Quantcast
Channel: Study Rankers
Viewing all 6190 articles
Browse latest View live
↧

NCERT Solutions for Class 8th: Ch 10 Visualising Solid Shapes Geometry

February 2, 2020, 11:57 am
$
0
0

NCERT Solutions for Chapter 10 Visualising Solid Shapes Class 8 Mathematics

Page No: 157

Exercise 10.1

1. For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.


Answer

(a) → (iii) → (iv)
(b) → (i) → (v)
(c) → (iv) → (ii)
(d) → (v) → (iii)
(e) → (ii) → (i)

2. For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.



Answer

(a) (i) → Front
(ii) → Side
(iii) → Top view

(b) (i) → Side
(ii) → Front
(iii) → Top view

(c) (i) → Front
(ii) → Side
(iii) → Top view

(d) (i) → Front
(ii) → Side
(iii) → Top view

3. For each given solid, identify the top view, front view and side view.


Answer

(a) (i) → Top view 
(ii) → Front view
(iii) → Side view

(b) (i) → Side view
(ii) → Front view
(iii) → Top view

(c) (i) → Top view
(ii) → Side view
(iii) → Front view

(d) (i) → Side view
(ii) → Front view
(iii) → Top view

(e) (i) → Front view
(ii) → Top view
(iii) → Side view

4. Draw the front view, side view and top view of the given objects:

Answer




Page No. 163

Exercise 10.2

1. Look at the given map of a city.
Answer the following.
(a) Colour the map as follows: Blue-water, red-fire station, orange-library, yellow
- schools, Green - park, Pink - College, Purple - Hospital, Brown - Cemetery.
(b) Mark a green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green ‘Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?

Answer

(a) The required map in colour is as follow:

(b) The marks can be put at the given point is indicated in the map in green X:


(c) The shortest route from library to the bus stop is indicated in the map in red colour:

(d) Between the market and the city park, the city park is further east.
(e) Between the primary school or the Sr. Secondary School, the Sr. Secondary School is further south.

Page No. 166

Exercise 10.3

1. Can a polygon have for its faces:
(i) 3 triangles
(ii) 4 triangles
(iii) a square and four triangles

Answer

(i) No, a polyhedron cannot have 3 triangles for its faces.
(ii) Yes, a polyhedron can have four triangles which is known as pyramid on triangular base.
(iii) Yes, a polyhedron has its faces a square and four triangles which makes a pyramid on square base.

2. Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid)

Answer

It is possible, only if the number of faces are greater than or equal to 4.

3. Which are prisms among the following:


Answer

Figure (ii) unsharpened pencil and figure (iv) a box are prisms.

4. (i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?

Answer

(i) A prism becomes a cylinder as the number of sides of its base becomes larger and larger.
(ii) A pyramid becomes a cone as the number of sides of its base becomes larger and larger.

5. Is a square prism same as a cube? Explain.

Answer

Yes, a square prism is same as a cube, it can also be called a cuboid. A cube and a square prism are both special types of a rectangular prism. A square is just a special type of rectangle! Cubes are rectangular prisms where all three dimensions (length, width and height) have the same measurement.

6. Verify Euler’s formula for these solids.


Answer

(i) Here, figure (i) contains 7 faces, 10 vertices and 15 edges.
Using Eucler’s formula, we see
 F + V – E = 2
Putting F = 7, V = 10 and E = 15,
 F + V – E = 2
⇒ 7 + 10 – 15 = 2
⇒ 17 – 15 = 2
⇒ 2 = 2
⇒ L.H.S. = R.H.S. Hence Eucler’s formula verified.

(ii) Here, figure (ii) contains 9 faces, 9 vertices and 16 edges.
Using Eucler’s formula, we see
F + V – E = 2
 F + V – E = 2
 9 + 9 – 16 = 2
⇒ 18 – 16 = 2
⇒ 2 = 2
⇒ L.H.S. = R.H.S.
Hence Eucler’s formula verified.

7. Using Euler’s formula, find the unknown:
Faces?520
Vertices6?12
Edges129?

Answer

In first column, F = ?, V = 6 and E = 12
Using Eucler’s formula, we see
F + V – E = 2
 F + V – E = 2
⇒ F + 6 – 12 = 2
⇒ F – 6 = 2
⇒ F = 2 + 6 = 8
Hence there are 8 faces.
In second column, F = 5, V = ? and E = 9
Using Eucler’s formula, we see
F + V – E = 2
F + V – E = 2
⇒ 5 + V – 9 = 2
⇒ V – 4 = 2
⇒ V = 2 + 4 = 6
Hence there are 6 vertices.
In third column, F = 20, V = 12 and E = ?
Using Eucler’s formula, we see
F + V – E = 2
F + V – E = 2
⇒ 20 + 12 – E = 2
⇒ 32 – E = 2
⇒ E = 32 – 2 = 30
Hence there are 30 edges.

8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?

Answer

If F = 10, V = 15 and E = 20.
Then, we know Using Eucler’s formula,
F + V – E = 2
L.H.S. = F + V – E
= 10 + 15 – 20
= 25 – 20
= 5
R.H.S. = 2
∵ L.H.S. ≠ R.H.S.
Therefore, it does not follow Eucler’s formula.
So polyhedron cannot have 10 faces, 20 edges and 15 vertices.

↧
Search

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.3

February 2, 2020, 11:23 pm
$
0
0

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.3

Chapter 9 Sequence and Series Exercise 9.3 NCERT Solutions for Class 11 Maths is very useful for completing your hometask as these Class 11 Maths NCERT Solutions prepared by Studyrankers experts are detailed and accurate. NCERT Solutions are best way through which you can look around the chapter easily and cover all the important topics.

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.3

1. Find the 20th and n th terms of the G.P. 5/2, 5/4, 5/8, ...........

Answer

Tn = arn-1, we have, a = 5/2, r = 5/4÷ 5/2 = 1/2;
T20 = 5/2(1/2)20 – 1 = 5/2 . 1/219 = 5/220
Tn = 5/2 (1/2)n -1 = 5/2 . 1/2n-1 = 5/2n

2. Find the 12th term of a G..P. whose 8th term is 192 and the common ratio is 2.

Answer

Let a be the first term and r be the common ratio of a G.P.
Then T8 = ar8–1 = ar7 = 192 … (i)
Now T12 = ar12 – 1 = ar11 = (ar7 )r 4
= 192 × 24 [ ∵ r = 2]
= 192 × 16 = 3072

3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q 2 = ps.

Answer

Let a be the first term and r the common ratio of G.P. then
T5 = p => ar4 = p … (i)
T8 = q => ar7 = q … (ii)
T11 = s => ar10 = s … (iii)
Now, q2 = (ar7)2 [Using (ii)]
=> q2 = a2r14
=> q2 = (ar4)(ar10)
=> q2 = ps [Using (i) and (iii)]

4. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7 th term.

Answer

Let a be the first term and r the common ratio of G.P. then
T4 of G.P. = (T2 of G.P.)2 and a = – 3
ar3 = (ar) 2
=> ar3 = a2r2
=> r = a = -3
Now, T7 = ar6 = (–3)(–3)6 = (–3)7 = –2187.

5. Which term of the following sequences:
(a) 2, 2√2, 4.... is 128?
(b) √3, 3, 3√3,.... is 729?
(c) 1/3, 1/9, 1/27 …….. is 1/19683?

Answer

(a) The G.P, is 2, 2√2, 4, .....
The first term = a = 2,
The common ration r = √2
n th term = arn – 1 = 128
or 2.(√2)n -1 = 128 or 2 n-1/2 = 64 = 26
=> (n – 1)/2 = 6 or n -1 = 12 or n = 13
(b) (b) Here a= √3, r = √3. Let Tn = 729
=> arn-1 = 729
=> (√3)(√3)n-1 = 729
=> (√3)n = (9)3
=> 3n/2 = (32)3 = 36
=> n/2 = 6 => n = 12
Hence, the 12th term is 729
(c) (c) a = 1/3, r = (1/9)/(1/3) = 1/9 × 3/1 = 1/3
let Tn = 1/19683 => arn-1 = 1/19683
=> 1/3(1/3)n-1 = 1/19683
=> (1/3)n = (1/3)9 => n = 9

6. For what values of x, the numbers –(2/7), x, -(2/7) are in G.P?

Answer

The numbers -2/7, x, -7/2 will be in GP

If x/-(2/7) = -7/(2/x) => x2 = -(7/2) × –(2/7) = 1 => x = ±1

7. Find the sum to indicated number of terms in each of the geometric progressions is 0.15, 0.015, 0.0015, .......... 20 terms.

Answer

We have a = 0.15, r = .015/.15 = 0.1 < 1, n = 20
Sn = (a(1 –rn))/(1 – r), r < 1;
S20 = (0.15[1 – (0.1)20])/1 = 0.1 = (0.15[1 – (0.1)20])/0.9
= 1/6 [1 – (0.1)20]

8. Find the sum to indicated number of terms in the following geometric progression:
√7, √21, 3√7,....... n terms.

Answer

Here a = √7
R = √21/√7 = √(21/7) = √3 > 1, n = n
Sn = (a(rn – 1))/(r – 1) = (√7[(√3)n – 1])/(√3 – 1)
= (√7(√3 + 1)[(√3)n – 1])/((√3 – 1)(√3 + 1))
= (√7(√3 + 1)[(√3)n – 1])/2

9. Find the sum to indicated number of terms in the following geometric progression: 1, – a, a2, – a3, ....... n terms (if a ≠ – 1)

Answer

G.P. is 1, – a, a2, – a 3 , .......
Now first term A = 1, r = – a
Sn = (a(1 – rn))/(1 –r)
= (1[1-(-a)n]/(1-(-a)) = (1-(-a)n)/(1 + a)

10. Find the sum to indicated number of terms in the following geometric progression x 3, x 5, x 7, ..... n terms (if x ≠ ± 1).

Answer

Here a = x3, r= x5/x3 = x2, x ≠ ± 1
Sn = (a(1 – rn))/(1 – r)
Sn = (x3(1 – x2n)/(1 – x2)

11.


Answer



12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Answer

Let the first three terms of a G.P. be a/r, a, ar.
Product of three terms = a/r × a × ar = 1
a3 = 1 => a = 1
Sum of these term1/r + 1 + r = 39/10 (put a = 1)
Multiplying by 10r; 10 + 10r + 10r2 = 39r
=> 10r2 + 10r – 39r + 10 = 0
=> 10r2 – 29r + 10 = 0
=> (2r – 5)(5r – 2) = 0 => r = 5/2 or 2/5
When r = 5/2 the term of G.P. are 2/5, 1, 5/2
When r = 2/5, the terms of G.P. are 5/2, 1, 2/5.

