MCQ Questions for Class 12 Political Science: Ch 8 Environment and Natural Resources

October 16, 2020, 6:03 am

≫ Next: NCERT Solutions for Class 10 Maths Ch 10 Circles

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 7 Security in the Contemporary World

MCQ Questions for Class 12 Political Science: Ch 8 Environment and Natural Resources

1. The example of global common is

(a) Gasoline pumps.

(b) Giant oil refineries.

(d) Chain of hotels.

► (c) The ocean floor.

2. According to Kyoto Protocol, the major nations abide to reduce concentration of green house gases was by

(a) 2008.

(b) 2010.

(d) 2018.

3. Progress that meets the needs of the present without compromising the ability of future generations to meet their own needs is:

(a) The tragedy of global commons.

(b) Sustainable development.

(d) The politics of the developed nations.

► (b) Sustainable development.

4. Smog is a combination of

(a) Air and water vapours.

(b) Water and smoke.

(d) Smoke and fog.

► (d) Smoke and fog.

5. The multinational company Western Mining Corporation (WMC) belongs to

(a) USA.

(b) Australia.

(d) New Zealand.

► (b) Australia.

6. The Constitution of India provides for the safeguard of cultural and educational rights of minorities under fundamental rights of Article

(a) 20.

(b) 23.

(d) 29.

► (d) 29.

7. WWF stands for

(a) World Wild Foundation.

(b) World Wildlife Federation.

(d) World Wildlife Fund.

► (d) World Wildlife Fund.

8. The Government of India refers indigenous people as

(a) Scheduled tribes.

(b) Refugee people.

(d) Outclass people.

► (a) Scheduled tribes.

9. The animal that is the symbol of the World Wide wildlife Fund is:

(a) Tiger.

(b) Panda.

(d) White Bear.

► (b) Panda.

10. Main cause of extinction of species from tropics is

(a) Soil erosion.

(b) Pollution.

(d) Afforestation.

► (c) Deforestation.

11. What was the other name for the 1987 Brundtland Report?

(a) Our Common Present

(b) Our Common Future

(d) None of the above

► (b) Our Common Future

12. Expand UNFCCC?

(a) United Nations Framework Convention on Climate Change

(b) United Nations Framework Counter on Climate Change

(d) United Nations Fast Convention on Climate Change

► (a) United Nations Framework Convention on Climate Change

13. India signed and ratified the 1997 Kyoto Protocol in:

(a) July 2002.

(b) August 2002.

(d) October 2002.

► (b) August 2002.

14. The World Council of Indigenous Peoples was formed in:

(a) 1973

(b) 1974

(d) 1976

↧

NCERT Solutions for Class 10 Maths Ch 10 Circles

October 16, 2020, 9:45 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 9 Globalisation

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 8 Environment and Natural Resources

NCERT Solutions for Class 10 Maths Ch 10 Circles

You will find NCERT Solutions for Class 10 Maths Ch 10 Circles here which will be useful in preparing your answers in taking help. These NCERT Solutions for Class 10 are prepared buy Studyrankers experts who have large experience of teaching. These will be helpful in preparing yourself before examinations.

Page No: 209

Exercise: 10.1

1. How many tangents can a circle have?

Answer

A circle can have infinite tangents.

2. Fill in the blanks :

(i) A tangent to a circle intersects it in ............... point(s).

(ii) A line intersecting a circle in two points is called a .............

(iii) A circle can have ............... parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called ............

Answer

(i) one

(ii) secant

(iii) two

(iv) point of contact

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) √119 cm

Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,

OQ² = OP² +PQ²
⇒ (12)²= 5² + PQ²

⇒PQ² = 144 - 25

⇒PQ² = 119

⇒PQ = √119 cm

(D) is the correct option.

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the

other, a secant to the circle.

Answer

AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.

Page NO: 213

Exercise: 10.2

In Q.1 to 3, choose the correct option and give justification.

1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm

Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem in ΔOPQ,

OQ² = OP² +PQ²
⇒ (25)²= OP² + (24)²

⇒ OP² = 625 - 576

⇒ OP² = 49

⇒ OP = 7 cm

The radius of the circle is option (A) 7 cm.

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Answer

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

∠PTQ is equal to option (B) 70°.

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°

Answer

OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°

In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°

⇒ ∠AOB = 100°

Now,

In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)

Thus ∠POB = ∠POA

∠AOB = ∠POB + ∠POA

⇒ 2 ∠POA = ∠AOB

⇒ ∠POA = 100°/2 = 50°

∠POA is equal to option (A) 50°

Page No: 214

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,

∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.

Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer

Let AB be the tangent to the circle at point P with centre O.

We have to prove that PQ passes through the point O.

Suppose that PQ doesn't passes through point O. Join OP.

Through O, draw a straight line CD parallel to the tangent AB.

PQ intersect CD at R and also intersect AB at P.

AS, CD // AB PQ is the line of intersection,

∠ORP = ∠RPA (Alternate interior angles)

but also,

∠RPA = 90° (PQ ⊥ AB)

⇒ ∠ORP = 90°

∠ROP + ∠OPA = 180° (Co-interior angles)

⇒∠ROP + 90° = 180°

⇒∠ROP = 90°

Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.

Hence, our supposition is wrong.

∴ PQ passes through the point O.

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer

AB is a tangent drawn on this circle from point A.

∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA² =AB²+ BO²
⇒ 5²= 4²+ BO²
⇒ BO² = 25 - 16
⇒ BO² = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.

Answer

Let the two concentric circles with centre O.

AB be the chord of the larger circle which touches the smaller circle at point P.

∴ AB is tangent to the smaller circle to the point P.

⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA² = AP² + OP²
⇒ 5² = AP² + 3²
⇒ AP² = 25 - 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer

From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)

BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC

9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer

We joined O and C

A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,

ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º

∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer

ABCD is a parallelogram,
∴ AB = CD ... (i)
∴ BC = AD ... (ii)

From the figure, we observe that,

DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x

We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x

Now semi perimeter of triangle (s) is,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x

Area of ΔABC = √s (s - a)(s - b)(s - c)

= √(14 + x) (14 + x -14)(14 + x - x - 6)(14 + x - x - 8)

= √(14 + x) (x)(8)(6)

= √(14 + x) 48 x ... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

= 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)

Equating equation (i) and (ii) we get,

√(14 + x) 48 x = 56 + 4x

Squaring both sides,

48x (14 + x) = (56 + 4x)²

⇒ 48x = [4(14 + x)]²/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7 cm

Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS

⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7

Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Go Back To NCERT Solutions for Class 10th Math

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Chapter 10 Class 10 Maths is very important for the purpose of examinations. There are total 3 topics in this chapter through you will learn more about circles.

• Introduction: A circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre). You will learn about various key terms such as related to a circle like chord, segment, sector, arc etc.

• Tangent to a Circle: A tangent to a circle is a line that intersects the circle at only one point. There is

only one tangent at a point of the circle. The are various problems related to this topics that includes fill in the blanks and finding the length of a given line.

• Number of Tangents from a Point on a Circle: There is one theorem in this topics that states the lengths of tangents drawn from an external point to a circle are equal. The questions includes MCQs and related proving and finding.

If you want to access other study resources prepared by Studyrankers then you can follow the provided links.

CBSE Notes for Class 10 Maths Chapter 10 Circles

MCQ Questions for Classs 10 Maths Chapter 10 Circles

There are total 2 exercises in the chapter which will check you basic understanding about the circles. These NCERT Solutions for Class 10 are best way to encourage students to learn new topics. You can also find exercisewise solutions just by clicking on the links given below.

Exercise 10.1
Exercise 10.2

NCERT Solutions for Class 10 Maths Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry

Chapter 9 Some Applications of Trigonometry

Chapter 11 Constructions

Chapter 12 Areas Related To Circles

Chapter 13 Surface Areas And Volumes

Chapter 14 Statistics

Chapter 15 Probability

FAQ on Chapter 10 Circles

How many exercises in Chapter 10 Circles

There are only 2 exercise in the Chapter 10 Class 10 Maths which will helpful in completing the homework on time and increasing concentration among students.

What is point of contact?

A line which touches a circle only in one point is called a tangent line and the point at which it touches the circle is called the point of contact.

The ______ at any point of a circle is perpendicular to the radius, through the point of contact.

The tangent at any point of a circle is perpendicular to the radius, through the point of contact.

What do you mean by Tangent?

A tangent to a circle is a line that touches the circle at only one point.

↧

MCQ Questions for Class 12 Political Science: Ch 9 Globalisation

October 16, 2020, 10:20 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 1 Challenges of Nation Building

≪ Previous: NCERT Solutions for Class 10 Maths Ch 10 Circles

MCQ Questions for Class 12 Political Science: Ch 9 Globalisation

1. Globalization has led to the flow of ideas across

(a) National boundaries.

(b) State.

(d) Different planets.

► (a) National boundaries.

2. With the globalisation of markets, the tastes and preferences of consumers world-wide are

(a) Becoming similar to the tastes and preferences of American consumers.

(b) Being encouraged by multinational organizations to become increasingly similar.

(d) Converging upon a global norm.

► (d) Converging upon a global norm.

3. According to the rightist view in India, globalisation tends to

(a) Benefits the weaker section of the society.

(b) Divides the State into rich and poor.

(d) Reduces political party competition.

► (c) Weakens the State.

4. The seventh WSF meeting was held in:

(a) Delhi, January 2007

(b) Nairobi, January 2007

(d) New York, January 2007

► (b) Nairobi, January 2007

5. Which of the following is available in India due to globalisation?

(a) Foreign TV channels

(b) Coca Cola and Pepsi

(d) All of the above

► (d) All of the above

6. India implemented the New Economic Policy in the year

(a) 1980.

(b) 1981.

(d) 1991.

► (d) 1991.

7. This type of globalisation refers to global markets and the flow of capital, technology & goods is

(a) Political globalisation.

(b) Cultural globalisation.

(d) Opposing globalisation.

► (c) Economic globalisation.

8. In 1986-87, the overall fiscal deficit of India touched an all time high of

(a) 5% of GDP.

(b) 7% of GDP.

(d) 10% of GDP.

► (c) 9% of GDP.

9. During the colonial period, India became an

(a) importer of primary goods and raw materials

(b) exporter of finished goods and also raw materials.

(d) exporter of primary goods and raw materials.

► (d) exporter of primary goods and raw materials.

10. Liberalisation means:

(a) integration among economies.

(b) reduced government controls and restrictions.

(d) competitive market.

► (b) reduced government controls and restrictions.

11. Which of the statements are true about the impact of globalisation?

(a) Globalisation has been uneven in its impact on states and societies.

(b) Globalisation has had a uniform impact on all states and societies.

(d) Globalisation inevitably results in cultural homogeneity.

► (a) Globalisation has been uneven in its impact on states and societies.

12. Which of the statements are true about globalisation?

(a) Globalisation is only about movement of commodities.

(b) Globalisation does not involve a conflict of values.

(d) Globalisation is about worldwide interconnectedness.

► (d) Globalisation is about worldwide interconnectedness.

13. Which of the statements are true about globalisation?

(a) Globalisation is purely an economic phenomenon.

(b) Globalisation began in 1991.

(d) Globalisation is a multidimensional phenomenon.

► (d) Globalisation is a multidimensional phenomenon.

14. Which of the statements are true about the causes of globalisation?

(a) Technology is an important cause of globalisation.

(b) Globalisation is caused by a particular community of people.

