.	136 - 144	9
4.	145 - 153	12
5.	154 - 162	5
6.	163 - 171	4
7.	172 - 180	2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii)Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Answer

(i) The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

S.No.	Length (in mm)	Number of leaves
1.	117.5 - 126.5	3
2.	126.5 - 135.5	5
3.	135.5 - 144.5	9
4.	144.5 - 153.5	12
5.	153.5 - 162.5	5
6.	162.5 - 171.5	4
7.	171.5 - 180.5	2

(ii) Yes, the data can also be represented by frequency polygon.

(iii) No, it is incorrect to conclude that the maximum number of leaves are 153 mm long because maximum number of leaves are lying between the length of 144.5 - 153.5

5. The following table gives the life times of 400 neon lamps:

Life Time (in hours)	Number of lamps
300 - 400	14
400 - 500	56
500 - 600	60
600 - 700	86
700 - 800	74
800 - 900	62
900 - 1000	48

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?

Answer

(i)

(ii) 74 + 62 + 48 = 184 lamps have a life time of more than 700 hours.

Page No. 260

6. The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Answer

The class mark can be found by (Lower limit + Upper limit)/2.
For section A,

Marks	Class Mark	Frequency
0-10	5	3
10-20	15	9
20-30	25	17
30-40	35	12
40-50	45	9

For section B,

Marks	Class Mark	Frequency
0-10	5	5
10-20	15	19
20-30	25	15
30-40	35	10
40-50	45	1

Now, we draw frequency polygon for the given data.

7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]

Answer

The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

Number of balls	Team A	Team B
0.5-6.5	2	5
6.5-12.5	1	6
12.5-18.5	8	2
18.5-24.5	9	10
24.5-30.5	4	5
30.5-36.5	5	6
36.5-42.5	6	3
42.5-48.5	10	4
48.5-54.5	6	8
54.5-60.5	2	10

Now, we draw frequency polygon for the given data.

Page No. 261

8. A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Answer

The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 1 is selected and the length of the rectangle is proportionate to it.

Age (in years)	Number of children (frequency)	Width of class	Length of rectangle
1-2	5	1	(5/1)×1 = 5
2-3	3	1	(3/1)×1 = 3
3-5	6	2	(6/2)×1 = 3
5-7	12	2	(12/2)×1 = 6
7-10	9	3	(9/3)×1 = 3
10-15	10	5	(10/5)×1 = 2
15-17	4	2	(4/2)×1 = 2

Taking the age of children on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn

9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Answer

(i) The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 2 is selected and the length of the rectangle is proportionate to it.

The proportion of the surnames per 2 letters interval can be calculated as:

Number of letters	Number of surnames	Width of class	Length of rectangle
1-4	6	3	(6/3)×2 = 4
4-6	30	2	(30/2)×2 = 30
6-8	44	2	(44/2)×2 = 44
8-12	16	4	(16/4)×2 = 8
12-20	4	8	(4/8)×2 = 1

(ii) The class interval in which the maximum number of surnames lie is 6-8.

Page No. 269

Exercise 14.4

1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.

Answer

Mean = Sum of all the observations/Total number of observations
= (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8

For Median, we will arrange the given data in ascending order,
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations (n) = 10
Number of observations are even so we will calculate median as,

= (3+3)/2 = 6/2 = 3

For Mode, we will arrange the given data in ascending order, we have

0, 1, 2, 3, 3, 3, 3, 4, 4, 5.

Here, 3 occurs most frequently (4 times)

∴ Mode = 3

2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.

Answer

Mean = Sum of all the observations/Total number of observations
= (41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/15 = 822/15 = 54.8

For Median, we will arrange the given data in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations (n) = 15
Number of observations are odd so we will calculate median as,

For Mode, we will arrange the given data in ascending order, we have

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

Here, 52 occurs most frequently (3 times)

∴ Mode = 52

3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

29, 32, 48, 50, x, x+2, 72, 78, 84, 95

Answer

Number of observations (n) = 10 (even)

According to question, Median = 63

∴ x + 1 = 63

⇒ x = 63−1 = 62

Hence, the value of x is 62.

4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Answer

The given data is,

14,25,14,28,18,17,18,14,23,22,14,18

Arranging the data in ascending order,

14,14,14,14,17,18,18,18,22,23,25,28

Here, 14 occurs most frequently (4 times). Mode = 14

5. Find the mean salary of 60 workers of a factory from the following table:

Answer

Salary (x_i)	Number of workers (f_i)	f_ix_i
3000	16	48000
4000	12	48000
5000	10	50000
6000	8	48000
7000	6	42000
8000	4	32000
9000	3	27000
10000	1	10000
Total	Σf_i = 60	Σf_ix_i = 305000

Hence, the mean salary is ₹5083.33

6. Give one example of a situation in which

(i) the mean is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Answer

(i) Mean marks in a test in mathematics.

(ii) Average beauty

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

Chapter 14 Statistics Class 9 Maths NCERT Solutions will help you in understanding the concepts in a better way. Statistics deals with collection, organisation, analysis and interpretation of data. It deals with the collection, presentation, analysis of data as well as drawing of meaningful conclusions from the data.

• Collection of Data: When the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data. In the second case, when the information was gathered from a source which already had the information stored, the data obtained is called secondary data.

• Presentation of Data: Presenting data in this form simplifies and condenses data and enables us to observe certain important features at a glance. This is called a grouped frequency distribution table.

• Graphical Representation of Data: The graphical representation makes data easier to understand than the actual data. Bar graphs, Histograms of uniform width, and of varying widths and Frequency polygons are some of the graphical representations in this section.

• Measures of Central Tendency: The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations. The median is that value of the given number of observations, which divides it into exactly two parts.

There are total 4 exercises in the Chapter 14 Statistics NCERT Solutions which will useful in developing your problem solving skills and learning diverse topics. We have also prepared exercisewise NCERT Solutions which you can find them below.

Studyrankers subject matter experts have curated these questions which will help in revising the chapter properly. These NCERT Solutions will guide student in a better way and useful in learning efficiently.

NCERT Solutions for Class 9 Maths Chapters:

Chapter 13 Surface Areas and Volumes