13. How many terms of G.P. 3, 32, 33, .......... are needed to give the sum 120?

Answer

Let n be the number of terms of the G.P. 3, 32, 3 3, .......... makes the sum = 120
We have a =3, r = 3
S = (a(rn -1))/(r – 1), r > 1;
Sum = (3(3n – 1))/(3 – 1) = 120
Or 3/2 (3n – 1) = 120
Multiplying both sides by 3/2
∴ 3n – 1 = 80
∴ 3n = 80 + 1= 81 = 34 => n = 4
∴ Required number of terms of given G. P. is 4.

14. The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer

Let a be the first term and r be the common ratio.
S3 = 16
(a(1 – r3))/(1- r) = 16 …….(i)
S6 – S3 = 128 => (a(1 – r6))/ (1 – r) – 16 = 128,
i.e., (a(1 –r6))/(1 – r) = 144 ….(ii)
(ii) ÷ (i) gives (1 – r6)/(1 – r3) = 144/16 = 9/1
=> ((1 – r3)(1 + r3))/(1 – r3) = 9/1 => 1 + r3 = 9
=> r3 = 8 => r3 = 23 => r = 2
Thus, common ratio = 2
As, r = 2, S3 = (a(r3 – 1))/(r – 1) = 16
=> (a(23 – 1)/(2 – 1) = 16 => (a(8 – 1))/1 = 16
=> a(7) = 16 => a = 16/7
Sn = (a(rn – 1))/r – 1 = 16/7 . (2n – 1)/(2 – 1) = 16/7 .(2n – 1).

15. Given a G.P. with a = 729 and 7th term 64, determine S7

Answer

a = 729, T7 = 64.Let common ratio = r
=> ar6 = T7 = 64 => 729 r 6 = 64 => r 6 = 64/729
=> r6 = (2/3)6 => r = 2/3 < 1
Sn = (a(1 – rn))/(1 – r)
S7 = (729[1 – (2/3)7])/1 – 2/3
= 729 × (3(37– 27))/37
= (36.3(37 – 37)/37 = 37 – 27 = 2187 – 128 = 2059.

16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Answer

Let a be the first term and r be the common ratio.
Also, S2 = –4, T5 = 4T3
=> (a(1 – r2))/(1 – r) = -4 => a(1 + r) = -4
=> a( 1 – r) = -4, ar4 = 4ar2 => r2 = 4
=> r = 2
Now, when r = 2 : a (1 + 2) = -4 => a = -4/3
Thus, the G.P. is -4/3 , -8/3, -16/3 ,………..
When r = -2 : a(1 – 2) = -4 => a(-1) = -4
=> a = 4
Thus, the G.P. is 4, -8, 16, -32, 64, ……….

17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Answer

Let a be the first term and r be the common ratio.
T4 = x => ar3 = x … (i)
T10 = y => ar9 = y … (ii)
T16 = z = ar15 = z …(iii)
Now, x, y, z will be in G.P.
If ar3, ar9, ar15 are in G.P.
i.e., ar9/ar3 = ar15/ar9 => r6 = r6, which is true.

18. Find the sum of n terms of the sequence 8, 88, 888, 8888, .........

Answer

Let S be the sum of n-terms of the series,
8 + 88 + 888 + 8888 + .......
S = 8 + 88 + 888 + 8888 + ............... to n terms
= 8(1 + 11 + 111 + 1111 + ....... to n terms)
= 8/9 (9 + 99 + 999 + 9999 + ......... to n terms)
= 8/9 [(10 – 1) + (102 – 1) + (103 – 1) + …………… to n terms]
= 8/9 [(10 + 102 + 103 + …………….. to n terms ) – (1 + 1+ 1 ……….. to n terms)]
= 8/9 [(10(10n – 1))/(10 – 1) - n]
S = 8/9 [10/9 (10n – 1) – n].

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

Answer

Sum of product of corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 is
S = 2.128 + 4.32 + 8.8 + 16.2 + 32 where S denotes the sum of these term.
S = 256 + 128 + 64 + 32 + 16 =
It is a G.P
a = 256, r = 1/2,
S = (a(1 –r5))/(1- r) = 256 × 2 × (1 – 1/32)
= 256 × 2 × 31/32 = 16 × 31 = 496.

20. Show that the products of the corresponding terms of the sequences a, ar, ar2 .......... arn – 1 and A, AR, AR2, ...... ARn – 1 form a G. P, and find the common ratio.

Answer

The two sequences are a, ar, ar2 ........ arn – 1 and A, AR, AR2, ..........ARn – 1
∴ Sum of the products of the corresponding term of these sequence
aA + aA(rR)+ aA(r2R2) + .......+ a Arn– 1R n – 1
= aA + aA (rR) + aA (rR) 2 + ........ + aA(rR) n– 1
This is a G.P. whose first term is aA and the common ratio = aA(rR) ÷ aA = rR

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Answer

Let a be the first term and r the common ratio
Tn =arn –1, T2 = ar, T3 = ar2 and T4 = ar3
Since third term is greater than the first by 9
T3 = T1 + 9 => ar2 = a + 9 …(i)
Also the second term is greater than the fourth by 18
∴ T2 = T4 + 18 => ar = ar3 + 18 …(ii)
Multiplying (i) by r, we get,
ar3 = ar + 9r …,(iii)
from (ii) & (iii)
ar = ar + 9r + 18 => 0 = 9r + 18
=> r = -18/9 = -2
Putting r = -2 in (i)
a(-2)2 = a + 9 => 4a = a + 9 => 3a = 9
=> a = 3
Now, T2 =ar = 3(-2) = - 6;
T3 = ar2 = 3(-2)2 = 12;
T4 = ar3 = 3(-2)3 = -24
∴ The required terms are 3, -6, 12 and -24.

22. If the p th, q th and r th terms of a G..P. are a, b, and c respectively. Prove that
aq-r br-p cp-q = 1.

Answer

Let A be the first term and R the common ratio of G.P.
Tp = a => ARp – 1= a …(i)
Tq = b => ARq – 1= b …(ii)
Tr = c => ARr – 1= c …(iii)
aq – r. br – p. cp – q
= (ARp – 1) q – r. (ARq – 1) r – p. (ARr – 1) p – q
= A q – r + r – p + p – q × R pq – pr – q + r + qr – pq –r + p + pr – qr – p + q
=A0R0 = 1. 1 = 1.

23. If the first and the n th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n

Answer

Let a be the first term and r the common ratio of a G.P.
T1 = a and Tn = b
arn – 1 = b … (i)
Also, P = T1. T2. T3. .............Tn
= a. ar. ar2, ..........arn – 1
= a n. r 1 + 2 + 3 + .........+ (n –1)
Since, 1 + 2 + 3 + .........+ (n – 1) = (n(n -1))/2
∴ P = an . rn(n – 1)/2
∴ P2 = a2n . rn(n – 1) = [a2 rn-1]n
= [a. arn -1]n = (a . b)n [using (i)]
Hence, P2 = (ab)n

24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n) th term is 1/rn

Answer

Sum of the first n terms of a G.P. = a + ar + ar2 + ..............+ arn – 1 = (a(1- rn))/(1 – r) ………(i)
 Tn + 1 = arn, Tn + 2 = arn +1, ..........T2n = ar2n – 1
∴ Sum of terms from (n + 1)th to (2n) th terms
=Tn + 1 + Tn + 2 + Tn + 3 + ..........+ T2n
= arn + arn + 1 + arn + 2 + .........+ar2n – 1
= (arn(1 – rn))/(1 – r)…(ii)
No. of terms = n
Dividing (i) by (ii),
(sum of first n terms of a G.P.)/(Sum of terms from (n + 1)th to (2n)th term)
= 

25. If a, b, c and d are in G.P. show that
(a 2 + b 2 + c 2) (b 2 + c 2 + d 2) = (ab + bc + cd) 2

Answer

Let r be the common ratio of the G.P a, b, c, d.
Then b = ar, c = ar2 and d = ar3
LHS = (a 2 + b 2 + c 2) (b 2 + c 2 + d 2)
= (a 2 + a 2 r 2 + a 2 r 4) (a 2 r 2 + a 2 r 4 + a 2 r 6)
= a 4 r 2 (1 + r 2 + r 4) 2
RHS = (ab + bc +cd) 2 = (a 2 r + a 2 r 3 + a 2 r 5) = a 4 r 2 (1 + r 2 + r 4) 2
Hence, (a 2 + b 2 + c 2) (b 2 + c 2 + d 2)
= (ab + bc + cd)2

26. Insert two number between 3 an 81 so that the resulting sequence is G.P.

Answer

Let G1, G2 be the two numbers such that 3, G1, G2, 81 are in G..P.
Let r be the common ratio,
T4 = arn – 1 = 81 or 3. r 3 = 81
∴ r3 = 81/3 = 27 = 33∴ r = 3
G1 = ar = 3 × 3 = 9; G2 = ar2 = 3 × 32 = 27
9, 27 are the required number.

27. Find the value of n so that (an +1 + bn+1)/(an + bn) may be the geometric mean between a and b.

Answer

The G.M. between a and b = √ab
(an + 1 + b n+1)/(an + bn) = √ab = a1/2b1/2
Cross – multiplying, we get
an+1 + bn+1 = an + 1/2 b1/2 + a1/2 b n+1/2
cross – multiplying, we get
an+1 + bn+1 = a n + 1/2 b1/2 + a1/2 b n + 1/2
=> a n+ 1 – an+1b1/2 = a1/2 bn + 1/2 – bn+ 1
=> a n + 1/2 (a1/2 – b1/2) = b n + 1/2 (a1/2 – b1/2)
Since a and b are different.
Cancelling (a1/2 – b1/2) from both sides, we get
a n + 1/2 = b n + 1/2 => (a n + 1/2)/(b n + 1/2) = 1
=> (a/b)n + 1/2 = 1 = (a/b)0
n + 1/2 = 0 => n = -(1/2).

28. The sum of two numbers is 6 times their geometric means. Show that numbers are in the ratio (3 + 2√2):(3 - 2√2).

Answer

Let numbers be a and b.
Sum of two numbers = 6 × their G.M.
=> a + b = 6√ab => (a + b)/(2√ab) = 3/1
Applying componendo and dividendo
(a + b + 2√ab)/(a + b – 2√ab) = (3 + 1)/(3 – 1)
=> (√a + √b)2/(√a – √b)2 = (√4)2/(√2)2
=> (√a + √b)/(√a – √b) = 2/√2 = √2/1
Again applying componendo and dividend
((√a + √b) +(√a – √b))/((√a + √b) – (√a – √b)) = (√2 + 1)/(√2 – 1)
=> 2√a/2√b = (√2 + 1)/(√2 – 1) => √a/√b = (√2 + 1)/(√2 – 1)
Squaring we get
a/b = (2 + 1 + 2√2)/(2 + 1 – 2√2) => a/b = (3 + 2√2)/(3 – 2√2).