(d) Economic interdependence alone causes globalisation.

► (a) Technology is an important cause of globalisation.

↧

MCQ Questions for Class 12 Political Science: Ch 1 Challenges of Nation Building

October 16, 2020, 11:30 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 2 Era of One Party Dominance

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 9 Globalisation

MCQ Questions for Class 12 Political Science: Ch 1 Challenges of Nation Building

1. The States Reorganisation Commission was appointed in which year?

(a) 1950

(b) 1951

(d) 1954

2. The Cities that were divided into ‘communal zones’ during the partition violence were

(a) Lahore, Amritsar and Calcutta.

(b) Kashmir, Lucknow and Allahabad.

(d) Delhi, Mumbai and Gwalior.

► (a) Lahore, Amritsar and Calcutta.

3. The interim government formed under the cabinet mission plan was headed by

(a) Dr. Rajendra Prasad.

(b) Jawaharlal Nehru.

(d) Rajagopalachari.

► (b) Jawaharlal Nehru.

4. The states created in 1960 were________.

(a) Maharashtra and Gujarat

(b) Orissa and West Bengal

(d) Punjab and Haryana

► (a) Maharashtra and Gujarat

5. Which state was not created in 2000?

(a) Jharkhand

(b) Chhattisgarh

(d) Bihar

► (d) Bihar

6. The ''communal zones'' exclude

(a) Lahore.

(b) Amritsar.

(d) Jammu & Kashmir.

► (d) Jammu & Kashmir.

7. Mohammed Ali Jinnah addressed the constituent assembly of Pakistan in Karachi on:

(a) 11th August, 1947

(b) 12th August, 1947

(d) 14th August, 1947

► (a) 11th August, 1947

8. Amrita pritam was a prominent poet from

(a) Punjab.

(b) Gujarat.

(d) Madras.

► (a) Punjab.

9. Name the original state from which Chattisgarh was carved out.

(a) Uttar Pradesh

(b) Bihar

(d) Madhya Pradesh

► (d) Madhya Pradesh

10. Which one of the following leaders played a vital role in the integration of princely states with India?

(a) Mahatma Gandhi

(b) Sardar Vallabhbhai Patel

(d) Dr. Bhimrao Ambedkar

► (b) Sardar Vallabhbhai Patel

11. How many princely states existed at the time of independence of India?

(a) 560

(b) 562

(d) 565

► (d) 565

12. When Meghalaya was carved out of Assam?

(a) 1970

(b) 1971

(d) 1973

13. When Gujarat was carved out of Bombay?

(a) 1950

(b) 1960

(d) 1980

► (b) 1960

14. Which one of the following is a princely states of India initially resisted to join the Indian Union?

(a) Baroda

(b) Hyderabad

(d) Gwalior

► (b) Hyderabad

↧

MCQ Questions for Class 12 Political Science: Ch 2 Era of One Party Dominance

October 16, 2020, 9:27 pm

≫ Next: MCQ Questions for Class 12 Political Science: Ch 3 Politics of Planned Development

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 1 Challenges of Nation Building

MCQ Questions for Class 12 Political Science: Ch 2 Era of One Party Dominance

1. Electronic Voting Machine (EVM) is used to record

(a) politicians’ preferences.

(b) voters’ preferences.

(d) pressure groups' preferences.

► (b) voters’ preferences.

2. The First General Elections in India were held in the year

(a) 1951.

(b) 1952.

(d) 1954.

► (b) 1952.

3. The Indian leader who is the first recipient of the Bharat Ratna Award is

(a) Balraj Madhok.

(b) Rajagopalachari.

(d) K.M. Munshi.

► (b) Rajagopalachari.

4. The socialists advocated ideology of

(a) Democratic Socialism.

(b) Socialism.

(d) Hindutva.

► (a) Democratic Socialism.

5. The constitution of India came into effect on the

(a) 26 January 1948.

(b) 12 March 1948.

(d) 26 January 1950.

► (d) 26 January 1950.

6. One of the guiding principles of the ideology of Swatantra Party was:

(a) Interests of the working class

(b) Protection of princely states

(d) Autonomy of states within the union

► (c) Economy free from state control

7. The party that won the second largest number of Lok Sabha seats in the first General elections was the

(a) Praja Socialist Party.

(b) Bhartiya Jana Sangh.

(d) Bhartiya Janta party.

► (c) Communist Party of India.

8. Institutional Revolutionary Party exercised power in:

(a) Mexico

(b) Finland

(d) Brazil

► (a) Mexico

9. In democracy, people use their power through the

(a) political parties.

(b) co-operatives.

(d) non-governmental organizations.

► (c) elections.

10. The first three Lok Sabha elections of 1952, 1957 and 1962 were dominated by the Congress Party under the leadership of:

(a) Mahatma Gandhi

(b) Indira Gandhi

(d) Jawahar Lal Nehru

► (d) Jawahar Lal Nehru

11. Who was the first Chief Election Commissioner of India?

(a) Shayama Prasad Mukherjee

(b) Sukumar Sen

(d) Balraj Madhok.

► (b) Sukumar Sen

12. Which one of the following is a state where Congress was not in power even after huge victory in the first general elections?

(a) Bihar

(b) Orissa

(d) Maharashtra

► (b) Orissa

13. Who was the founder of Bharatiya Jana Sangh?

(a) Mahatma Gandhi

(b) Indira Gandhi

(d) Sukumar Sen

► (c) Shayama Prasad Mukherjee

14. Name the founder President of Congress Socialist Party.

(a) Acharya Narendra Dev

(b) K.M. Munshi.

(d) Sukumar Sen

► (a) Acharya Narendra Dev

↧

MCQ Questions for Class 12 Political Science: Ch 3 Politics of Planned Development

October 17, 2020, 3:12 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 4 India’s External Relations

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 2 Era of One Party Dominance

MCQ Questions for Class 12 Political Science: Ch 3 Politics of Planned Development

1. Which of the following is not a contributory factor for development?

(a) Abundance of national resources

(b) Character of people

(d) Level of technology

► (b) Character of people

2. During 1965-67 which state faced a near famine situation?

(a) Tamil Nadu.

(b) Bihar

(d) Manipur.

► (c) Bihar

3. Socialist model of development was related to

(a) U.S.A.

(b) Europe.

(d) Pakistan.

► (c) USSR.

4. Tribals of Odisha fear their displacement due to the extraction of which mineral?

(a) Copper

(b) Zinc

(d) Iron ore

► (d) Iron ore

5. The state which has achieved nearly total literacy in India is

(a) Tamil Nadu.

(b) Andhra Pradesh.

(d) Manipur.

► (c) Kerala.

6. The idea of Five Year Plan was adopted in India from

(a) United States of America.

(b) Japan.

(d) Australia.

► (c) United Soviet Socialist Republics.

7. What was the name given to the model of decentralised planning adopted by southern state of India?

(a) Kerala Model

(b) Tamil Nadu Model

(d) Karnataka Model

► (a) Kerala Model

8. What is the name of document that the government of India prepares that has a plan for all its income and expenditure?

(a) Census Report

(b) Statistical Abstract

(d) Union Addresses

► (c) Budget

9. The Second Five Year Plan was launched in

(a) 1955.

(b) 1956.

(d) 1958.

► (b) 1956.

10. The principal obstacle in the way of agricultural growth was the

(a) people's attitude.

(b) status of Zamindars.

(d) heath of farmers.

► (c) pattern of land distribution.

11. Which of the following ideas did not form part of the early phase of India’s development policy?

(a) Planning

(b) Liberalisation

(d) Self-sufficiency

► (b) Liberalisation

12. Which of these statements about the Bombay Plan is incorrect?

(a) It was a blueprint for India’s economic future.

(b) It supported state-ownership of industry.

(d) It supported strongly the idea of planning.

► (a) It was a blueprint for India’s economic future.

13. Which are the two models of development?

(a) Capitalist and Socialist

(b) Capitalist and Marxist

(d) None of the above

► (a) Capitalist and Socialist

14. _______ was introduced to bring about revolutionary changes in agriculture.

(a) WhiteRevolution

(b) Green Revolution

(d) Pink Revolution

► (b) Green Revolution

↧

MCQ Questions for Class 12 Political Science: Ch 4 India’s External Relations

October 17, 2020, 4:57 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 5 Challenges to and Restoration of Congress System

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 3 Politics of Planned Development

MCQ Questions for Class 12 Political Science: Ch 4 India’s External Relations

1. Who accompanied the Tibetan spiritual leader Dalai Lama during the official Chinese visit to India in 1956?

(a) Hua Guofeng

(b) Zhao Ziyang

(d) Li Peng

► (c) Chinese Premier Zhou Enlai

2. Apartheid is the practice of

(a) religious discrimination.

(b) racial discrimination.

(d) human behaviour.

► (b) racial discrimination.

3. India convened the Asian Relations Conference in March 1947 under the leadership of

(a) Dr B.R. Ambedkar

(b) Dr. Radha Krishnan.

(d) Rajendra Prasad.

► (c) Jawaharlal Nehru.

4. In the Nehru cabinet who was apprehensive of the future attack from china?

(a) Vallabhbhai Patel

(b) C. Rajagopalachari

(d) Jagjivan Ram

► (a) Vallabhbhai Patel

5. Who exercised profound influence in the formulation and implementation of India’s foreign policy from 1946 to 1964?

(a) Motilal Nehru

(b) Sardar Patel

(d) Jawaharlal Nehru

► (d) Jawaharlal Nehru

6. After second World War, the world was divided into the blocks of

(a) U.S.A and U.K.

(b) USSR and China.

(d) U.S.A and USSR.

► (d) U.S.A and USSR.

7. Article 51 of the Indian Constitution lays down some provisions regarding...

(a) Powers of Prime Minister.

(b) Fundamental Rights.

(d) Fundamental Duties.

► (c) Promotion of international peace and security.

8. The basis of India's foreign policy is

(a) non alignment.

(b) promoting US bloc.

(d) alligning with China.

► (a) non alignment.

9. After Nehru, the first Prime Minister to visit China was

(a) Lal Bahadur Shastri.

(b) Indira Gandhi.

(d) V.P. Singh.

► (c) Rajiv Gandhi.

10. For the early freedom of which country India made earnest effort by convening international conference?

(a) Malaysia

(b) Indonesia

(d) Singapore

► (b) Indonesia

11. Why Britain attacked Egypt in 1956?

(a) over Suez Canal issue

(b) over Suzhou Canal issue

(d) over Bruges Canal issue

► (a) over Suez Canal issue

12. Which conference marked the zenith of India’s engagement with the newly independent Asian and African nations?

(a) Bandung Conference

(b) Singapore Conference

(d) Malaysian Conference

► (a) Bandung Conference

13. When China annexed Tibet?

(a) 1930

(b) 1940

(d) 1960

14. Expand NEFA.

(a) North Eastern Far Agency

(b) North Eastern Frontier Agency

(d) North Eastern Farther Agency

► (b) North Eastern Frontier Agency

↧

MCQ Questions for Class 12 Political Science: Ch 5 Challenges to and Restoration of Congress System

October 17, 2020, 5:45 am

≫ Next: NCERT Solutions for Class 10 Maths Ch 11 Constructions

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 4 India’s External Relations

MCQ Questions for Class 12 Political Science: Ch 5 Challenges to and Restoration of Congress System

1. The socialist leader who gave the strategy ‘non-Congressism’ was

(a) Giani Zail Singh.

(b) V.V. Giri.

(d) V.P. Singh.

► (c) Ram Manohar Lohia.

2. The Indian leader who gave the slogan “Garibi hatao” was

(a) Lal Bahadur shastri.

(b) Indira Gandhi.

(d) B.R. Ambedkar.

► (b) Indira Gandhi.

3. The chairman of drafting committee of the Constituent Assembly was

(a) Lal Bahadur shastri.

(b) Indira Gandhi.

(d) Dr. B.R. Ambedkar.

► (d) Dr. B.R. Ambedkar.

4. The era of coalition government began in India in

(a) 1971.

(b) 2004.

(d) 1967.

5. The policy initiatives of Mrs.Indira Gandhi's government did not include

(a) abolition of privy purse.

(b) nationalisation of banks.

(d) ban on nuclear tests.

► (d) ban on nuclear tests.

6. Lal Bahadur Shastri's famous slogan was

(a) Garibi Hatao.

(b) Jai Jawan Jai Kisan.

(d) Indira Hatao.

► (b) Jai Jawan Jai Kisan.

7. The Congress Socialist party was founded by

(a) B. R. Ambedkar.

(b) Jawaharlal Nehru.

(d) Ram Manohar Lohia.

► (d) Ram Manohar Lohia.

8. After the death of Shastri, there was an intense competition between Indira Gandhi and

(a) Ram Manohar Lohia

(b) K. Kamraj.

(d) Karpoori Thakur.

► (c) Morarji Desai.

9. The slogan of 'garibi hatao’ was used before the

(a) 1952 elections.

(b) 1967 elections.

(d) 1977 elections.

10. Use of English language was strongly opposed by the

(a) Gulzari Lal Nanda.

(b) J. L. Nehru.

(d) Karpoori Thakur.

► (d) Karpoori Thakur.

11. Shastri was the country’s Prime Minister from _____.

(a) 1962 to 1965

(b) 1964 to 1966

(d) 1964 to 1969

► (b) 1964 to 1966

12. Indira Gandhi got the Congress Working Committee to adopt a Ten Point Programme in ______.

(a) April1967

(b) May 1967

(d) July1967

► (b) May 1967

13. The factional rivalry between the Syndicate and Indira Gandhi came in the open ______.

(a) Presidential Election, 1969

(b) Parliamentary Election, 1969

(d) None of the above

► (a) Presidential Election, 1969

14. The fifth general election to Lok Sabha were held in _______.

(a) October 1970

(b) November 1970

(d) February 1971

► (d) February 1971

↧

NCERT Solutions for Class 10 Maths Ch 11 Constructions

October 17, 2020, 7:45 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 6 The Crisis of Democratic Order

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 5 Challenges to and Restoration of Congress System

NCERT Solutions for Class 10 Maths Ch 11 Constructions

We have provided NCERT Solutions for Class 10 Maths Ch 11 Constructions in this page which will be helpful in completing your homework in less time. It will encourage students to learn new topics and give better understanding of the chapter. Class 10 NCERT Solutions is the best way to revise whole chapter properly and learning efficiently. These solutions are updated according to the latest pattern of CBSE.

Page No: 219

Exercise 11.1

In each of the following, give the justification of the construction also:

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer

Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with AB is drawn.
Step III: After that, a ray BY is drawn parallel to AX making equal acute angle as in the previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked on AX and point B1, B2.... to B8 is marked on BY such that AA1 = A1A2 = A2A3 =....BB1= B1B2 = .... B7B8
Step V: A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8

Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8

2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer

Steps of Construction:

Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4 cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2....A5
Step VII: A5B is joined. A2B' is drawn parallel to A5B and B'C' is drawn parallel to BC.
ΔAB'C' is the required triangle

Justification:
∠A(Common)
∠C = ∠C' and ∠B = ∠ B' (corresponding angles)

Thus ΔAB'C' ~ ΔABC by AAA similarity condition

From the figure,

AB'/AB = AA2/AA5 = 2/3

AB' =2/3 AB

AC' = 2/3 AC

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.

Answer

Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.

Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C' is drawn parallel to B5C and C'A' is parallel to CA.
Thus A'BC' is the required triangle.

Justification
ΔAB'C' ~ ΔABC by AAA similarity condition
∴ AB/A'B = AC/A'C' = BC/BC'
and BC/BC' = BB5/BB7 = 5/7
∴A'B/AB = A'C'/AC = = BC'/BC = 7/5

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer

Steps of Construction:

Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersect BC at O.
Step III: At a distance of 4 cm, a point A is marked on perpendicular bisector of BC.
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.

Step VIII: B3C' is drawn parallel to B2C and C'A' is drawn parallel to CA.
Thus, A'BC' is the required triangle formed.

Justification:

ΔAB'C' ~ ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
also,
AB/AB' = AA2/AA3 = 2/3
⇒ AB' = 3/2 AB, B'C' = 3/2 BC and AC' = 3/2 AC

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer

Steps of Construction:

Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VII: B3 is joined with C' to form B3C'.
Step VIII: C'A' is drawn parallel CA.
Thus, A'BC' is the required triangle.

Justification:

∠A = 60° (Common)
∠C = ∠C'
ΔAB'C' ~ ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
also,
AB/AB' = AA3/AA4 = 4/3
⇒ AB' = 3/4 AB, B'C' = 3/4 BC and AC' = 3/4 AC

6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

Answer

Sum of all side of triangle = 180°
∴ ∠A + ∠B + ∠C = 180°
∠C = 180° - 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45° with BC.

Step III: At C, a ray making an angle of 30° with BC is drawn intersecting the previous ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VI: B3C is joined and B4C' is made parallel to B3C.
Step VII: C'A' is made parallel CA.
Thus, A'BC' is the required triangle.

Justification:

∠B = 45° (Common)
∠C = ∠C'
ΔAB'C' ~ ΔABC by AA similarity condition.
∴ BC/BC' = AB/A'B' = AC/A'C'
also,
BC/BC' = BB3/BB4 = 34
⇒ AB = 4/3 AB', BC = 4/3 BC' and AC = 4/3 A'C'

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer

Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is made on previous ray intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is marked on BX.
Step VII: B3C is joined B5C' is made parallel to B3C.
Step VIII: A'C' is joined together.
Thus, ΔA'BC' is the required triangle.

Justification:
As in the previous question 6.

Page No. 221

Exercise 11.2

In each of the following, give also the justification of the construction:

1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is drawn.
Step II: A point P at a distance of 10 cm from the centre O is taken. OP is joined.
Step III: Perpendicular bisector OP is drawn and let it intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is drawn intersecting previous circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.

Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.

Similarly, PR is a tangent of the circle.

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Answer

Steps of Construction:

Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.

Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.

Step III: P be any point on the circle of radius 6 cm and OP is joined.

Step IV: Perpendicular bisector of OP is drawn which cuts it at M

Step V: With M as a centre and OM as a radius, a circle is drawn which intersect the the circle of radius 4 cm at Q and R

Step VI: PQ and PR are joined.

Thus, PQ and PR are the two tangents.

Measurement:

OQ = 4 cm (Radius of the circle)

PQ = 6 cm ( Radius of the circle)

∠PQO = 90° (Angle in the semi circle)

Applying Pythagoras theorem in ΔPQO,

PQ² + QO² = PO²
⇒ PQ² + 4²= 6²

⇒ PQ² + 16 = 36
⇒ PQ² = 36 - 16
⇒ PQ² = 20
⇒ PQ = 2√5

Justification:

∠PQO = 90° (Angle in the semi circle)

∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.

Similarly, PR is a tangent of the circle.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus, PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.

Justification:

∠PMO = 90° (Angle in the semi circle)

∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.

Similarly, PN, QR and QT are tangents of the circle.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Answer

We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O) = 360° - (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.

Justification:

Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° - 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer

Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4 cm, a circle is drawn which cut the line at point O.
Step III: With B as a centre and radius equal to 3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a radius, a circle is drawn which intersect the previous two circles at P, Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.

Justification:
∠BPA = 90° (Angle in the semi circle)

∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.

Similarly, AQ, BR and BS are tangents of the circle.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer

Steps of Construction:
Step I: A ΔABC is drawn.
Step II: Perpendicular to AC is drawn to point B which intersected it at D.

Step III: With O as a centre and OC as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.

Justification:

∠OEA = 90° (Angle in the semi circle)

∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.

Similarly, AB is tangent of the circle.

Go Back To NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Class 11 Constructions NCERT Solutions will be very useful in preparation of the examinations and understanding the concepts properly. There are three exercises given in Chapter 11 Constructions which will embed with you knowledge about various types of mathematical constructions.

• Introduction: In earlier class, we have done certain constructions such as bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles etc. and in this chapter, we will be doing more constructions like that and will give mathematical reasoning behind them.

• Division of a Line Segment: We will be dividing a line segment in a given ratio and construct a triangle similar to a given triangle as per given scale factor.

• Construction of Tangents to a Circle: We have to construct the tangents to a circle from a point outside it and provide justification of the construction also.

If you want to access other Study material for Class 10 then you can find them below.

CBSE Notes for Class 10 Maths Chapter 11 Constructions

MCQ Questions for Class 10 Maths Chapter 11 Constructions

There are only two exercises in the Chapter 11 Constructions which will improve your logical and thinking skills. We have also provided NCERT Solutions for Class 10 Maths exercisewise which you can get from below.

NCERT Solutions for Class 10 Maths Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry

Chapter 9 Some Applications of Trigonometry

Chapter 10 Circles

Chapter 12 Areas Related To Circles

Chapter 13 Surface Areas And Volumes

Chapter 14 Statistics

Chapter 15 Probability

FAQ on Chapter 11 Constructions

How many exercises in Chapter 11 Constructions

There are 2 exercise in the Chapter 11 Constructions Class 10 Maths which are prepared by Studyrankers which will give in depth learning about the concepts.

If a point lies inside a circle, then there cannot be a ______ to the circle through this point.

Tangent

What is Scale Factor?

The ratio of the corresponding sides of two similar geometric figures is called the scale-factor.

Tangents to a circle from a point lying outside the circle

Two tangents can be drawn to a circle through a point outside the circle and pair of these tangents are always equal in length.

↧

MCQ Questions for Class 12 Political Science: Ch 6 The Crisis of Democratic Order

October 17, 2020, 8:12 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 7 Rise of Popular Movements

≪ Previous: NCERT Solutions for Class 10 Maths Ch 11 Constructions

MCQ Questions for Class 12 Political Science: Ch 6 The Crisis of Democratic Order

1. Shah Commission was appointed to inquire

(a) abuse of authority during emergency.

(b) domestic violence.

(d) conditions of dalits.

► (a) abuse of authority during emergency.

2. The ''state of emergency'' was declared on

(a) 13th May, 1971.

(b) 23rd May, 1972.

(d) 11th March, 1975.

► (c) 25th June, 1975.

3. The Prime Minister of India during 1977-79 was

(a) Indira Gandhi.

(b) Charan Singh.

(d) Morarji Desai

► (d) Morarji Desai

4. The Janata Party made which election into a referendum on the Emergency?

(a) 1977

(b) 1980

(d) 1984

► (a) 1977

5. Mandal Commission was set up for the issue of

(a) reservation of other backward classes.

(b) temple conflict.

(d) child labour.

► (a) reservation of other backward classes.

6. The Bihar movement was led by

(a) Charu Majumdar.

(b) J. P. Narayan.

(d) C. Natarajan.