29. If A and G be A.M and G.M respectively between two positive numbers. Prove that the numbers are A ± √(A + G)(A - G).

Answer


30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and n th hour?

Answer

Number of bacteria present in the culture form a G.P.
Whose first term a = 30 and common ratio r = 2
∴ Bacteria present after 2 hours
= ar2 = 30 × 22 = 30 × 4 = 120
Bacteria present after 4 hours = ar4 = 30 × 24
= 30 × 16 = 480
Bacteria present after n hours = arn = 30 × 2n
= 30. 2n

31. What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Answer

Let A, P, r, T be the amount, principal, rate of interest percent per annum and period in years respectively. The amount A is given by,
A = P(1 + r/100)T
P = ₹500, r = 10% p.a, T = 10 years;
A = 500(1 + 10/100)10 = 500 × (1.1)10

32. If A.M. and G.M of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Answer

Let α, β be the roots of the quadratic equation.
A.M. of α, β = (α + β)/2 = 8;
G.M. ofα,β = αβ = 5 => √αβ = 52
Equation whose roots are α,β is x 2 – (α + β)
x + αβ = 0
x 2 – 16x + 25 = 0
↧
↧

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.4

February 3, 2020, 12:35 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.4

If you're in search of NCERT Solutions of Chapter 9 Sequence and Series Exercise 9.4 then you can find them here. We at Studyrankers have prepared detailed and accurate Class 11 Maths NCERT solutions of every questions so you can easily get them at one place without wasting your precious time. These will help you a lot if you want to improve your marks in the examinations.

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.4

1. Find the sum of the following series upto n terms:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ..... 

Answer
Let Tn denotes the nth term of the given series
Tn = [nth term of 1, 2, 3, ......] × [nth term of 2, 3, 4, .....]
= [1 + (n – 1) 1] [2 + (n – 1)1]
= n (n + 1)
Tn = n 2 + n

= (n(n +1)(2n + 4))/6
= (2n(n + 1)(n + 2))/6 = (n(n + 1)(n + 2))/3

2. Find the sum of the following series upto n terms: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + .......

Answer

Let Tn denote the nth term of the given series.
Then
Tn = [nth term of 1, 2, 3, ......] [n th term of 2, 3, 4, .....] [n th term of 3, 4, 5, .....]
= [1 + (n – 1)1] [2 + (n – 1).1] [3 + (n –1).1]
= n (n + 1) (n + 2)
= n(n 2 + 2n + 3) = n 3 + 3n 2 + 2n
∴ Sn = ∑n3 + 3∑n2 + 2∑n
= (n2(n + 1)2)/4 + (3n(n + 1)(2n +1))/6 + 2.(n(n+1))/2
= (n(n + 1))/4 [n(n + 1) + 2(2n + 1) + 4]
= (n(n + 1))/4 [n2 + n + 4n + 2 + 4]
= (n(n + 1))/4 (n2 + 5n + 6)
= (n(n + 1)(n + 2)(n + 3))/4

3. Find the sum of n terms of the following series.
3 × 12 + 5 × 22 + 7 × 32 + .....

Answer

Let Tn denote the nth term of the given series.
Then,
Tn = [nth term of 3, 5, 7,...] [nth term of 1, 2, 3, ...]2
= {3 + 2 (n – 1)2}{1 + (n – 1).1]2
= (2n + 1) (n)2
= n 2 (2n + 1) = 2n 3 + n 2
Sn = 2∑n3 + ∑n2
= 2. (n2(n + 1)2)/4 + (n(n + 1)(2n + 1))/6
= (n(n + 1))/6 . [3n(n + 1) + (2n + 1)]
= (n(n + 1))/6 (3n2 + 3n + 2n + 1)
= (n(n+1)(3n2 + 5n + 1))/6

4. Find the sum of n terms of the following series. 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + …………

Answer

Tn = 1/([nth term of 1, 2, 3, …..][n th term of 2, 3, 4, ….])
= 1/([1 + (n -1)1][2 + (n – 1)1]) = 1/(n(n + 1))
Let 1/(n(n + 1)) = A/n + B/(n + 1) ……….(1)
=> 1 = A(n + 1) + Bn …………(2)
[on multiplying both sides by n(n + 1)]
To find A: Put n = 0 in (2), we get
1 = A(0 + 1) => A = 1
To find B: put n = -1 in (2), we get
1 = B(-1) => B = -1
Put these values of A and B in (1), the partial fractions are
1/(n(n + 1)) = 1/n – 1/(n + 1)
∴ Tn = 1/n – 1/(n + 1)
Putting n = 1, 2, 3, ………n, we get,
T1 = 1/1 – 1/2
T2 = 1/2 – 1/3
T3 = 1/3 – 1/4
……………………..
Tn = 1/n – 1/(n + 1)
Adding vertically, we get,
Sn = 1 – 1/(n + 1) = (n + 1 – 1)/(n + 1) = n/(n + 1)

5. Find the sum of the following series upto n terms:
5 2+ 62 + 72 +.... 202

Answer

Tn of given series
= (nth term of 5, 6, 7, ....)2
= [5 + (n + 1) × 1]2
= (n + 4)2 = n 2 + 8n + 16
Sn = ∑n2 + 8∑n + 16 × n
= (n(n + 1)(2n + 1))/6 + 8 × (n(n + 1))/2 + 16n
= n/6[(n + 1)(2n + 1) + 24(n + 1) + 96]
= n/6[2n2 + n + 2n + 1+ 24n + 24 + 96]
= n/6 (2n2 + 27n + 121)
Put n = 16, then
S = 16/6 ( 2 × 162 + 27 × 16 + 121)
8/3 (512 + 432 + 121) = 8/3 × 1065
= 8 × 355 = 2840
Let Tn = 20, a = 5
d = 1
=> 20 = 5 + (n –1).1
or 20 = 5 + n – 1
or 20 – 4 = n
or n = 16

6. Find the sum of the following series upto n terms
3 × 8 + 6 × 11 + 9 × 14 +....

Answer

Here the series is formed by multiplying the corresponding terms of two series both of which are A.P.
viz. 3, 6, 9, ...... and 8, 11, 14, ......
Tn of given series = (nth term of 3, 6, 9,...) × (nth term of 8, 11, 14,...)
= [3 + (n –1)3] [8 + (n – 1)3]
= [3n] [3n + 5] = 9n 2 + 15n
Sn = 9∑n2 + 15 ∑n
= 9 × (n(n + 1)(2n + 1))/6 + (15.n(n + 1))/2
= 3/2 . n(n + 1)[2n + 1 + 5]
= 3/2 . n(n + 1)(2n + 6) = 3n(n + 1)(n + 3)

7. Find the sum to n terms of given series, 1 2 + (12 + 22) + (12 + 22 + 32) + ................

Answer

Let Tn denote the n th term, then
Tn = 12 + 22 + 32 + ..........+ n 2
= ∑n2 = (n(n + 1)(2n + 1))/6
= 1/6 (2n3 +3n2 + n)
Sn = 1/6 [2 ∑n3 + 3 ∑n2 + ∑n]
= 1/6 [ 2. (n2 (n + 1)2)/4 + 3. (n(n +1 )(2n + 1))/6 + (n(n + 1))/2]
= (n(n + 1))/12 [n(n + 1) + (2n + 1) + 1]
= (n(n + 1))/12 [n2 + n + 2n + 1+ 1]
= (n(n + 1)(n2 + 3n + 2))/12 = (n(n + 1)(n + 1)(n + 2))/12
= (n(n + 1)2(n + 2))/12

8. Find the sum to n terms of the following series whose nth term is given by:
n(n + 1) (n + 4)

Answer

Tn = n(n + 1) (n + 2) = n(n 2 + 5n + 4)
= n 3 + 5n 2 + 4n
Sn =
= [(n(n + 1)2)/2] + 5. (n(n + 1)(2n + 1))/6 + 4 . (n(n + 1))/2
= (n2(n + 1)2)/4 + (5n(n + 1)(2n + 1))/6 + 2n(n + 1)
= (n(n + 1))/12 [3n(n + 1) + 10(2n + 1) + 2n]
= (n(n + 1))/12 [3n2 + 3n + 20n + 10 + 2n]
= (n(n + 1))/12 [3n2 + 23n + 3n]
= 1/12 . n(n + 1)[3n2 + 23n + 3n]

9. Find the sum to n terms of the following series whose n th term is given by:
n 2 + 2n

Answer

Tn = n 2 + 2n
Put n = 1, 2, 3, ....., n, we get,
T1 = 12 + 21
T2 = 22 + 22
T3 = 32 + 23
.........................
Tn = n 2 + 2n
Adding vertically, we have,
Sn = (12 + 22 + 32 +... + n 2) + (21 + 22 + 23 +... + 2n)
= ∑n2 + (2(2n – 1))/(2 - 1)
= (n(n + 1)(2n + 1))/6 + 2(2n – 1)

10. Find the sum to n terms of the given series whose n th term is (2n –1)2.

Answer

Tn = (2n – 1)2 = 4n 2 – 4n + 1
Sn = 
= (4n(n + 1)(2n + 1))/6 - (4n(n + 1))/2 + n
= n/3 [2(n + 1)(2n + 1) – 6(n + 1) + 3]
= n/3 [2(2n2 + 3n + 1) – 6n – 6 + 3]
Sn = n/3 [4n2 – 1] = (n(2n – 1)(2n + 1))/3
↧

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

February 3, 2020, 2:49 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

NCERT Solutions of Chapter 10 Straight Lines Exercise 10.1 is given here that are detailed and accurate so you can easily solve difficult questions without wasting your time. NCERT Solutions for Class 11 Maths will increase an individual's problem solving skills and increase his efficiency. These NCERT Solutions will improve your marks and solve complex problems easily.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, –5) and (– 4, – 2). Also find its area.

Answer

Given points (– 4, 5), (0, 7), (5, –5) and (– 4, 2) are plotted on a graph paper. They are denoted by A, B, C and D respectively.
Now, divide the quadrilateral into two triangle viz. △ABD and △BDC.
Area of a triangle is
= 1/2 I x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)I
The verticals of △ABD are (-4, 5), (0, 7) and (-4, -2)
Area of △ABD
= 1/2 I -4(7 + 2) + 0.(-2 – 5)+ (-4)(5 – 7) I
= 1/2 I -36 + 8 I = 28/2 = 14        ….(i)
The vertices of △BCD are (0, 7), (5, -5), (-4, -2)
Area of △BCD
= 1/2 I 0(-5 + 2) + 5(-2 -7) – 4(7 + 5) I
= 1/2  I -45 – 48 I = 93/2     ……(ii)
Area of quadrilateral ABCD
= Area of △ABD + area of △BCD
= 14 + 93/2           [from (i) & (ii)]
= (28 + 93)/2 = 121/2 = 60.5 sq. units.

2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Answer

BC is the base of △ABC such that BO = OC = a
AB = AC = 2a
Now, AO2 = AB2 - BO2
∴ AO = √(4a2 – a2) = √3a
∴ The point A lying on x-axis is (√3a, 0)

The point B is (0, a) and the point C is (0, – a), when A lies on the left of y-axis the vertices of triangle are (–√3a, 0) a (0, a) and (0, – a).

3. Find the distance between P(x1, y 1) and Q(x2, y 2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Answer

(i) When PQ is parallel to y-axis, every point on this line has the same abscissa, let
x 2 = x 1.
The points P and Q are P(x 1, y 1 ),Q(x 2, y 2)
∴ PQ = |y 2 – y1|


(ii) When PQ is parallel to x-axis. Every point on the line has the same ordinate.
Let y 2 = y 1
∴ The coordinates of points P and Q are
P(x 1, y 1), Q(x 2, y 2)
∴ PQ = |x 2 – x 1|


4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer

Let the point on the x-axis be (x 1, 0). The other two points A and B are A(7, 6), B (3, 4)
we have
PA = PB
or PA2 = PB2
or (x 1 – 7)2 + 62 = (x 1 – 3)2 + 42
or x12 – 14x1 + 49 + 36 = x12 – 6x1 + 9 + 16
or –14 x1 + 85 = –6x1 + 25
or 8x 1 = 60
or x1 = 60/8
∴ x1 = 15/2
∴ The point on x-axis equidistant from A and B is (15/2 , 0).

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, –4) and B(8, 0).

Answer

Coordinates of mid-point of two points (x 1, y 1) and (x 2, y 2) are ((x1 + x2)/2 , (y1 + y2)/2 )
The mid-point of B(8, 0) and P(0, –4) is
((8 + 0)/2 , (0 – 4)/2) i.e., (4, -2)

The slope of line joining the points (x 1, y 1) and (x 2, y 2) = (y2 – y1)/(x2 – x1)
The two points are O(0, 0) and A(4, –2)
∴ slope = (-2 – 0)/(4 – 0) = -1/2

6. Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Answer

We have three points A(4, 4), B(3, 5) and C (–1, –1)

∴ Slope of AB = (y2 – y1)/(x2 – x1) = (5 – 4)/(3 – 4) = -1
Slope of AC = (-1 -4)/(-1 – 4) = -5/-5 = 1
Product of slopes of AB and AC = 1× -1 = -1
∴ AB ⊥ AC
Hence, △ABC is a right angled triangle.

7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Answer

The line OP makes an angle of 30° with y-axis measured anti-clock-wise.

∴ OP makes an angle of 90 + 30 = 120 with positive direction of x-axis
∴ Slope of OP = tan 120° = tan (180° – 60°)
= –tan 60° = –√3

8. Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.

Answer

We have the points A(x, – 1), B(2, 1), C(4, 5).
A, B, C are collinear if the
Slope of AB = Slope of BC.
Slope of AB = (1 + 1)/(2 – x) = 2/(2 – x);
Slope of BC = (5 – 1)/(4 – 2) = 4/2 = 2
∴ 2/(2 – x) = 2 or 2 – x = 1 or x = 1

9. Without using distance formula, show that points (–2, 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Answer


The given points are A(–2, –1), B(4, 0), C(3, 3), D(–3, 2)
Slope of AB = 1/(4 + 2) = 1/6;
Slope of DC = (3 – 2)/(3 + 3) = 1/6
∴ Slope of AB = Slope of DC
=> AB || DC
Slope of AD = (2 + 1)/(-3 + 2) = 3/-1;
Slope of BC = (3 – 0)/(3 – 4) = 3/-1
∴ Slope of AD = Slope of BC
=> AD parallel BC Hence, ABCD is a parallelogram.

10. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Answer

Slope of the line joining the point P(3, –1) and Q (4, –2) is equal to (-2 + 1)/(4 – 3) = -1/1 = -1
tan θ = –1 ∴ θ = 135°

11. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1 3, find the slopes of the lines.

Answer

Let m1 and m2 be the slopes of the two lines
Given that m1 = 2m2
Let θ be the angle between the line,
tan θ = 1/3 = (m1 – m2)/(1 +m1m2)
putting m1 = 2m2, 1/3 = (2m2 – m2)/(1 + 2m2m2) = m2/(1 + 2m22)
or 3m2 = 1 + 2m22 or 2m22 – 3m2 + 1 = 0
or (2m2 – 1)(m2 -1) = 0 m2 = 1 or 1/2
since, m1 = 2m2
∴ for m2 = 1, m1 = 2, for m2 = 1/2, m1 = 1
∴ slopes of these lines are 1, 1/2 or 2, 1

12. A line passes through (x1, y 1) and (h, k). If slope of the line is m, show that
k – y 1 = m(h – x1).

Answer

Slope of the line joining the points A(x 1, y 1) and B(h, k)
= (k – y1)/(h – x1) = m (Given)
Cross-multiplying, k – y 1 = m(h – x 1) Hence proved.

13. If three points (h, 0), (a, b) and (0, k) lies on a line, show that a/h + b/k = 1.

Answer

The given points are A(h, 0), B(a, b), C(0, k), they lie on the same plane.
∴ Slope of AB = Slope of BC
∴ Slope of AB = (b – 0)/(a – h) = b/(a – h);
Slope of BC = (k – b)/(0 – a) = (k – b)/-a
∴ b/(a – h) = (k – b)/ -a or by cross-multiplication
–ab = (a – h)(k – b)
or –ab = ak – ab – hk + hb
or 0 = ak – hk + hb
or ak + hb = hk
Dividing by hk => ak/hk + hb/hk = 1 or a/h + b/k = 1
Hence proved.

14. Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?


Answer

Slope of the line joining the point A(1985, 92) and B(1995, 97)
= (97 – 92)/(1995 – 1985) = 5/10 = 1/2
Let P be the population in the year 2010
∴ The point P(2010, P) lies on it
∴ Slope of BP = Slope of AB
Slope of BP = (P – 97)/(2010 – 1995) = 1/2
=> 2(P – 97) = 2010 – 1995 = 15
P – 97 = 16/2 => P = 97 + 7.5
∴ P = 104.5 crores
↧

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

February 3, 2020, 11:52 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

Chapter 10 Straight Lines Exercise 10.2 Class 11 Maths NCERT Solutions is prepared by Studyrankers experts that will be useful guide in the journey of preparation of exams. You can improve your problem solving skills just by practicing the questions. NCERT Solutions for Class 11 Maths prepared by Studyrankers are updated as per the latest CBSE syllabus.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

1. Find the equation of the line, which satisfy the given condition. Write the equations for x and y-axis.

Answer

Equation of the x-axis is y = 0
Equation of the y-axis is x = 0

2. Find the equation of the line, which satisfy the given condition. Passing through (– 4, 3) with slope 1/2.

Answer

Equation of a line passing through (x, y) with slope m is (y – y 1) = m(x – x 1)
Here x 1 = – 4, y = 3, m = 1/2
∴ Equation of the line required is
(y – 3) = 1/2 (x + 4)
or 2y – 6 = x + 4 or x – 2y + 10 = 0

3. Find the equation of the line passing through (0, 0) with slope m.

Answer

The line passes through (0, 0) and its slope = m
Equation of the line is
y – y 1 = m(x – x 1)
=> y – 0 = m(x – 0)
=> y = mx

4. Find the equation of the line passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.

Answer

We have x 1 = 2, y 1 = 2√3
m = tan 75°= tan (45° + 30°)
= (tan 45° + tan 30°)/(1 – tan 45° tan 30°) = (1 + 1/√3)/(1 – 1/√3) = (√3 + 1)/( √3 – 1)
Equation of the required line is,
y – 2√3 = (√3 + 1)/( √3 – 1) (x – 2)
=> (√3 – 1)y – 2√3(√3 – 1)
= (√3 + 1)x – 2(√3 + 1)
=> (√3 + 1)x – (√3 – 1)y
= -6 + 2√3 + 2√3 + 2
= 4√3 – 4 = 4(3√ – 1)
Then, equation of required line is
(√3 + 1)x – (√3 – 1)y = 4(√3 – 1)

5. Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of origin with slope – 2.

Answer

The line passes through (– 3, 0) and has slope – 2.
Now, equation of the line is point-slope form is
y – y 1 = m(x – x 1)
y – 0 = –2(x + 3)
y = – 2x – 6
2x + y + 6 = 0
which is the required equation.

6. Find the equation of the line intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.

Answer

The line PQ intersects Y-axis at (0, 2)
Slope of PQ = tan 30° 1/√3

Equation of PQ is y – 2 = 1/√3 (x – 0)
=> √3y – 2√3 = x => x –√3y + 2√3 = 0

7. Find the equation of the line passing through the points (–1, 1) and (2, –4).

Answer

Let the line passes through the points A(–1, 1) and B(2, –4).
Equation of the line passing through (x 1, y 1) and (x 2, y 2) is y – y1 = (y2 – y1)/(x2 – x1) (x – x1)
We have x 1 = –1, y 1 = 1, x 2 = 2, y 2 = – 4
∴ Equation of AB is y -1 = (-4 -1)/(2 + 1) (x + 1)
=> y – 1 = -5/3 (x + 1) => 3(y – 1) = –5(x + 1)
=> 3y – 3 = –5x – 5 => 5x + 3y – 3 + 5 = 0
=> 5x + 3y + 2 = 0

8. Find the equation of the line perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Answer

If a line is at a distance p from the origin and perpendicular make an angle ω with positive direction of x-axis then its equation is
xcosω + ysinω = p

We have p = 5, ω= 30°
∴ Equation of the required line is,
xcos 30° + ysin 30° = p
√3/2 x + 1/2 y = 5 => √3x + y = 10

9. The vertices of △PQR are P(2, 1), Q(–2, 3) and R(4, 5). Find equation of the median through the vertex R.

Answer

The middle point of PQ is S = ((2 – 2)/2 , (1 + 3)/2) or (0, 2)
∴ Equation of the median RS where R is (4, 5) and S is the point (0, 2) is

y-5 = (2 – 5)/(0 – 4) . (x – 4)
or y – 5 = -3/-4  (x – 4)
or 4(y – 5) = 3(x – 4)
or 4y – 20 = 3x - 12
i.e. 3x – 4y = –8
∴ Equation of median RS is 3x – 4y + 8 = 0

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Answer

Slope of the line joining A(2, 5) and B(–3, 6)
= (6 – 5)/(-3 -2) = 1/-5
∴ m = slope of any line perpendicular to AB = 5
The equation of the line passing through (–3, 5) and perpendicular to AB is
y – 5 = 5(x + 3) => 5x – y + 20 = 0

11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Answer

Slope of the line joining the points A(1, 0) and B(2, 3) = ( 3 – 0)/(2 – 1) = 3/1 = 3
∴ Slope of the CD perpendicular to
AB= - (1/3)

[∵ product of slopes of perpendicular lines = –1]
The point P divides AB in the ratio 1: n
∴ Coordinates of P are,
((1 × 2 + 1 × n)/(1 + n) , (1 × 3 + 0 × n)/(1 + n)) or ((n + 2)/(n + 1) , 3/(n + 1))
Equation of the line CD which is perpendicular to AB passes through P is
y –  3/(n + 1) = -(1/3) (x – (n + 2)/(n + 1))
[Using the formula y – y 1 = m(x – x 1)]
3(n + 1)y – 9 = –(n + 1)x + (n + 2)
=> (n + 1)x + 3(n + 1)y = n + 2 + 9
=> (n + 1)x + 3(n + 1)y = n + 11

12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Answer

The line making equal intercepts has the slope either 1 or –1

tan 45° = 1 and tan 135° = –1
Given that, a = b
∴ The equation become x/a + y/a = 1
=> x + y = a
Passing through the point (2, 3)
∴ 2 + 3 = a => a = 5
∴ The equation become x + y  = 5.

13. Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Answer

Let the intercepts along x-axis and y-axis be a and b.
a + b = 9
or b = 9 – a … (i)
∴ Equation of the line in intercepts form is
x/a + y/b = 1 …..(ii)
=> x/a + y/(9 – a) = 1 [using (i)]
Since, point A(2, 2) lies on it
∴ 2/a + 2/(9 – a) = 1
=> (2(9 –a) + 2a)/(a(9 – a)) = 1
=> 18 – 2a + 2a = a(9 – a)
=> 18 = 9a – a2
=> a2 – 9a + 18 = 0 => (a – 3)(a – 6) = 0
a = 3 or a = 6
b = 9 – a b = 9 – a
b = 9 – 3 = 6 b = 9 – 6 = 3
=> line is x/a + y/b = 1 or line is x/a + y/b = 1
x/3 + y/6 = 1 or x/6 + y/3 = 1
2x + y = 6 or x + 2y = 6
Hence, the required equation of straight lines are 2x + y = 6 and x + 2y = 6.

14. Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Answer

Slope of the line = tan 2π/3
= tan (π - π/3) = tan(-(π/3) = - √3

Equation of the line AC passing through (0, 2) and having the slope - √3 is
y - 2 = -√3(x – 0) => √3x + y – 2 = 0
Since, another line BD is parallel to AC
∴ Slope of BD = Slope of AC = - √3
BD is passing through the point B(0, –2)
∴  Equation of BD is y + 2 = - √3 (x – 0)
=> √3x + y + 2 = 0
Thus, equation of AC and BD are √3x + y – 2 = 0 and √3x + y + 2 = 0.

15. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Answer

Let ON be perpendicular to AB.

The point is (–2, 9)
∴ Slope of ON = (9 – 0)/(-2 – 0) = -9/2
Slope of AB which is perpendicular to ON = 2/9
(∵ Lines are perpendicular if m1m2 = –1)
Now, AB passes through (–2, 9) and has the slope 2/9 .
∴ Equation of AB, y – 9 = 2/9 (x + 2).
Or, 9y – 81 = 2x + 4 or 2x – 9y + 85 = 0.

16. The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C. 

Answer

L is linear function of C
∴ L = a + bC
For L = 124.942, C = 20,
124.942 = a + 20 b … (i)
For L = 125.134, C = 110,
125.134 = a + 110b … (ii)
Subtracting (i) from (ii), 0.192 = 90b
∴ b = 0.192/90 = 0.00213
from (i) 124.942 = a + 20 × 0.00213
=> 124.942 = a + 0.0426
a = 124.942 – 0.0426 = 124.8994
Now L in term of C is L = a + bC
=> L = 124.8994 + 0.00213 C.

17. The owner of milk store finds that, he can sell 980 litres of milk each week at ₹14/ litre and 1220 litres of milk each week at ₹16/ litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/ litre?

Answer

Let y litre milk be sold at ₹ x/litre and x, y have linear relationship.
y = a + bx i.e.
it is a straight line
Now, y 1 = 980 litre, x 1 = ₹14/ litre, y2 = 1220 litre, x 2 = ₹16/ litre
Slope of the line = (1220 – 980)/(16 – 14) = 240/2 = 120
Equation of the line, y – 980 = 120 (x – 14)
when x = 17, y = 980 + 120 (17 – 14)
= 980 + 120 × 3 = 980 + 360 = 1340
Hence, y = 1340 litre milk may be sold at ₹17 litre.

18. P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b = 2.

Answer

Let the line AB makes intercepts p, q on the axes.

∴ A is (p, 0) and B is (0, q)
Now, P(a, b) is the mid-point of AB
∴ (p + 0)/2 = a, (0 + q)/2 = b
∴ p = 2a, q = 2b
Intercept form of the line AB is
x/p + y/q = 1 => x/2a + y/2b = 1
=> x/a + y/b = 2

19. Point R(h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Answer

Let the line AB makes intercepts a and b on the axes. The point R(h, k) divides it in the ratio 1:2

∴ h = (1 × 0 + 2 × a)/(1 + 2) = 2a/3
∴ a = 3/2 h and k = (1 × b + 2 × 0)/(1 + 2) = b/3
∴ b = 3k
Intercept form of the line AB is
x/a + y/b = 1 => x/3h/2 + y/3k = 1
=> 2x/3h + y/3k = 1 => 2kx + hy = 3kh.

20. By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.

Answer

Equation of the line passing through (3, 0) and (–2, –2) is
y – 0 = (-2 – 0)/(-2 -3) (x – 3) => y = 2/5 (x – 3)
=> 5y = 2x – 6 => 2x – 5y = 6
If the point (8, 2) lies on it, (8, 2) will satisfy the equation of the lines.
LHS = 2× 8 – 5 × 2 = 16 – 10 = 6 = RHS
Hence (8, 2) lies on it
∴ (3, 0), (–2, –2) and (8, 2) are collinear.
↧
↧

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

February 3, 2020, 11:19 pm
$
0
0

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

If you are facing trouble finding the correct NCERT Solutions of Chapter 10 Straight Lines Exercise 10.3 then you we have provided these for you here. It will help you in finding the NCERT Solutions for Class 11 Maths difficult questions. The solutions provided here are updated as per the latest pattern of CBSE which will improve your problem solving skills.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

1. Reduce the following equations into slope intercept form and find their slopes and intercepts on the axes.
(i) x + 7y = 0
(ii) 6x + 3y – 5 = 0
(iii) y = 0

Answer

(i) x + 7y = 0 or y = - (1/7) x + 0
∴ Slope = -(1/7)  and y-intercept = 0

(ii) 6x + 3y – 5 = 0 or 3y = – 6x + 5
y = – 2x + 5/3
∴ Slope = – 2 and y-intercept = 5/3

(iii) y = 0 or y = 0.x + 0
∴ Slope = 0, y-intercept = 0.

2. Reduce the following equations into intercepts form and their intercepts on the axes.
(i) 3x + 2y – 12 = 0
(ii) 4x – 3y = 6
(iii) 3y + 2 = 0

Answer

(i) 3x + 2y – 12 = 0
or 3x + 2y = 12
or 3x/12 + 2y/12 = 1
or x/4 + y/6 = 1 (intercepts form)
where a = 4, b = 6

(ii) 4x + 3y = 6
4x/6 – 3y/6 = 1
x/3/2 + y/-2 = 1 (intercepts form)
where a = 3/2, b = -2

(iii) 3y + 2 = 0
3y = 2
3y/-2 = 1 => y/-2/3 = 1
Now b = -(2/3) and no intercept along x- axis.

3. Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis.
(i) x – √3y + 8 = 0
(ii) y – 2 = 0
(iii) x – y = 4

Answer

(i) x – √3y + 8 = 0
-x + √3y = 8
Now a = 1, b = √3
Here √a2 + b2 = √1 + 3 = √4 = 2
Thus, -(x/2) + √3/2 y = 8/2
-(x/2) + √3/2 y = 4
Or x cos 120° + y sin 120° = 4 (normal form)
Where p = 4 and ω = 120°
(ii) y – 2 = 0
y = 2
now a = 0, b = 1
here √a2 + b2 = √0 + 1 = 1
thus, 0x + 1y = 2
or x cos 90° + ysin 90° = 2 (normal form)
where ω = 90° and p = 2
(iii) x – y = 4
Now a = 1, b = -1
Here, √a2 + b2 = √1 + 1 = √2
Thus, 1/√2 x – 1/√2y = 4
Or xcos 315° + ysin(315°) = 4(normal form)
Where ω = 315° and p = 4

4. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5 (y – 2)

Answer


5. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

Answer

The given line is x/3 + y/4 = 1
Multiplying by 12
4x + 3y = 12 or 4x + 3y – 12 = 0
Any point on the x-axis is (x, 0)
Perpendicular distance from (x, 0) to the given line
(4x – 12)/√42 + 32 = 4
Or | 4x - 12 |/√25 = 4 or | 4x - 12 | = 5 × 4 = 20
Taking (+ve)sign
4x – 12 = 20 or 4x = 20 + 12 = 32
=> x = 32/4 = 8
Taking (-ve) sign 4x – 12 = -20
4x = -20 + 12 = -8
x = -2
∴ required points on the x – axis are (8, 0) and (-2, 0)

6. Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l(x + y) – r = 0.

Answer

(i) One of the lines is 15x + 8y – 34 = 0
Putting y = 0, 15x = 34 or x = 34/15
∴ The point (34/15, 0) lies on
15x + 8y – 34 = 0
∴ perpendicular distance from (34/15 , 0)
to 15x + 8y + 31 = 0 is
= |15 . 34/15 + 0 + 31|/√152 + 82 =| 34 + 31/√225 + 64
= 65/√289 = 65/17
(ii) l(x + y) + p = 0
lx + ly – r = 0
Now A = l, B = l, C1 = p, C2 = –r
Distance between parallel lines
d = |C1 – C2|/√A2 + B2
= |p + r|/√l2 + l2 = |p + r|/√2 l = 1/√2 |(p + r)/l|

7. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

Answer

The given line is 3x – 4y + 2 = 0
Slope of the line = 3/4
Let equation of the parallel line to the given line
3x – 4y + 2 is 3x – 4y + λ = 0
Required line passing through point (–2, 3)
∴ 3 × (–2) – 4 × 3 + λ = 0
=>  – 6 – 12 + λ = 0
∴  λ = 18
∴ Required equation is 3x – 4y + 18 = 0

8. Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.

Answer

Equation of required line is 7x + y + λ = 0

It passes through the point (3, 0)
∴ 7 × 3 + 0 + λ = 0
λ=-21
∴ Required equation is 7x + y – 21 = 0.