► (b) J. P. Narayan.

7. Who left Congress party and founded Bharatiya Kranti Dal in 1967?

(a) Sanjay Gandhi

(b) Jagjivan Ram

(d) Morarji Desai

► (c) Chaudhary Charan Singh

8. After 1977 general elections, which government which came into power?

(a) Bharatiya Jana Sangh.

(b) Congress.

(d) Socialist Party.

► (c) Janata Party.

9. The Bihar Movement in 1974 was led by

(a) Charu Majumdar.

(b) Jayaprakash Narayan.

(d) C. Natarajan.

► (b) Jayaprakash Narayan.

10. Congress was defeated in the lok sabha elections of

(a) January 1975.

(b) February 1976.

(d) April 1977.

► (c) March 1977.

11. When JP led a peoples’ march to the Parliament?

(a) 1975.

(b) 1976.

(d) 1978.

► (a) 1975.

12. Find the odd one out in the context of proclamation of Emergency.

(a) The call for ‘Total Revolution’

(b) The Railway Strike of 1974

(d) The Allahabad High Court Verdict

► (c) The Naxalite Movement

13. Through which Article of the Constitution, government declared emergency in 1975?

(a) Article 350

(b) Article 351

(d) Article 354

► (c) Article 352

14. Railway strike of 1974 was called by ________ led by George Fernandes.

(a) National Coordination Committee

(b) National Coordination Corporation

(d) National Coordination Association

► (a) National Coordination Committee

↧

MCQ Questions for Class 12 Political Science: Ch 7 Rise of Popular Movements

October 17, 2020, 5:51 pm

≫ Next: MCQ Questions for Class 12 Political Science: Ch 8 Regional Aspirations

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 6 The Crisis of Democratic Order

MCQ Questions for Class 12 Political Science: Ch 7 Rise of Popular Movements

1. In early 1990s, the BKU pressurised the state to accept its

(a) political demands.

(b) social demands.

(d) economic demands.

► (d) economic demands.

2. Mazdoor Kisan Shakti Sangathan started in 1990 in

(a) Rajasthan.

(b) Madhya Pradesh.

(d) Maharashtra.

► (a) Rajasthan.

3. Chipko Movement started in which of the following states?

(a) Uttar Pradesh

(b) Madhya Pradesh

(d) Punjab

► (c) Uttarakhand

4. Dalit Panthers is a:

(a) Social Organisation.

(b) Student federation.

(d) Pressure Group.

► (a) Social Organisation.

5. Anti-Arrack Movement demanded the ban on the

(a) falling of trees.

(b) practice of untouchability.

(d) construction of dam.

► (c) sale of alcohol.

6. The non-political organization ‘Bharatiya Kisan Union’ was formed by

(a) Brinda Karat.

(b) George Fernandes.

(d) Sanal Edamaruku.

► (c) Mahendra Singh Tikait.

7. On what issue the villagers related to Chipko movement protested?

(a) Deforestation and commercial logging.

(b) Construction of big dams.

(d) Exploitation of adivasis.

► (a) Deforestation and commercial logging.

8. In which state the Chipko movement started?

(a) Uttarakhand

(b) Kerala

(d) Orissa

► (a) Uttarakhand

9. What was the very novel aspect of the Chipko movement?

(a) Womens' active Participation

(b) Rise of independent economic movement

(d) The social discontent in Indian society

► (a) Womens' active Participation

10. One of the most successful social movement of eighties was

(a) farmer's movement.

(b) anti-arrack movement.

(d) fish worker's movement.

► (a) farmer's movement.

11. Which of these statements are incorrect regarding The Chipko Movement.

(a) Was an environmental movement to prevent cutting down of trees.

(b) Raised questions of ecological and economic exploitation.

(d) Demanded that local communities should have control over their natural resources.

► (c) Was a movement against alcoholism started by the women.

12. Dalit Panthers, a militant organisation was formed in which state?

(a) Uttar Pradesh

(b) Maharashtra

(d) Bihar

► (b) Maharashtra

13. Which popular movement led to the empowerment of women?

(a) Chipko Movement

(b) Anti-Arrack Movement

(d) None of the above

► (b) Anti-Arrack Movement

14. What was the main demand of the Chipko Movement?

(a) To prevent child marriages

(b) To fight against discrimination

(d) To discontinue child labour

► (c) To prevent trees from being cut and fell.

↧

MCQ Questions for Class 12 Political Science: Ch 8 Regional Aspirations

October 17, 2020, 8:39 pm

≫ Next: MCQ Questions for Class 12 Political Science: Ch 9 Recent Developments in Indian Politics

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 7 Rise of Popular Movements

MCQ Questions for Class 12 Political Science: Ch 8 Regional Aspirations

1. Nagaland state was created in the year

(a) 1980.

(b) 1963.

(d) 1970.

► (b) 1963.

2. Which two states were created in 1966?

(a) Delhi and Haryana

(b) Rajasthan and Gujarat

(d) Maharashtra and Gujarat

► (c) Punjab and Haryana

3. On the question of diversity which approach is adopted by India?

(a) Totalitarian approach

(b) Dictatorial approach

(d) Democratic approach

► (d) Democratic approach

4. Rajiv Gandhi joined active politics after

(a) 1950.

(b) 1960.

(d) 1980.

► (d) 1980.

5. During the period between 1953 and 1974, lots of influence on the politics of Kashmir was exercised by the

(a) Congress.

(b) Bhartiya Janata Party.

(d) Akali Dal.

► (a) Congress.

6. The first democratic election to Sikkim assembly in 1974 was swept by the

(a) Sikkim Congress.

(b) Indian National Congress

(d) Communist Party of India (CPI).

► (a) Sikkim Congress.

7. The '' Kashmir issue'' is a dispute between India and

(a) Bangladesh.

(b) Sri Lanka.

(d) China.

► (c) Pakistan.

8. Master Tara Singh was a leader of

(a) SGPC.

(b) AIADMK.

(d) PDP.

► (a) SGPC.

9. Which region is referred to as the ‘seven sisters’?

(a) East-West Region

(b) North-West Region

(d) North-East Region

► (d) North-East Region

10. ‘Bodo’ is a community in

(a) Manipur.

(b) Assam.

(d) Arunachal Pradesh.

► (b) Assam.

11. Why Operation Blue Star was launched by army and Government of India?

(a) 1982

(b) 1983

(d) 1985

12. Which students group led the anti-foreigner movement in Assam?

(a) All Assam Workers’ Union

(b) All Assam Students’ Union

(d) All Assam Women’ Union

► (b) All Assam Students’ Union

13. What was the anther name of Rajiv Gandhi - Longowal Accord?

(a) Maharashtra Accord

(b) Punjab Accord

(d) Kerala Accord

► (b) Punjab Accord

14. Anandpur Sahib Resolution was passed at the conference of Akali Dal in _______.

(a) 1971

(b) 1972

(d) 1974

↧

MCQ Questions for Class 12 Political Science: Ch 9 Recent Developments in Indian Politics

October 18, 2020, 1:27 am

≫ Next: MCQ Questions for Class 11 History: Ch 1 From the Beginning of Time

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 8 Regional Aspirations

MCQ Questions for Class 12 Political Science: Ch 9 Recent Developments in Indian Politics

1. Which political party was joined by B.P. Mandal in 1977?

(a) Bahujan Samaj Party.

(b) Rashtriya Swamsewak Sangh

(d) Janata Party.

► (d) Janata Party.

2. The reason for 2002 Gujarat riots is

(a) earthquake relief was not provided by the Centre.

(b) A bogey of a train returning from Ayodhya carrying Karsevaks was set on fire.

(d) state emergency.

► (b) A bogey of a train returning from Ayodhya carrying Karsevaks was set on fire.

3. The single largest party in the 1996 parliamentary elections was

(a) Bhartiya Janta Party.

(b) Janta Dal.

(d) Congress.
► (a) Bhartiya Janta Party.

4. Who headed the central government in India after 1998 parliamentary elections?

(a) Atal Bihari Vajpayee

(b) P.V. Narsimha Rao.

(d) I. K. Gujral.

► (a) Atal Bihari Vajpayee

5. Which party pursued the politics of ‘Hindutva’ and adopted the strategy of mobilising the Hindus?

(a) Bhartiya Janata Party.

(b) Congress.

(d) Bahujan Samaj Party.

► (a) Bhartiya Janata Party.

6. The recommendations of the Mandal Commission were implemented in the year

(a) 1990.

(b) 1991.

(d) 1993.

► (a) 1990.

7. In February and March 2002, large scale violence against Muslims took place in

(a) Rajasthan.

(b) Gujarat.

(d) Uttar Pradesh.

► (b) Gujarat.

8. The General Elections of 2004 resulted in

(a) dominance of Congress.

(b) coalition government.

(d) emergence of Akali Dal.

► (b) coalition government.

9. Prime Minister Rajiv Gandhi was assassinated in the year

(a) 1989.

(b) 1990.

(d) 1992.

10. The political developments after the 1990s reveals the emergence of

(a) coalition government.

(b) consolidation of BJP.

(d) regional politics.

► (a) coalition government.

11. Who was the chairperson of Mandal Commission?

(a) Bindeshwari Prasad Mandal.

(b) Nandeshwari Prasad Mandal.

(d) None of the Above

► (a) Bindeshwari Prasad Mandal.

12. Which alliance formed the government at the centre in 1989?

(a) National Front

(b) United Front

(d) Real Front

► (a) National Front

13. Choose the correct term for coming together of two or more political parties to form government.

(a) United Government

(b) Coalition Government

(d) None of the above

► (b) Coalition Government

14. Elections in _____led an era of coalitions with the defeat of Congress Party

(a) 1988

(b) 1989

(d) 1991

► (b) 1989

↧

MCQ Questions for Class 11 History: Ch 1 From the Beginning of Time

October 18, 2020, 4:02 am

≫ Next: MCQ Questions for Class 11 History: Ch 2 Writing and City Life

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 9 Recent Developments in Indian Politics

MCQ Questions for Class 11 History: Ch 1 From the Beginning of Time

1. The Hominids survived for about:

(a) 1 Million years

(b) 2 Million years

(d) 4 Million years

► (a) 1 Million years

2. In the Eastern Hadza community, the lands belonged to

(a) Women.

(b) Headmen.

(d) No one.

► (d) No one.

3. An archaeological site, which was used for quarrying limestone was:

(a) Chesowanja, Kenya.

(b) Schoningen, Germany.

(d) Neander Valley, Germany.

► (d) Neander Valley, Germany.

4. The term ‘pathological idiot’ has been used with reference to

(a) Australopithecus.

(b) Homo Habilis.

(d) Neanderthal.

► (d) Neanderthal.

5. Zagros mountains, which form one limit of the Fertile Crescent, is situated in:

(a) Egypt.

(b) Lebanon.

(d) Syria.

► (c) Iran.

6. The last Ice Age ended around:

(a) 15,000 years ago.

(b) 14, 000 years ago.

(d) 12,000 years ago.

► (c) 13,000 years ago.

7. The main reason that led to a change from foraging to farming occupation was:

(a) Domestication.

(b) End of Ice Age.

(d) Warmer and wetter conditions.

► (a) Domestication.

8. Dexterity of the hands refers to:

(a) Roughness of the hands.

(b) Shape of the hands.

(d) Flexibility of the hands.

► (d) Flexibility of the hands.

9. Population pressure in some areas increased due to the

(a) end of ice age.

(b) change in climate.

(d) domestication.

► (c) increase in food availability.

10. Genus is the branch of the:

(a) Hominids.

(b) Hominoids.

(d) Australopithecus.

► (a) Hominids.

11. ______ are the remains or impressions of a very old plant, animal or human which have turned into stone.

(a) Fossils

(b) Neanderthal

(d) None of the above

► (a) Fossils

12. When did Primates originate in Asia and Africa?

(a) Around 32 million years ago.

(b) Around 36 million years ago.

(d) Around 44 million years ago.

► (b) Around 36 million years ago.

13. From which organism group the hominids were evolved?

(a) Neanderthal.

(b) Hominoids.

(d) Australopithecus.

► (b) Hominoids.

14. What is the base of classification of fossils of early humans?

(a) Bone Structure

(b) DNA

(d) None of the above

► (a) Bone Structure

15. The Hominids belong to a family which is known as _____.

(a) Hominidae

(b) Neanderthal

(d) Australopithecus

► (a) Hominidae

16. Long forelimbs, Curved hand and foot bones and mobile ankle joints helped ______ to live on trees.

(a) Hominidae

(b) Neanderthal

(d) Australopithecus

► (d) Australopithecus

17. Fossils found in the ______ were called as Homo neanderthalensis.

(a) Barun Valley

(b) Hunza Valley

(d) Lidder Valley

► (c) Neander Valley

18. Fossils of Hominid found in Asia were not older as compared to fossils found in ______.

(a) Europe

(b) Africa

(d) America

► (b) Africa

↧

MCQ Questions for Class 11 History: Ch 2 Writing and City Life

October 18, 2020, 5:15 am

≫ Next: MCQ Questions for Class 11 History: Ch 3 An Empire Across Three Continents

≪ Previous: MCQ Questions for Class 11 History: Ch 1 From the Beginning of Time

MCQ Questions for Class 11 History: Ch 2 Writing and City Life

1. Mesopotamian weapons were prominently made of

(a) bronze.

(b) copper.

(d) iron.

► (a) bronze.

2. The technological landmark witnessed by the urban economy of the city of Uruk was

(a) bronze tools.

(b) construction of brick columns.

(d) oil pressing technique.

► (c) potter's wheel.

3. In the 2000 BCE, the city that flourished as the royal capital of Mesopotamia was

(a) Ur.

(b) Mari.

(d) Kalibangan.

► (b) Mari.

4. The Mesopotamian ruler who became the king of Akkad, in 2370 BCE, was

(a) Enmerkar

(b) Gilgamesh

(d) Sargon

► (d) Sargon

5. Inanna was the Goddess of

(a) the Moon

(b) love and War

(d) fire

► (b) love and War

6. The earliest temples in Southern Mesopotamia were built in

(a) c.4000 BCE

(b) c.5000 BCE

(d) c.7000 BCE

► (b) c.5000 BCE

7. The Mesopotamian city, which was systematically excavated in the 1930s, was

(a) Uruk.

(b) Ur.

(d) Nineveh.

► (b) Ur.

8. The Achaemenids of Iran conquered Babylon in

(a) 639 BCE.

(b) 539 BCE.

(d) 339 BCE.

► (b) 539 BCE.

9. The King who released Babylonia from Assyrian domination in 625 BCE was

(a) Alexander.

(b) Nabonidus.

(d) Sargon.

► (c) Nabopolassar.

10. War captives and local people who were put to work for the temple or for the ruler were paid

(a) bronze tools.

(b) cattle.

(d) rations.

► (d) rations.

11. In ______ part of South Iran, the first cities and writing emerged.

(a) plain

(b) deserted

(d) hilly

► (b) deserted

12. Agriculture of southern ______ was the most productive.

(a) Iran

(b) China

(d) None of the above

► (c) Mesopotamia

13. Which region is important to Europeans because of references to it in the Old Testament,

(a) Iran

(b) Turkey

(d) Mesopotamia

► (d) Mesopotamia

14. From which word Cuneiform is derived from?

(a) Latin

(b) Greek

(d) Mayan

► (a) Latin

15. Which language replaced the Sumerian language after 2400 BCE?

(a) Mesopotamian language

(b) Akkadian language

(d) Middle Eastern language

► (b) Akkadian language

16. With which person the first event of Sumerian trade is associated?

(a) The ancient ruler of Uruk City, Enmerkar.

(b) The ancient ruler of Lebanon City, Enmerkar.

(d) The ancient ruler of Aral City, Enmerkar.

► (a) The ancient ruler of Uruk City, Enmerkar.

↧

MCQ Questions for Class 11 History: Ch 3 An Empire Across Three Continents

October 18, 2020, 6:50 am

≫ Next: MCQ Questions for Class 11 History: Ch 4 The Central Islamic Lands

≪ Previous: MCQ Questions for Class 11 History: Ch 2 Writing and City Life

MCQ Questions for Class 11 History: Ch 3 An Empire Across Three Continents

1. The Ostrogoths established their kingdom in Italy in the year

(a) 493 CE

(b) 494 CE

(d) 496 CE

► (a) 493 CE

2. The religion of Islam arose during the

(a) 5th century CE

(b) 8th century CE

(d) 7th century CE

► (d) 7th century CE

3. The Roman emperor who consolidated the rise of provincial upper classes so as to exclude the senators from military command was

(a) Augustus

(b) Constantine

(d) Tiberius

► (c) Gallienus

4. Augustus, the first Roman Emperor was called the ‘leading citizen’ whose Latin term is

(a) 'Basileus'.

(b) 'Dominus'.

(d) 'Res gestae'.

► (c) 'Princeps'.

5. Christianity became the state religion of Roman Empire in the

(a) 1st century CE.

(b) 2nd century CE.

(d) 4th century CE.

► (d) 4th century CE.

6. The emperor who made Christianity the official religion in the Roman Empire was

(a) Alexander

(b) Augustus

(d) Nero

► (c) Constantine

7. In the Roman urban life, the entertainment shows called “spectacula’ happened for at least

(a) 150 days

(b) 160 days

(d) 176 days

► (d) 176 days

8. The Roman Empire got best kind of wine from the city of

(a) Byzaciuma

(b) Campania

(d) Sicily

► (b) Campania

9. The Roman silver coin, known as denarius, weighed _________ gm of pure silver.

(a) 2½

(b) 3½

(d) 5½

► (c) 4½

10. Saint Augustine was bishop of the North African city of

(a) Annaba

(b) Algeria

(d) Numidia

► (c) Hippo

11. ______ and Greek languages were used in administration of the Roman Empire.

(a) Chinese

(b) Mayan

(d) Turkish

► (c) Latin

12. Roman ruler _____ was considered as the ‘leading citizen’ only to show that he was not the absolute ruler.

(a) Augustus

(b) Constantine

(d) Tiberius

► (a) Augustus

13. What were Amphorae?

(a) A type of army

(b) A type of container

(d) None of the above

► (b) A type of container

14. The best kind of wine came to Rome from _______

(a) Fayum

(b) Byzacium

(d) Campania

► (d) Campania

15. Which one of the following is a rive that form the boundary of Roman Empire?

(a) Mekong River

(b) Rhine River

(d) Yangtze River

► (b) Rhine River

16. The ______ System of the Roman Empire was the system of government in which the real power vested in the Senate.

(a) Democrat

(b) Republican

(d) Marxist

► (b) Republican

↧

MCQ Questions for Class 11 History: Ch 4 The Central Islamic Lands

October 18, 2020, 8:05 am

≫ Next: NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

≪ Previous: MCQ Questions for Class 11 History: Ch 3 An Empire Across Three Continents

MCQ Questions for Class 11 History: Ch 4 The Central Islamic Lands

1. The Umayyad Emperor who introduced Arabic as the language of administration was

(a) Abd al-Malik

(b) Ali

(d) Muawiya

► (a) Abd al-Malik

2. The followers and non-followers of Ali were called

(a) Arabs and Iranians.

(b) Shias and Sunnis.

(d) Byzantines and Sasanids

► (b) Shias and Sunnis.

3. The Caliph during whose reign the Hijri was established was:

(a) Abu Bakr

(b) Muhammad

(d) Umar

► (d) Umar

4. What was the name given to nomadic Arab tribes?

(a) Sanam

(b) Qabila

(d) Umma

► (c) Bedouins

5. Zakat was an/a

(a) agriculture tax

(b) alms tax

(d) trade tax

► (b) alms tax

6. The semi-histrical works sira in the Islamic history are

(a) biographies

(b) chronicles

(d) inscriptions

► (a) biographies

7. Caliph Muawiya shifted his capital to

(a) Rome

(b) Medina

(d) Damascus

► (d) Damascus

8. The woman-saint sufi who, in her poems, preached instense love for God by uniting with God was

(a) Rabia.

(b) Aisha

(d) Khadija

► (a) Rabia.

9. After the Arabs and Iranians, the group that was rising to power on the Islamic scene was the

(a) Buyids.

(b) Iranians.

(d) Turks.

► (d) Turks.

10. The sub-sect of Shiism, from which the Fatimids descended, is known as the

(a) Ismaili

(b) Shia

(d) Umma

► (a) Ismaili

11. Prophet Muhammad declared himself to be the messenger of God

(a) around 610 CE.

(b) around 611 CE.

(d) around 613 CE.

► (c) around 612 CE.

12. Prophet Muhammad’s migration from Mecca to Medina with his followers in 622 CE is called hijra:

(a) Hijra

(b) Kaba

(d) Calipha

► (a) Hijra

13. By whom Ali assassinated?

(a) Muawiya

(b) Kharji

(d) None of the above

► (b) Kharji

14. Who introduced an islamic coinage.

(a) Abu Bakr

(b) Abd-al-Malik

(d) Umar

► (b) Abd-al-Malik

↧

NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

October 18, 2020, 8:48 am

≫ Next: NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes

≪ Previous: MCQ Questions for Class 11 History: Ch 4 The Central Islamic Lands

NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

In this page, you will find Chapter 12 Areas Related to Circles NCERT Solutions for Class 10 Maths which are very helpful in finding solutions of difficult problems given in the exercise. Class 10 NCERT Solutions are very beneficial in increasing concertation among students. These NCERT Solutions are updated as per the latest syllabus of CBSE.

Page No: 230

Exercise: 12.1

Unless stated otherwise, use π =22/7.

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer

Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm

⇒ R = 28 cm

The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer

Let the radius of the third circle be R.
Area of the circle with radius R = πR²
Area of the circle with radius 8 cm = π × 8²= 64π cm²
Area of the circle with radius 6 cm = π × 6²= 36π cm²
Sum of the area of two circles = 64π cm²+36π cm²= 100πcm²
Area of the third circle = πR² = 100πcm²
⇒ πR² = 100πcm²
⇒ R² = 100cm²
⇒ R = 10cm
Thus, the radius of the new circle is 10 cm.