9. Find the angles between the lines √3x + y = 1 and x + √3y =1.

Answer

The first line is √3x +y = 1 or y = -√3 . x + 1
The slope m1 of the line √3x + y = 1 is - √3
or, m1 = - √3
The other line is
x + √3y = 1 => y = -(1/√3) x + 1/√3
The slope m2 of the line x + √3y = 1 is –(1/√3)
∴ m2 = -(1/√3)
if θ is the angle between two lines, then
tan θ = 

tanθ = 2/√3 . 1/2 = 1/√3 => θ = π/6

10. The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

Answer

Slope of the line PQ passing through P(h, 3) and Q (4, 1) is 2/(h – 4)
Slope of the line AB 7x – 9y – 19 = 0 is 7/9.

The lines AB and PQ are perpendicular to each other
∴ Product of their slopes = –1
Or 2/(h – 4) × 7/9 = -1    ∴ 14 = -9(h – 4)
Or 9h = 36 – 14 or 9h = 22
Or h = 22/9.

11. Prove that the line through the point (x1, y 1) and parallel to the line Ax + By + C = 0 is A(x – x1) + B(y – y 1 )= 0.

Answer

The given line is Ax + By + C = 0
or y = -(A/B)x – C/B

Slope of the parallel line to the line
Ax + By + C = 0 is –A/B
The parallel line passes through (x 1, y 1)
∴ Equation of the parallel line passing through (x 1, y 1) is
∴  y – y1 = -(A/B)(x – x1)
=> B(y – y1) = -A(x – x1)
A(x – x1) + B(y – y1) = 0 Hence proved.

12. Two lines passing through the point (2, 3) intersects each other at an angle 60°. If slope of one line is 2, find equation of the other line.

Answer

Let m be the slope of the other line and given slope of one line = 2
and angle between them = 60°
∴ tan 60 = ± (m – 2)/(1 + 2m)
=> (m – 2)/(1 + 2m) = ± √3

For +ve sign,
m – 2 = √3(1 + 2m) = √3 + 2√3m
(2√3 – 1)m = -2-√3
∴ m = -((2 + √3)/(2√3 – 1))
Equation of the line passing through P(2, 3) with slope m is
y – 3 = -((2 + √3)/(2√3 – 1))( x – 2)
=> (2√3 – 1)y – 3(2√3 – 1)
= -(2 + √3)x + 2(2 + √3)
=> (2√3 – 1)y – 6√3 + 3
= -(2 + √3)x + 4 + 2√3
=> (2 + √3)x + (2√3 – 1)y – 6√3 + 3 – 4 – 2√3 = 0
=> (2 + √3)x + (2√3 – 1)y – 8√3 – 1 = 0
For –ve sign (m – 2)/(1 + 2m) = - √3
=> m – 2 = -√3(1 + 2m) = -√3 – 2√3m
(2√3 + 1)m = 2 – √3
∴ m = - ((2 – √3)/(1 + 2√3))
Equation of the line passing through P(2, 3) with slope m is
y – 3 = -((2 – √3)/(1 + 2√3))(x – 2)
=> (1 + 2√3)y – 3(1 + 2√3)
= (2 – √3)x – 2(2 – √3)
=> (1 + 2√3)y – 3 – 6√3 = (2 – √3)x – 4+ 2√3
=> (2 – √3)x – (1 + 2√3)y – 4 + 2√3 + 3 + 6√3 = 0
=> (2 – √3)x – (1 + 2√3)y – 1 + 8√3 = 0
∴ Required lines are
(√3 + 2)x + (2√3 – 1)y – 8√3 – 1 = 0
And (2 – √3)x – (1 + 2√3)y + 8√3 – 1 = 0.

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Answer

Slope of the line joining the points A(–1, 2) and B(3, 4)
= (4 – 2)/(3 + 1) = 2/4 = 1/2

Right bisector PQ is perpendicular to AB
∴ Slope of PQ = – 2
middle point of AB is ((-1 + 3)/2 , (2 + 4)/2 ) i.e. (1, 3)
Right bisector passes through M(1, 3)
Equation of right bisector PQ is y – 3
= – 2 (x – 1) = –2x + 2
=>  2x + y – 3 – 2 = 0 Þ 2x + y – 5 = 0.

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x –4y – 16 = 0.

Answer

The equation of the given line is
3x – 4y – 16 = 0
Let the equation of a line perpendicular to the given line is 4x + 3y + k = 0, where k is a constant .If this line passes through the point (–1, 3), then
– 4 + 9 + k = 0 => k = –5
∴ The equation of a line passing through the point (–1, 3) and perpendicular to the given line is
4x + 3y – 5 = 0
∴ The required point of the foot of the perpendicular is the point of the intersection of the lines
3x – 4y – 16 = 0 … (i)
and 4x + 3y – 5 = 0 … (ii)
Solving (i) and (ii) by cross-multiplication, we have
x/(20 + 48) = y/(-64 + 15) = 1/(9 + 16)
=> x/68 = y/-49 = 1/25 => x = 68/25 , y = -(49/25)
∴ The required point is (68/25 , -49/25).

15. The perpendicular from the origin to the line y = mx + c meets it at the point (– 1, 2). Find the values of m and c.

Answer

Let the perpendicular OM is drawn from the origin to AB.

M is the foot of the perpendicular
Slope of OM = (2 – 0)/(-1 – 0) = 2/-1 ;
Slope of AB = m
OM ⊥ AB ∴ m × (-2) = -1  ∴ m = 1/2
M(–1, 2) lies on AB whose equation is
y = mx + c or y = 1/2 x + c
2 = 1/2 × (-1) + c => c = 2 + 1/2 = 5/2
∴ m = 1/2 or c = 5/2.

16. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively, prove that p2+ 4q2= k2 .

Answer

The perpendicular distance from the origin to the line x cosθ – y sinθ = k cos2θ …(i)
p = (kcos 2θ)/cos2θ + sin2θ = kcos 2θ,
the other line is x secθ + y cosecθ = k
or x/cosθ + y/sinθ = k
xsinθ + ycosθ = ksinθ cosθ
xsinθ + ycosθ = k/2 sin 2θ …….(ii)
∴ the perpendicular distance q from the origin to the line (ii)
q = (k/2 sin 2θ)/(sin2θ+ cos 2θ) = k/2 sin 2θ
now p2 + 4q2 = k2 cos22 + 4(k/2 sin 2)2
=> p2 + 4q2 = k2 cos2 2θ + 4. K2/4 sin2 2θ
p2 + 4q2 = k2(cos22θ + sin 22θ),
Hence, p2 + 4q2 = k2

17. In the triangle ABC with vertices A(2, 3), B(4, –1) and C(1, 2), find the equation and length of altitude from the vertex A.

Answer

The vertices of △ABC are A(2, 3), B(4, –1) and C(1, 2) and AM is the altitude.
Slope of BC = (2 + 1)/(1 – 4) = 3/-3 = -1
∴ Slope of altitude AM = 1
Now, altitude passes through A(2, 3) and has the slope 1.


∴ Equation of AM is y – 3 = (x – 2)
x – y + 3 – 2 = 0
or y – x = 1
Equation of BC passing through B(4, –1) and C(1, 2)
y + 1 = (2 – 4)/(1 + 1) (x – 4) => y + 1 = -2/2 (x – 4)
y + 1 = -(x – 4)
or x + y = 3 or x + y – 3 = 0
∴ length of altitude = AM
= perpendicular distance from A(2, 3) to BC.
= (2 + 3 – 3)/(√12 + 12) = 2/√2 = √2
∴ Length of altitude = √2
Equation of altitude is y – x = 1 and length of altitude = √2.

18. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p2 = 1/a2 + 1/b2.

Answer

Equation of the line which makes intercepts a, b on the axes is x/a + y/b = 1

∴ The perpendicular distance p from the origin
p =
=> 1/p = 
↧

NCERT Solutions for Class 8th: Ch 11 Mensuration Geometry

February 4, 2020, 12:04 am
$
0
0

NCERT Solutions for Chapter 11 Mensuration Class 8 Mathematics

Page No: 171

Exercise 11.1

1. A square and a rectangular field with measurements as given in the figure have the same perimeter.

Which field has a larger area?
Answer

Given:
The side of a square = 60 m and the length of rectangular field = 80 m
According to question,
Perimeter of rectangular file = Perimeter of square field
⇒ 2(l+b) = 4 × Side
⇒ 2(80 + b) = 4 × 60
⇒ (80 + b) = 240/2
⇒ (80 + b) = 120
⇒ b = 120 - 80
⇒ b = 40 m
Hence, the breadth of the rectangular field is 40 m.
Now,
Area of Square field= (Side)2
= (60)2 sq.m = 3600 sq.m
Area of Rectangular field = (length × breadth)
= 80 × 40 sq. m = 3200 sq. m
Hence, area of square field is larger.

2. Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs. 55 per m2.
Answer

Side of a square plot = 25 m
∴ Area of square plot = (Side)2 = (25)2 = 625 m2
Length and Breadth of the house is 20 m and 15 m respectively
∴ Area of the house = (length x breadth )
= 20 × 15 = 300 m2
Area of garden = Area of square plot – Area of house
= (625 – 300) = 325 m2
∵ Cost of developing the garden around the house is Rs.55
∴ Total Cost of developing the garden of area 325 sq. m = Rs.(55 × 325)
= Rs.17,875

2. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden
[Length of rectangle is 20 – (3.5 + 3.5 meters]

Answer

Given:
Total length of the diagram = 20 m
Diameter of semi circle on both the ends = 7 m
∴ Radius of semi circle = diameter/ 2 = 7/2 = 3.5 m
Length of rectangular field = [Total length - (radius of semicircle on both side)]
={20 – (3.5 + 3.5)}
= 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = ( l x b)
= (13×7) ⇒ 91 m2
Area of two semi circles = 2 × 1/2 πr2
= 2 × 1/2 × 22/7 × 3.5 × 3.5 = 38.5 m2
Total Area of garden = (91 + 38.5)⇒129.5 m2
Perimeter of two semi circles = 2 × πr = 2 × 22/7 × 3.5 = 22 m
Hence, Perimeter of garden = (22 + 13 + 13)m = 48 m

3. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2 ? [If required you can split the tiles in whatever way you want to fill up the corners]

Answer

Base of flooring tile = 24 cm ⇒ 0.24 m
height of a flooring tile = 10 cm
⇒ 0.10 m [1cm = 1/100 m]
Now, Area of flooring tile= Base × Altitude
= 0.24 × 0.10 sq. m
= 0.024 m2
∴ Number of tiles required to cover the floor = Area of floor/area of one tile
= 1080/0.024
= 45000 tiles
Hence 45000 tiles are required to cover the floor.

4. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.
(a)
(b)
(c)

Answer

(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr = 22/7 × 1.4 = 4.4 cm

Total distance covered by the ant= (Circumference of semi circle + Diameter)
=( 4.4 + 2.8 )cm = 7.2 cm

(b) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2 = 1.4cm
Circumference of semi circle = πr = 22/7 × 1.4 = 4.4 cm
Total distance covered by the ant = (1.5 + 2.8 + 1.5 + 4.4) 10.2 cm

(c) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr = 22/7 × 1.4 = 4.4 cm
Total distance covered by the ant= (2 + 2 + 4.4) = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.

Page No. 177

Exercise 11.2

1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
 
Answer

Parallel side of the trapezium AB =1m , CD = 1.2 m and height (h) of the trapezium (AM) = 0.8 m
Area of top surface of the table = 1/2 (sum of parallel sides) Height
= 1/2 × (AB + CD) × AM
= 1/2 × ( 1 + 1.2) × 0.8
= 1/2 × 2.2 × 0.8
= 0.88 m2
Thus surface area of the table is 0.88 m2

2. The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm.

Find the length of the other parallel side.

Answer

Let the length of the other parallel side be = b cm
Length of one parallel side = 10 am and height (h) = 4 cm
Area of trapezium = 1/2 (sum of parallel sides) Height
⇒ 34 = 1/2 ( a + b)h
⇒ 34 = 1/2 (10 + b) × 4
⇒ 34 = ( 10 + b) × 2
⇒ 34 = 20 + 2b
⇒ 34 - 20 = 2b
⇒ 14 = 2b
⇒ 7 = b
⇒ b = 7
Hence another required parallel side is 7 cm.

3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Answer

Given: BC = 48 m, CD = 17 m,
AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD = Sum of all sides
120 =(AB + BC + CD + DA)
120 = AB + 48 + 17 + 40
120 = AB + 105
(120 – 105) = AB
AB = 15 m
Now Area of the field = 1/2 × ( Sum of parallel sides) × Height
= 1/2 × (BC + AD) × AB
= 1/2 × (48 + 40) × 15 m2
= 1/2 ×(88) × 15 m2
= 1/2 × (1320) m2
= 660 m2
Hence area of the field ABCD is 660  m2

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Answer

Here h1 = 13 m, h2 = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of △ABC + Area of △ADC
= 1/2 b × h1+ 1/2 b × h2
= 1/2 b ( h1 + h2)
= 1/2 × 24 ( 13 + 8) m2
= 1/2 × 24 (21) m2
= 12 × 21 m2
= 252 m2
Hence required area of the field is 252  m2

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer

Given: d1 =7.5 cm and d2 = 12 cm
Area of rhombus =12×(Product of digonals)
= 12 × (d1 × d2)
= 12×(7.5 × 12) cm2
= 45 cm2
Hence area of rhombus is 45 cm2

6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.

Answer

Rhombus is also a kind of Parallelogram.
∴ Area of rhombus= Base × Altitude
= (6 × 4) cm2
= 24 cm2
Also Area of rhombus = 12× (d1× d2)
⇒ 24 = 1/2 ×( 8 × d2)
⇒ 24 = 4 d2
⇒ 244 cm = d2
⇒ d2 = 6 cm
Hence, the length of the other diagonal is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹4.

Answer

Here, d1 = 45 cm and d2 = 30 cm
∵ Area of one tile 
= 12×(d1× d1)
= 12×(45 × 30 )
= 12×(350)
= 675 cm2
So, the area of one tile is 675 cm2

Area of 3000 tiles = 675 × 3000 cm2
= 2025000 cm2
= 2025000100∗100 m2
[1 cm = 1/100 m, Here cm2 = cm×cm = 1/100×1/100 m2]
= 202.50 m2
∵ Cost of polishing the floor per sq. meter = Rs. 4
∴ Cost of polishing the floor per 202.50 sq. meter = Rs. 4 × 202.50 = Rs. 810
Hence the total cost of polishing the floor is Rs. 810.

8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the sidealong the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer


Given: Perpendicular distance (h) AM = 100 m
Area of the trapezium shaped field = 10500 m2
Let side along the road AB= x m
side along the river CD = 2x m
∴ Area of the trapezium field = 12× (AB + CD) x AM
10500 = 1/2 ( x + 2x) × 100
10500 = 3x × 50
3x = 10500/50
x = 10500/(50 × 3)
x = 70 m
Hence the side along the river = 2m = (2×70) = 140 m.

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Answer

Given: Octagon having eight equal sides, each 5 m.
Construction: Join HC and GD It will divide the octagon into two equal trapezium.
And AM is perpendicular on HC and EN is perpendicular on GD
Area of trap. ABCD = Area of trap. GDFE ---(1)
Area of two trapeziums = (area of trap. ABCH + area of trap. GDFE)
= (area of trap. ABCH + area of trap. ABCH) (by statement 1).
= (2×area of trap. ABCH )
= (2×1/2×(sum of parallel sides×height)
= (2×1/2×(AB + CH) x AM)
=(11 + 5)×4 m2
=(16) × 4
= 64 m2

Area of rectangle (HCDG ) = length × breadth
= HC×HG = 11×5 = 55 m2

∴ Total area of octagon = Area of 2 Trapezium + Area of Rectangle
= 64 m2 + 55 m2 = 119 m2

10. There is a pentagonal shaped park as shown in the figure. For finding its are a Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Answer

First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= 1/2 (AP + BC) x CP + 1/2 (ED + AP) × DP
= 1/2 (30 + 15 ) x CP + 1/2(15 + 30) × DP
= 1/2 (30 + 15) (CP + DP)
= 1/2 × 45 × CD
= 1/2 × 45 × 15
= 337.5 m2
Second way: By Kavita’s diagram
Here, a perpendicular AM drawn to BE. AM = 30 – 15 = 15 m
Area of pentagon = Area of △ABE + Area of square BCDE

={12×15×15}+(15 × 15) m2
= (112.5 + 225.0) m2
= 337.5 m2
Hence total area of pentagon shaped park = 337.5 m2

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of theframe, if the width of each section is same.


Answer

Here two of given figures (I) and (II) are similar in dimensions. And also figures (III) and (IV) are similar in dimensions.
∴ Area of figure (I) = Area of trapezium
= 1/2 (a+b) × h = 1/2 (28 + 20) × 4
= 1/2 × 48 × 4 = 96 cm2
Also, Area of figure (II) = 96 cm2
Now Area of figure (III)
Area of trapezium = 1/2 (a + b) × h
= 1/2 (24 + 16) × 4
= 1/2 × 40 × 4
= 80 cm2
Also Area of figure (IV) = 80 cm2

Page No. 186

Exercise 11.3

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?



Answer

(a) Length of cuboidal box (l)= 60 cm
Breadth of cuboidal box (b) = 40 cm
Height of cuboidal box (h) = 50 cm
∴ Total surface area of cuboidal box = 2(lb+bh+hl)
= 2 (60 × 40 + 40 × 50 + 50 × 60) cm2
= 2 (2400 + 2000 + 3000) cm2
= 2 × 7400 cm2
= 14800 cm2

(b) Length of the cube is 50 cm
∴ Total surface area of cuboidal box =6(side)2
= 6 (50)2 cm2
= 6 (2500) cm2
= 15000 cm2
Thus, the cuboidal box (a) requires the lesser amount of material.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Answer

Given:
Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm
And Height of cuboidal box (h) = 24 cm
∴ Total surface area of suitcase box= 2(lb+bh+hl)
= 2 (80 × 48 + 48 × 24 + 24 × 80) cm2
= 2 (3840 + 1152 + 1920)
= 2 × 6912 = 13824 cm2
Area of Tarpaulin cloth = Surface area of suitcase
⇒ l × b = 13824
⇒ l × 96 = 13824
⇒ l = 13824/96
= 144 cm
Required tarpaulin for 100 suitcases = (144 × 100) cm
= 14400 cm
= 144 m [ 1cm = 100 m]
Thus, 144 m tarpaulin cloth required to cover 100 suitcases.

3.Find the side of a cube whose surface area id 600 cm2

Answer

Here, Surface area of cube = 600 cm2
⇒ 6l2 = 600 cm2
⇒ l2 = 100 cm2
⇒ l = √100 cm
⇒ l = 10 cm
Hence the side of cube is 10 cm

4. Rukshar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?


Answer

Length of cabinet (l) = 2 m
Breadth of cabinet (b) = 1 m
Height of cabinet (h) = 1.5 m
∴ Surface area of cabinet = {Area of Base of cabinet (Cuboid) + Area of four walls}
= lb+2(l+b)h
={2×1 + 2(1+2) 1.5} m2
= 2 + 2 (3) 1.5 m2
= 2+6 (1.5) m2
= (2 + 9.0) m2
= 11 m2
Hence required surface area of cabinet is 11 m2

5. Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room?

Answer

Length of wall (l) = 15 m
Breadth of wall (b) = 10 m
Height of wall (h) = 7 m
∴ Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)
= lb+2(l+b)h
= {15 × 10 + 2 (10 +15) (7)} m2
= {150 + 2 (25) (7)} m2
= (150 + 350) m2
= 500 m2
Area of one can is 100 m2
Now Required number of cans = Area of hall/ Area of one can = 500/100 = 5 cans
Hence 5 cans are required to paint the room.

6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?



Answer

Diameter of cylinder = 7 cm
∴ Radius of cylinder (r) = 7/2 cm
Height of cylinder (h) = 7 cm
Lateral surface area of cylinder = 2πrh
= 2 × 22/7 × 7/2 × 7
= 154 cm2
Now lateral surface area of cube = 4(Side)2 = 4(7)2 cm2
= (4×49) cm2
= 196 cm2
Hence the cube has larger lateral surface area.

7.A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. Howmuch sheet of metal is required?


Answer

Radius of cylindrical tank (r) = 7 m
Height of cylindrical tank (h) = 3 m
Total surface area of cylindrical tank = (Curved surface area + Area of upper end (circle)+ Area of Lower (circle) end)
= (2πrh+πr2+πr2)
= (2πrh+2πr2)
=2πr(h+r)
= 2 × 22/7 × 7(3 + 7) m2
= 44 × 10 m2
= 440 m2
Hence 440 m2 metal sheet is required.

8. The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Answer

Lateral surface area of hollow cylinder = 4224 cm2
Height of hollow cylinder = 33 cm
Curved surface area of hollow cylinder = 2πrh
⇒ 4224 = 2 × 22/7 × r × 33
⇒ r = (4224 × 7)/(2 × 22 × 33)
= (64 × 7)/22 cm
Now Length of rectangular sheet = 2πr
l = 2 × 22/7 × (64 × 7)/22
= 128 cm
Perimeter of rectangular sheet = 2(l + b)
= 2 (128 + 33)
= 2 × 161
= 322 cm
Hence perimeter of rectangular sheet is 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.