3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Answer

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, r₁ = 21/2 cm = 10.5 cm
Each of the other bands is 10.5 cm wide,
∴ Radius of second circle, r₂ = 10.5 cm + 10.5 cm = 21 cm
∴ Radius of third circle, r₃ = 21 cm + 10.5 cm = 31.5 cm
∴ Radius of fourth circle, r₄ = 31.5 cm + 10.5 cm = 42 cm
∴ Radius of fifth circle, r₅ = 42 cm + 10.5 cm = 52.5 cm
Area of gold region = π r₁²= π (10.5)²= 346.5 cm²
Area of red region = Area of second circle - Area of first circle = π r₂² - 346.5 cm²

=π(21)² - 346.5 cm² = 1386 - 346.5 cm²= 1039.5 cm²

Area of blue region = Area of third circle - Area of second circle = π r₃²- 1386 cm²

=π(31.5)² - 1386 cm²= 3118.5 - 1386 cm²= 1732.5 cm²

Area of black region = Area of fourth circle - Area of third circle = π r₃² - 3118.5 cm²

= π(42)² - 1386 cm²= 5544 - 3118.5 cm²= 2425.5 cm²

Area of white region = Area of fifth circle - Area of fourth circle = π r₄² - 5544 cm²

= π(52.5)² - 5544 cm²= 8662.5 - 5544 cm²= 3118.5 cm²

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer

Diameter of the wheels of a car = 80 cm

Circumference of wheels = 2πr = 2r × π = 80 π cm

Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s

No. of revolutions = Distance travelled by car/Circumference of wheels

= 1100000/80 π = (1100000 × 7)/(80×22) = 4375

4375 complete revolutions does each wheel make in 10 minutes.

5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units (B) π units (C) 4 units (D) 7 units

Answer

Let the radius of the circle be r.

∴ Perimeter of the circle = Circumference of the circle = 2πr

∴ Area of the circle = π r²

A/q,

2πr = π r²

⇒ 2 = r

Thus, the radius of the circle is 2 units. (A) is correct.

Page No: 230

Exercise: 12.2

Unless stated otherwise, use π =22/7.

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer

Area of the sector making angle θ = (θ/360°)×π r²

Area of the sector making angle 60° = (60°/360°)×π r²cm²

= (1/6)×6²π = 36/6 π cm² = 6 × 22/7 cm²= 132/7 cm²

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer

Quadrant of a circle means sector is making angle 90°.

Circumference of the circle = 2πr = 22 cm

Radius of the circle = r = 22/2π cm = 7/2 cm

Area of the sector making angle 90° = (90°/360°)×π r²cm²

= (1/4)×(7/2)²π = (49/16) π cm² = (49/16) × (22/7) cm²= 77/8 cm²
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer

Here, Minute hand of clock acts as radius of the circle.

∴ Radius of the circle (r) = 14 cm

Angle rotated by minute hand in 1 hour = 360°

∴ Angle rotated by minute hand in 5 minutes = 360° × 5/60 = 30°

Area of the sector making angle 30° = (30°/360°)×π r²cm²

= (1/12) × 14²π = 196/12 π cm² = (49/3) × (22/7) cm²= 154/3 cm²
Area swept by the minute hand in 5 minutes = 154/3 cm²
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

Answer

Radius of the circle = 10 cm

Major segment is making 360° - 90° = 270°

Area of the sector making angle 270°

= (270°/360°) × π r²cm²

= (3/4) × 10²π = 75 π cm²
= 75 × 3.14 cm²= 235.5 cm²
∴ Area of the major segment = 235.5 cm²

Height of ΔAOB = OA = 10 cm

Base of ΔAOB = OB = 10 cm

Area of ΔAOB = 1/2 × OA × OB

= 1/2 ×10 × 10 = 50 cm²

Major segment is making 90°

Area of the sector making angle 90° = (90°/360°) × π r²cm²

= (1/4) × 10²π = 25 π cm²
= 25 × 3.14 cm²= 78.5 cm²
Area of the minor segment = Area of the sector making angle 90° - Area of ΔAOB

= 78.5 cm²-50 cm²= 28.5 cm²

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

Answer

Radius of the circle = 21 cm

(i) Length of the arc AB = θ/360° × 2πr
= 60°/360° × 2 × 22/7 × 21
= 1/6 × 2 × 22/7 × 21 = 22

The length of the arc is 22 cm.

(ii) Angle subtend by the arc = 60°

Area of the sector making angle 60° = (60°/360°) × π r²cm²

= (1/6) × 21²π = 441/6 π cm²
= 441/6 × 22/7 cm²= 231 cm²
∴ Area of the sector formed by the arc is 231 cm²
(iii) Area of equilateral ΔAOB = √3/4 × (OA)²= √3/4 × 21²= (441√3)/4 cm²

Area of the segment formed by the corresponding chord

= Area of the sector formed by the arc - Area of equilateral ΔAOB

= 231 cm²-(441√3)/4 cm²

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

Answer

Radius of the circle = 15 cm

ΔAOB is isosceles as two sides are equal.

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

Triangle is equilateral as ∠A = ∠B = ∠C = 60°

∴ OA = OB = AB = 15 cm

Area of equilateral ΔAOB = √3/4 × (OA)²= √3/4 × 15²

= (225√3)/4 cm²= 97.3 cm²

Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × π r²cm²

= (1/6) × 15²πcm² =225/6 πcm²

= (225/6) × 3.14 cm²= 117.75 cm²

Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB

= 117.75 cm² - 97.3 cm²= 20.4 cm²

Angle made by Major sector = 360° - 60° = 300°

Area of the sector making angle 300° = (300°/360°) × π r²cm²

= (5/6) × 15²πcm² =1125/6 πcm²

= (1125/6) × 3.14 cm²= 588.75 cm²

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

= 588.75 cm² + 97.3 cm²= 686.05 cm²

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer

Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° - (90° + 60°) = 30°
cos 30° = AD/OA
⇒ √3/2 = AD/12
⇒ AD = 6√3 cm
⇒ AB = 2 × AD = 12√3 cm
sin 30° = OD/OA
⇒ 1/2 = OD/12
⇒ OD = 6 cm
Area of ΔAOB = 1/2 × base × height
= 1/2 × 12√3 × 6 = 36√3 cm
= 36 × 1.73 = 62.28 cm²

Angle made by Minor sector = 120°

Area of the sector making angle 120° = (120°/360°) × π r²cm²

= (1/3) × 12²πcm² =144/3 πcm²

= 48 × 3.14 cm²= 150.72 cm²
∴ Area of the corresponding Minor segment = Area of the Minor sector - Area of ΔAOB
= 150.72 cm²-62.28 cm²
= 88.44 cm²
8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m
long rope (see Fig. 12.11). Find

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Answer

Side of square field = 15 m

Length of rope is the radius of the circle, r = 5 m

Since, the horse is tied at one end of square field, it will graze only quarter of the field with radius 5 m.

(i) Area of circle = π r²= 3.14 × 5² = 78.5 m²

Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m²

(ii) Area of circle if the length of rope is increased to 10 m = π r² =3.14 × 10² = 314 m²

Area of that part of the field in which the horse can graze = 1/4 of area of the circle

= 314/4 = 78.5 m²

Increase in grazing area = 78.5 m² - 19.625 m² = 58.875 m²

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find:

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Answer

Number of diameters = 5
Length of diameter = 35 mm
∴ Radius = 35/2 mm

(i) Total length of silver wire required = Circumference of the circle + Length of 5 diameter
= 2π r + (5×35) mm = (2 × 22/7 × 35/2) + 175 mm
= 110 + 175 mm = 185 mm

(ii) Number of sectors = 10
Area of each sector = Total area/Number of sectors
Total Area = π r² = 22/7 × (35/2)²=1925/2 mm²
∴ Area of each sector = (1925/2) × 1/10 = 385/4 mm²
10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer

Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm
Area between the two consecutive ribs of the umbrella =
Total area/Number of ribs
Total Area = π r² = 22/7 × (45)²=6364.29 cm²
∴ Area between the two consecutive ribs = 6364.29/8 cm²
= 795.5 cm²
11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer

Angle of the sector of circle made by wiper = 115°

Radius of wiper = 25 cm

Area of the sector made by wiper = (115°/360°) × π r²cm²

=23/72 × 22/7 × 25²=23/72 × 22/7 × 625 cm²

= 158125/252 cm²

Total area cleaned at each sweep of the blades = 2 ×158125/252 cm² =158125/126 = 1254.96 cm²

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)

Answer

Let O bet the position of Lighthouse.
Distance over which light spread i.e. radius, r = 16.5 km
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × π r²km²
= 2/9 × 3.14 × (16.5)²km²
= 189.97 km²

13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm² . (Use √3 = 1.7)

Answer

Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = ₹ 0.35 per cm²
∠O= 360°/6 = 60°

ΔAOB is isosceles as two sides are equal. (Radius of the circle)

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

∠A + ∠B + ∠O = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

Triangle is equilateral as ∠A = ∠B = ∠C = 60°
Area of equilateral ΔAOB = √3/4 × (OA)²= √3/4 × 28²= 333.2 cm²
Area of sector ACB = (60°/360°) × π r²cm²

= 1/6 × 22/7 × 28 × 28 = 410.66 cm²

Area of design = Area of sector ACB - Area of equilateral ΔAOB

= 410.66 cm² - 333.2 cm²= 77.46 cm²

Area of 6 design = 6 × 77.46 cm²= 464.76 cm²

Total cost of making design = 464.76 cm²× ₹ 0.35 per cm² = ₹ 162.66

14. Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR (B) p/180 × π R²(C) p/360 × 2πR (D) p/720 × 2πR²

Answer

Area of a sector of angle p = p/360 × π R²

=p/360 × 2/2 × π R²

= 2p/720 × 2πR²

Hence, Option (D) is correct.

Page No: 234

Exercise: 12.3

Unless stated otherwise, use π =22/7

1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer

PQ = 24 cm and PR = 7 cm

∠P = 90° (Angle in the semi-circle)

∴ QR is hypotenuse of the circle = Diameter of the circle.

By Pythagoras theorem,

QR²= PR²+ PQ²

⇒ QR²= 7²+ 24²

⇒ QR²= 49+ 576

⇒ QR²= 625

⇒ QR= 25 cm

∴ Radius of the circle = 25/2 cm

Area of the semicircle = (π R²)/2

= (22/7 × 25/2 × 25/2)/2 cm²

= 13750/56 cm²= 245.54 cm²

Area of the ΔPQR = 1/2 × PR × PQ

= 1/2 × 7 × 24 cm²

= 84 cm²

Area of the shaded region = 245.54 cm² - 84 cm²= 161.54 cm²

2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Answer

Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × π r²cm²
= 1/9 × 22/7 × 14²= 68.44 cm²
Area of the sector OBD = (40°/360°) × π r²cm²

= 1/9 × 22/7 × 7²= 17.11 cm²

Area of the shaded region ABDC = Area of the sector OAC - Area of the sector circle OBD

= 68.44 cm² - 17.11 cm²= 51.33 cm²

3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer

There are two semicircles in the figure.

Side of the square = 14 cm

Diameter of the semicircle = 14 cm

∴ Radius of the semicircle = 7 cm

Area of the square = 14 × 14 = 196 cm²

Area of the semicircle = (π R²)/2

= (22/7 × 7 × 7)/2 cm²= 77 cm²
Area of two semicircles = 2 × 77 cm² = 154 cm²

Area of the shaded region = 196 cm²-154 cm² = 42 cm²

4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer

OAB is an equilateral triangle with each angle equal to 60°.

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = √3/4 × (OA)²= √3/4 × 12²= 36√3 cm²

Area of the circle = π R² = 22/7 × 6²= 792/7 cm²

Area of the sector making angle 60° = (60°/360°) × π r²cm²

= 1/6 × 22/7 × 6²cm²= 132/7 cm²

Area of the shaded region = Area of the equilateral triangle + Area of the circle - Area of the sector

= 36√3 cm² + 792/7 cm² - 132/7 cm²

= (36√3 + 660/7) cm²

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

Answers

Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)²= 4²= 16 cm²

Area of the quadrant = (π R²)/4 cm² = (22/7 × 1²)/4 = 11/14 cm²

∴ Total area of the 4 quadrants = 4 × (11/14) cm² = 22/7 cm²

Area of the circle = π R²cm² = (22/7 × 1²) = 22/7 cm²

Area of the shaded region = Area of square - (Area of the 4 quadrants + Area of the circle)

= 16 cm²- (22/7 + 22/7) cm²

= 68/7 cm²

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

Answer

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm
In ΔADB,

By Pythagoras theorem,

AB²= AD²+ BD²

⇒ AB²= 48²+ (AB/2)²

⇒ AB²= 2304+ AB²/4
⇒ 3/4 (AB²)= 2304

⇒ AB²= 3072

⇒ AB= 32√3 cm
Area of ΔADB = √3/4 × (32√3)²cm²= 768√3 cm²
Area of circle = π R² = 22/7 × 32 × 32 = 22528/7 cm²
Area of the design = Area of circle - Area of ΔADB

= (22528/7 - 768√3) cm²

Page No: 236

7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer

Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 14²= 196 cm²
Area of the quadrant = (π R²)/4 cm² = (22/7 × 7²)/4 cm²

= 77/2 cm²

Total area of the quadrant = 4 × 77/2 cm²= 154cm²

Area of the shaded region = Area of the square ABCD - Area of the quadrant

= 196 cm²- 154 cm²

= 42 cm²

8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If

the track is 10 m wide, find :

(i) the distance around the track along its inner edge

(ii) the area of the track.

Answer

Width of track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O'C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O'B

= 30 + 10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)

= 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m

= 212 + 1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) - 2 × (area of inner semicircle)

= (AB × CD) + (EF × GH) + 2 × (πr²/2) - 2 × (πR²/2) m²

= (106 × 10) + (106 × 10) + 2 × π/2 (r² -R²) m²

= 2120 + 22/7 × 70 × 10 m²

= 4320 m²

9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other
and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer

Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm²
Area of larger circle = πR²= 22/7 × 7² = 154 cm²

Area of larger semicircle = 154/2 cm²= 77 cm²

Area of smaller circle = πr² = 22/7 × 7/2 × 7/2 = 77/2 cm²

Area of the shaded region = Area of larger circle - Area of triangle - Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154 - 49 - 77 + 77/2) cm²

= 133/2 cm² = 66.5 cm²

10. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Answer

ABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

There are three sectors each making 60°.

Area of ΔABC = 17320.5 cm²

⇒ √3/4 × (side)² =17320.5

⇒ (side)² =17320.5 × 4/1.73205

⇒ (side)² = 4 × 10⁴

⇒ side = 200 cm

Radius of the circles = 200/2 cm = 100 cm

Area of the sector = (60°/360°) × π r²cm²

= 1/6 × 3.14 × (100)²cm²

=15700/3cm²

Area of 3 sectors = 3 × 15700/3 = 15700 cm²=

Area of the shaded region = Area of equilateral triangle ABC - Area of 3 sectors

= 17320.5 - 15700 cm²= 1620.5 cm²

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Answer

Number of circular design = 9

Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm

Area of the square = 42 × 42 cm² = 1764 cm²

Area of the circle = π r²= 22/7 × 7 × 7 = 154 cm²

Total area of the design = 9× 154 = 1386 cm²

Area of the remaining portion of the handkerchief = Area of the square - Total area of the design

= 1764 - 1386 = 378 cm²

12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB, (ii) shaded region.

Answer

Radius of the quadrant = 3.5 cm = 7/2 cm

(i) Area of quadrant OACB = (πR²)/4 cm²

= (22/7 × 7/2 × 7/2)/4 cm²

= 77/8 cm²

(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm²

= 7/2 cm²

Area of shaded region = Area of quadrant - Area of triangle BOD

= (77/8 - 7/2) cm²= 49/8 cm²

= 6.125 cm²

13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Answer

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB ,

OB²= AB²+ OA²

⇒ OB²= 20²+ 20²

⇒ OB²= 400+ 400
⇒ OB²= 800

⇒ OB= 20√2 cm
Area of the quadrant = (πR²)/4 cm²= 3.14/4 × (20√2)²cm²= 628cm²

Area of the square = 20 × 20 = 400 cm²

Area of the shaded region = Area of the quadrant - Area of the square

= 628 - 400 cm²= 228cm²

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.

Answer

Radius of the larger circle, R = 21 cm

Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30°

Area of the larger sector = (30°/360°) × π R²cm²

= 1/12 × 22/7 × 21²cm²

=231/2cm²

Area of the smaller circle = (30°/360°) × π r²cm²

= 1/12 × 22/7 × 7²cm²

=77/6cm²

Area of the shaded region = 231/2 - 77/6 cm²

= 616/6 cm² = 308/3cm²

15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC

as diameter. Find the area of the shaded region.

Answer

Radius of the the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC²= AB²+ AC²

⇒ BC²= 14²+ 14²

⇒ BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm²
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm²

Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm²

Area of the shaded region =Area of the semicircle + Area of ΔABC - Area of quadrant

= 154 + 98 - 154 cm²= 98cm²

Page No: 238

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

Answer

AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm²

Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 8²

= 352/7 cm²

Area of shaded region = (Area of quadrant AECB - Area of ΔABC) + (Area of quadrant AFCD - Area of ΔADC)

= (352/7 - 32) + (352/7 -32) cm²

= 2 × (352/7 -32) cm²

= 256/7 cm²
Go Back to NCERT Solutions for Class 10 Maths

NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

Chapter 12 Areas related to Circles is an important chapter that will help you in fetching more marks in the examinations. There are total 4 topics in the chapter which are important for improving your knowledge and for board exam purpose.

• Introduction: In our daily lives, we come across things that are circular in shape such as wheel of bicycles, cake and various others. Through this chapter, we will get to know about perimeter and area of circle and and will also learn terms like segment and sector.

• Perimeter and Area of a Circle — A Review: The distance covered by travelling once around a circle is its perimeter which is usually called its circumference. In the exercise, we are focused on finding radius and area of circles.

• Areas of Sector and Segment of a Circle: The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. We will be finding the areas of sector of circle.

• Areas of Combinations of Plane Figures: We are focused on finding areas of sector of circles with plane figures which is based on daily lives things like window designs, designs on table covers etc.

Other than NCERT Solutions, you can also find our study materials for Class 10:

CBSE Notes for Class 10 Maths Chapter 12 Areas Related to Circles

MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles

There are total 3 exercises in the chapter 12 Class 10 Maths which will help you in knowing application of formulas. Also, you can find exercisewise NCERT Solutions for Class 10 Maths Chapter 12 Areas related to Circles.

NCERT Solutions for Class 10 Maths Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry

Chapter 9 Some Applications of Trigonometry

Chapter 10 Circles

Chapter 11 Constructions

Chapter 13 Surface Areas And Volumes

Chapter 14 Statistics

Chapter 15 Probability

FAQ on Chapter 12 Areas related to Circles

How many exercises in Chapter 12 Areas related to Circles

There are 3 exercise in the Chapter 12 Areas related to Circles which will encourage students to learn topics. These NCERT Solutions are prepared by Studyrankers experts who have vast experience in teaching students.

What is Area of a circle?

Area of a circle = πr2

What is Perimeter of semicircle:?

Perimeter of a semicircle or protractor = πr+2r

What is Circumference of a circle or Perimeter of a circle?

The distance around the circle or the length of a circle is called its circumference or perimeter.

Circumference (perimeter) of a circle = πd or 2πr, where r is the radius of the circle and π=22/7 or 3.14.

↧

NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes

October 18, 2020, 10:15 am

≫ Next: MCQ Questions for Class 11 History: Ch 5 Nomadic Empires

≪ Previous: NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes

In this page, you will find Class 10 Maths NCERT Solutions Chapter 13 Surface Areas and Volumes which are very helpful in knowing the important topics in the chapter and completing your homework in no time. These NCERT Solutions for Class 10 Maths will give you step by step solutions of every questions. Our experts have updated these solutions according to the latest pattern of CBSE.

Page No: 244

Exercise 13.1

Unless stated otherwise, take π = 22/7.

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer

Volume of each cube(a³) = 64 cm³
⇒ a³ = 64 cm³
⇒ a = 4 cm
Side of the cube = 4 cm
Length of the resulting cuboid = 4 cm
Breadth of the resulting cuboid = 4 cm
Height of the resulting cuboid = 8 cm
∴ Surface area of the cuboid = 2(lb + bh + lh)
= 2(8×4 + 4×4 + 4×8) cm²
= 2(32 + 16 + 32) cm²
= (2 × 80) cm² = 160 cm²

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer

Diameter of the hemisphere = 14 cm
Radius of the hemisphere(r) = 7 cm
Height of the cylinder(h) = 13 - 7 = 6 cm
Also, radius of the hollow hemisphere = 7 cm
Inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part
= (2πrh+2πr²) cm²= 2πr(h+r) cm²
= 2 × 22/7 × 7 (6+7) cm²= 572 cm²

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer

Radius of the cone and the hemisphere(r) = 3.5 = 7/2 cm
Total height of the toy = 15.5 cm
Height of the cone(h) = 15.5 - 3.5 = 12 cm

Curved Surface Area of cone = πrl = 22/7 × 7/2 × 25/2
= 275/2 cm²

Curved Surface Area of hemisphere = 2πr²
= 2 × 22/7 × (7/2)²
= 77 cm²

Total surface area of the toy = CSA of cone + CSA of hemisphere
= (275/2 + 77) cm²
= (275+154)/2 cm²
= 429/2 cm² = 214.5cm²
The total surface area of the toy is 214.5cm²

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer

Each side of cube = 7 cm
∴ Radius of the hemisphere = 7/2 cm
Total surface area of solid = Surface area of cubical block + CSA of hemisphere - Area of base of hemisphere

TSA of solid = 6×(side)²+ 2πr²-πr²
= 6×(side)²+ πr²
= 6×(7)²+ (22/7 × 7/2 × 7/2)

= (6×49) + (77/2)
= 294 + 38.5 = 332.5 cm²

The surface area of the solid is 332.5 cm²

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer

Diameter of hemisphere = Edge of cube = l
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cube + CSA of hemisphere - Area of base of hemisphere
TSA of remaining solid = 6 (edge)²+ 2πr²- πr²
= 6l²+ πr²
= 6l²+ π(l/2)²
= 6l²+ πl²/4
= l²/4(24 + π) sq units

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Answer

Two hemisphere and one cylinder are given in the figure.

Diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Length of the capsule = 14 mm

∴ Length of the cylinder = 14 - (2.5 + 2.5) = 9mm

Surface area of a hemisphere = 2πr² = 2 × 22/7 × 2.5 × 2.5

= 275/7 mm²

Surface area of the cylinder = 2πrh
= 2 × 22/7 × 2.5 x 9
= 22/7 × 45
990/7 mm²

∴ Required surface area of medicine capsule
= 2 × Surface area of hemisphere + Surface area of cylinder
= (2 × 275/7) × 990/7
= 550/7 + 990/7
= 1540/7 = 220 mm²

Page No. 245

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)

Answer

Tent is combination of cylinder and cone.

Diameter = 4 m
Slant height of the cone (l) = 2.8 m
Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
Height of the cylinder (h) = 2.1 m

∴ Required surface area of tent = Surface area of cone+Surface area of cylinder
= πrl + 2πrh
= πr(l+2h)
= 22/7 × 2 (2.8 + 2×2.1)
= 44/7 (2.8 + 4.2)
= 44/7 × 7 = 44 m²

Cost of the canvas of the tent at the rate of ₹500 per m²
= Surface area × cost per m²
= 44 × 500 = ₹22000

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm².

Answer

Diameter of cylinder = diameter of conical cavity = 1.4 cm

∴ Radius of cylinder = Radius of conical cavity = 1.4/2 = 0.7

Height of cylinder = Height of conical cavity = 2.4 cm

TSA of remaining solid = Surface area of conical cavity+TSA of cylinder

= πrl + (2πrh + πr²)

= πr (l + 2h + r)

= 22/7 × 0.7 (2.5 + 4.8 + 0.7)

= 2.2 × 8 = 17.6 cm²

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

Answer

Given,

Height of the cylinder, h = 10cm and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm

Now, required total surface area of the article = 2 × Surface area of hemisphere + Lateral surface area of cylinder

Exercise 13.2

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Answer

Given, solid is a combination of a cone and a hemisphere.

Also, we have radius of the cone (r) = Radius of the hemisphere = 1cm and height of the cone (h) = 1cm

∴ Required volume of the solid = Volume of the cone + Volume of the hemisphere

2. Rachel, an engineering, student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel mode. (Assume the outer and inner dimensions of the model to be nearly the same.)

Answer

Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model,BC = ED = 3 cm.

3. A gulab jamun, contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Answer

Let r be the radius of the hemisphere and cylinder both. h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder.

Given, length = 5 cm, diameter = 2.8 cm

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Answer

Given, length of cuboid (l) = 15 cm

Breadth of cuboid (b) = 10cm

and height of cuboid (h) = 3.5 cm

∴ Volume of cuboid = l×b×h = 15 ×10 × 3.5 = 525³

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)

Answer

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be

345 cm³. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

Answer

She is not correct.

Exercise 13.3

1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Answer

Given, the radius of the sphere (r) = 4.2 cm

Radius of the cylinder (r1) = 6 cm

2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer

Let r1, r2 and r3 be the radius of given three spheres and R be the radius of a single solid sphere.

Given, r1 = 6 cm , r2 = 8 cm and r3 =10 cm

3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Answer

Given, the height of deep well which form a cylinder (h1) = 20 m

4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.

Answer

Given, the height of deep well which form a cylinder (h1) = 14 m

5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer

Let the height and radius of ice cream container (cylinder) be h1 and r1.

6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 10 3 5 . . cm cm cm × ?

Answer

We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1 respectively.

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer

Let the radius and slant height of the heap of sand are r and l.

Given, the height of the heap of sand h = 24 cm.

8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh⁻¹. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Answer

Given, speed of flow of water (l) = 10 kmh-1 = 10 × 1000 mh⁻¹

Area of canal = 6 × 1.5 = 9m²

9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kmh⁻¹, in how much time will the tank be filled?

Answer

Given, speed of flow of water = 3 kmh⁻¹ = 3×1000 mh⁻¹

∴ Length of water in 1 h = 3000 m

Exercise 13.4

1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Answer

Let the height of the frustum of a cone be h.

2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Answer

Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2.

3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Answer

Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container. If it costs ₹8 per 100 cm². (Take π = 3.14)

Answer

Let h be the height of the container, which is in the form of a frustum of a cone whose lower end is closed and upper end is opened. Also, let the radius of its lower end be r1 and upper end be r2.

5. A metallic right circular cone 20 cm high and whose vertical angle is 60º is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Answer

Let r1 and r2 be the radii of the frustum of upper and lower ends cut by a plane. Given, height of the cone = 20 cm.

∴ Height of the frustum = 10 cm

Exercise 13.5 (Optional)

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.

Answer

Since, the diameter of the wire is 3 mm.

When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.

Given, length of the cylinder = 12 cm = 120 mm

∴ Number of rounds to cover 120 mm = 120/3 = 40

Given, diameter of a cylinder is d = 10 cm

∴ Radius, r = 10/2 = 5 cm

∴ Length of wire required to complete one round = 2πr = 2π(5) = 10π cm

∴ Length of the wire in covering the whole surface

= Length of the wire in completing 40 rounds

= 10π × 40 = 400π cm

= 400 × 3.14 = 1256 cm

Now, radius of copper wire = 3/2 mm = 3/20 cm

2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).

Answer

Here, ABC is a right angled triangle at A and BC is the hypotenuse.

3. A cistern, internally measuring 150 cm × 120 cm× 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. how many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Answer

Given, internally dimensions of cistern = 150 cm×120 cm×110 cm

∴ Volume of cistern = 150×120×110

= 1980000 cm³

Volume of water = 129600 cm³

∴ Volume of cistern to be filled = 1980000 - 129600 = 1850400 cm³

Let required number of bricks = n

4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer

Given, area of the valley = 97280 km²

5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Answer

Given, oil funnel is a combination of a cylinder and a frustum of a cone.

Also, given height of cylindrical portion h = 10 cm

6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. Using the symbols as explained.

Answer

Leth be the height, l be the slant height and r₁ and r₂ be the radii of the bases (r₁>r₂) of the frustum of a cone. We complete the conical part OCD.

The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l₁ and its height be h₁ i.e.,OB = OA = l₁ and OP = h₁

Then in ∆ACE,

7. Derive the formula for the volume of the frustum of a cone given to you in the section 13.5 using the symbols as explained.

Answer

Leth be the height, l be the slant height and r₁ and r₂ be the radii of the bases (r₁>r₂) of the frustum of a cone. We complete the conical part OCD.

The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h₁ and its slant height be l₁.

i.e., OP = h₁ and OA = OB = l₁

Then, height of the cone OCD = h₁ −1

∆OQD ~ ∆OPB (AA similarity criterion)

Go Back To NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes

Chapter 13 Class 10 Maths NCERT Solutions plays very important role during the preparation of board exams as a lot of questions from this topic can be asked. There are total 5 topics in this chapter which will guide students in a better way.

• Introduction: We came across various types of solids made up of combinations of two or more of the basic solids such as water tank in our daily lives. We are going to learn about how to find surface areas and volumes of these types of objects.

• Surface Area of a Combination of Solids: We can breakdown complex figure into simpler one and can find their areas. The total surface area of the new solid is the sum of the curved surface areas of each of the individual parts. TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ respectively.

• Volume of a Combination of Solids: We will discuss how to calculate volumes of solids. The volume of the solid formed by joining two basic solids will be the sum of the volumes of the constituents.

• Conversion of Solid from One Shape to Another: We will learn to find the radius and volume when one shape got converted from one to another, or when a liquid which originally filled one container of a particular shape is poured into another container of a different shape or size.

• Frustum of a Cone: When we cut through a cone with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part that is now left over on the other side of the plane is called a frustum of the cone. In this topics, we will dealing with the questions based on finding the volume of a frustum of a cone.

If you want complete understanding of the chapter, you can see other Study Materials for Class 10 by clicking the links below.

CBSE Notes for Class 10 Maths Chapter 13 Surface Areas and Volumes

MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

If you want to find exercisewise NCERT Solutions of Chapter 13 Surface Areas and Volumes the you can find them below.

NCERT Solutions for Class 10 Maths Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry

Chapter 9 Some Applications of Trigonometry

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Areas related to Circles

Chapter 14 Statistics

Chapter 15 Probability

FAQ on Chapter 13 Surface Areas and Volumes

How many exercises in Chapter 13 Surface Areas and Volumes

There are total 5 exercise in the Chapter 13 Surface Areas and Volumes which will make students while solving any questions. These are also useful for competitive exams and higher grades.

What do you mean by Cube?

Cube is a special type of cuboid in which length = breadth = height which is also called an edge of a cube.

What is lateral surface area of cuboid?

Lateral surface area of cuboid = 2 (1 + b)h

If two identical solid cubes of side ‘x’ are joined end to end, then the total surface area of the resulting cuboid is 12x2. Is it true?

∵ The total surface area of a cube of side x is 6x²

When they are joined end to end, the length bcomes 2x

∴Total surface area

= 2[lh + bh + hl]

= 2 [(2x · x) + (x · x) + (2x · x)]

= 2 [2x² + x² + 2x²]

= 2 [5x²] = 10x²≠ 12x²

∴ False

↧

MCQ Questions for Class 11 History: Ch 5 Nomadic Empires

October 18, 2020, 9:47 pm

≫ Next: MCQ Questions for Class 11 History: Ch 6 The Three Orders

≪ Previous: NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes

MCQ Questions for Class 11 History: Ch 5 Nomadic Empires

1. During Genghis Khan’s attempt to conquer China, North China was ruled by

(a) Hsi Hsia.

(b) Jurchen.

(d) Ogodei.

► (a) Hsi Hsia.

2. Timur, a Barlas Turk who rose to power between 1370-1405 CE, claimed descent from Genghis Khan, through the lineage of:

(a) Chaghtay

(b) Toluy

(d) Ogedei

► (a) Chaghtay

3. Through the 1180s and 1190s, Temujin (Genghis Khan) remained an ally of

(a) Boghurchu

(b) Jamuqa

(d) Ong Khan

► (d) Ong Khan

4. What was known as the ‘yam’?

(a) Courier system

(b) Army system

(d) Administration system

► (a) Courier system

5. In 3rd century BCE in China, the fortifications started to be integrated into a common defensive outwork, known as the

(a) ‘Great Wall of China’.

(b) ‘Humen Weiyuan Fort of China’.

(d) 'Wanping Castle of China’.

► (a) ‘Great Wall of China’.

6. The childhood name of Genghis Khan was:

(a) Morin Khuur

(b) Nominjin

(d) Tamujin

► (c) Temujin

7. Genghis Khan died in

(a) 1224 CE.

(b) 1225 CE.

(d) 1227 CE.

► (d) 1227 CE.

8. The army of Genghis Khan was organised into

(a) clan.

(b) decimal units.

(d) tribal groups.

► (b) decimal units.

9. The Mongolian and Chinese narratives on Genghis Khan were translated as The Secret History of the Mongols by:

(a) Boris Yakovlevich Vladimirtsov

(b) Igor de Rachewiltz

(d) Vasily Vladimirovich Bartold

► (b) Igor de Rachewiltz

10. The Mongols travelled with their herds to pasture lands and lived in tents called:

(a) ‘Anda’

(b) ‘Gers’

(d) ‘Yasa’

► (b) ‘Gers’

11. Why was the ‘Great Wall of China’built?

(a) To store grains

(b) For protection of China

(d) None of the above

► (b) For protection of China

12. The Mongols imported _______ from China.

(a) Cotton

(b) Crude Oil

(d) Iron Utensils

► (d) Iron Utensils

13. Genghis Khan' mother name was:

(a) Oelun-leke

(b) Oelun-eke

(d) None of the above

► (b) Oelun-eke

14. Why Genghis Khan commanded to plunder and destroy Nishapur.

(a) Because a Mongol prince was killed.

(b) Because Nishapur refused to export agricultural produce.

(d) None of the above

► (a) Because a Mongol prince was killed.

15. Why were the Mongols defeated by the Egyptian army?

(a) Because Mongol rulers began to take more interest in China.

(b) Because Mongol army were less equipped

(d) None of the above

► (a) Because Mongol rulers began to take more interest in China.

↧