Answer

Diameter of road roller = 84 cm
∴ Radius of road roller (r) = d/2 = 84/2
= 42 cm
Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller = 2πrh
= 2 × 22/7 × 42 × 100
= 26400 cm2
∴ Area covered by road roller in 750 revolutions = 26400 × 750 cm2
= 1,98,00,000 cm2
= 1980 m2 [∵ 1 m2= 10,000 cm2]
Thus, the area of the road is 1980 m2.

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Answer

Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm
Height of cylindrical container = 20 cm
Height of the label (h) = (20 – 2 – 2) = 16 cm
Curved surface area of label = 2πrh
= 2×22/7×7×16
= 704 cm2
Hence the area of the label of 704 cm2.

Page No. 191

Exercise 11.4

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

Answer

(a) Volume (it is measure of the amount of space inside of a solild figures)
(b) Surface area (the outside part or uppermost layer of the soild figures)
(c) Volume

2. Diameter of cylinder A is 7 cm and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area.


Answer

Yes, we can say that volume of cylinder B is greater, Because radius of cylinder B is greater than that of cylinder A.
Diameter of cylinder A = 7 cm
⇒ Radius(r) of cylinder A = 7/2 cm and Height(h) of cylinder A = 14 cm
∴ Volume of cylinder A = πr2h
= 22/7 × 7/2 ×7/2 × 14
= 539 cm3
Now Diameter of cylinder B = 14 cm
⇒ Radius of cylinder B = 14/2 = 7 cm and Height of cylinder B = 7 cm
∴ Volume of cylinder A = πr2h
= 22/7 × 7 ×7 ×7 cm3
= 1078 cm3

Since the cylinder A and cylinder B is open from upper end then it will exclude from the Total surface area

Total surface area of cylinder A = ( Area of lower end circle + curved surface area of cyliner)
= πr2 + 2πrh)
= πr(r+2h)
= (22/7 × 7/2) (7/2 + 2×14)
= 11(7/2+28)
= 11(31.5) cm2 = 346.5 cm2
Total surface area of cylinder B = πr(2h + r)
= 22/7 × 7(2×7 + 7)
= 22 × (14 + 7)
= 22 × 21 = 462 cm2
Yes, cylinder with greater volume also has greater surface area.

3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?

Answer

Let the Length, breadth and height of the cuboid be l, b, h.
Base of the cuboid is form a Recatangle so,that the Base(Reactangle) Area is (Length x Breadth)
Base area of cuboid = 180 cm2
L x B = 180 cm2 --- (1)
Volume of cuboid = l ×b × h
Volume of cuboid = 900 cm2
(lb)h = 900 (From eq. 1)
(180) h = 900
h = 900/180
= 5 m
Hence the height of cuboid is 5 m.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Answer

Given: Length of cuboid (l) = 60 cm,
Breadth of cuboid (b) = 54 cm and
Height of cuboid (h) = 30 cm
We know that, Volume of cuboid = l × b ×h
= (60 × 54 × 30) cm3
And Volume of cube = (Side)3
= 6 × 6 × 6 cm3
∴ Number of small cubes = (Volume of cuboid)/( Volume of cube)
= (60×54×30)/(6×6× 6)
= 450
Hence required number of small cubes are 450.

5. Find the height of the cylinder whose volume if 1.54 m3 and diameter of the base is 140 cm.

Answer

Given: Volume of cylinder = 1.54 m3 and Diameter of cylinder = 140 cm
∴ Radius (r) = d/2 = 140/2 = 70 cm
= 70/100 m = 0.7m [1cm=1/100m]
Volume of cylinder = πr2h
⇒ 1.54 = 22/7 × 0.7 × 0.7 × h
⇒ h = (1.54 × 7)/(22 × 0.7 × 0.7)
⇒ h = ( 154 × 7 × 10 × 10)/( 22 × 7 × 7 × 100)
= 1 m
Hence height of the cylinder is 1 m.

6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank.

Answer

Given: Radius of cylindrical tank (r) = 1.5 m
Height of cylindrical tank (h) = 7 m

Volume of cylindrical tank = πr2h
= 22/7 × 1.5 × 1.5 × 7
= 49.5 m3
= 49.5 × 1000 liters [∵1m3 = 1000 liters]
= 49500 liters
Hence required quantity of milk is 49500 liters that can be stored in the tank.

7. If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?

Answer

Let, l units be the edge of the cube.
Surface area = 6l2
and Volume of the cube =  l3
When its edge is doubled = 2l

(i) 6(2l)2 = 6(4l2) = 4(6l2)
= 4 (Surface area) {∵6l2 = Surface area of the cube}
The surface area of the new cube will be 4 times that of the original cube.

(ii) Volume of cube (V) = l3
When edge of cube is doubled = 2l, then
Volume of new cube = (2l)3 = 8l3
Volume of new cube = 8× Volume of cube
Hence, volume will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108m3, find the number of hours it will take to fill the reservoir.

Answer

Volume of reservoir = 108m3
= 108×1000 litres [1m3=1000 l]
=108000 litres
Since water is pouring into reservoir @ 60 litres per minute and in
Time taken to fill the reservoir = 108000/(60×60) hours
= 30 hours
Hence, 30 hours it will take to fill the reservoir.

↧

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.1

February 4, 2020, 2:27 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.1

NCERT Solutions of Chapter 11 Conic Sections Exercise 11.1 is provided here which will help you solving difficult questions easily and completing your homework in no time. NCERT Solutions for Class 11 Maths are prepared by Studyrankers experts that are detailed and correct so you can improve your score in the examinations.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.1

1. Find the equation of the circle with centre (0, 2) and radius 2.

Answer

Here h = 0, k = 2 and r = 2. Therefore, the required equation of the circle is
(x – 0)2 + (y – 2)2 = (2)2
or x 2 + y 2 – 4y + 4 = 4
or x 2 + y 2 – 4y = 0

2. Find the equation of the circle with centre (–2, 3) and radius 4.

Answer

Here h = –2, k = 3 and r = 4. Therefore, the required equation of the circle is
[x – (–2)]2 + (y – 3)2 = (4)2
or (x + 2)2 + (y – 3)2 = 16
or x 2 + 4x + 4 + y 2 – 6y + 9 = 16.

3. Find the equation of the circle with centre (1/2 , 1/4) and radius 1/12.

Answer

Equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = 1/122 = 1/144
=> x2 + y2 – x – y/2 + 1/4 + 1/16 = 1/144
=> 36x 2 + 36y 2 – 36x – 18y + 11 = 0.

4. Find the equation of the circle with centre (1, 1) and radius √2.

Answer

Centre of circle is (1, 1), radius = √2
Equation of circle is
(x – 1)2 + (y – 1)2 = (√2)2 =2
or x2 + y2 – 2x – 2y + 2 = 2
or x2 + y2 – 2x – 2y = 0.

5. Find the equation of the circle with centre (–a, –b) and radius √a2 + b2

Answer

Centre of circle is (–a, –b), radius =√a2 + b2
∴ Equation of the circle is
(x + a)2 + (y + b)2 = a2 – b2
Or x2 + y2 + 2xa + 2yb + a2 + b2 = a2 – b2
Or x2 + y2 + 2ax + 2by + 2b2 = 0.

6. Find the centre and radius of the circle.
(x + 5)2 + (y – 3)2 = 36

Answer

Comparing the equation of the circle
(x + 5)2 + (y – 3)2 = 36
with (x – h)2 + (y – k)2 = 2
∴ –h = 5 or h = –5, k = 3, r 2 = 36, r = 6
∴ Centre of the circle is (–5, 3) and radius = 6

7. Find the centre and radius of the circle.
x2 + y2 – 4x – 8y – 45 = 0

Answer

The given equation is
x2 + y2 – 4x – 8y – 45 = 0
 or (x2 – 4x) + (y2 – 5y) = 45
Now completing the squares with in the parenthesis, we get
(x 2 – 4x + 4) + (y 2 – 8y + 16) = 4 + 16 + 45
or (x – 2)2 + (y – 4)2 = 65
Therefore, the given circle has centre at (2, 4) and radius √65.

8. Find the centre and radius of the circle.
x2 + y2 – 8x + 10y – 12 = 0.

Answer

The given equation is
x2 + y2 – 8x + 10y – 12 = 0
or (x 2 – 8x) + (y 2 + 10) = 12
or (x 2 – 8x + 16) + (y 2 + 10y + 25) = 12 + 16 + 25
or (x – 4)2 + (y + 5)2 = 53
Therefore, the given circle has centre at (4, –5) and radius √53.

9. Find the centre and radius of the given circle
2x 2 + 2y 2 – x = 0.

Answer

Equation of circle is 2x 2 + 2y 2 – x = 0
=> x2 + y2 – x/2 = 0 => (x2 – x/2) + y2 = 0
=> (x2 – x/2 + 1/16) + y2 = 1/16
=> (x – 1/4)2 + y2 = 1/16
Centre is (1/4 , 0)and radius is 1.

10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Answer

Let the equation of the circle be
(x – h)2 + (y – k)2 = r2… (i)
The points (4, 1) and (6, 5) lies on it
∴ (4 – h)2 + (1 – k)2 = r2
=> h2 + k2 – 8h – 2k + 17 = r2…..(ii)
and (6 – h)2 + (5 – k)2 = r2 …..(iii)
The centre (h, k) lies on
4x + y = 16
4h + k = 16 … (iv)
Subtracting (iii) from (ii),
∴ 4h + 8k – 44 = 0 Þh + 2k = 11 … (v)
Multiplying (v) by 4, 4h + 8k = 44
Subtracting eqn (iv) from it
7k = 44 – 16 = 28 ∴ k = 4
From (v) h + 8 = 11 ∴ h = 3
Putting h = 3, k = 4 in (ii)
9 + 16 – 24 – 8 + 17 = r2
=> 42 – 32 = r 2
∴ r 2 = 10
∴ Equation of the circle is
(x – 3)2 + (y – 4)2 = 10
=> x 2 + y 2 – 6x – 8y + 15 = 0

11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Answer

Let the equation of the circle be
(x – h) 2 + (y – k) 2 = r 2 … (i)
Since, the points (2, 3) and (–1, 1) lies on it.
∴ (2 – h) 2 + (3 – k) 2 = r 2
h 2 + k 2 – 4h – 6k + 13 = r 2 … (ii)
Centre (h, k) lies on x – 3y – 11 = 0
h – 3k – 11 = 0 … (iii)
Subtracting (ii) from (i) 6h + 4 k – 11 = 0 … (iv)
Multiply eqn (iii) by 6 6h – 18 k – 66 = 0 … (v)
Subtracting (v) from (iv) 22k + 55 = 0
∴ k = -(55/22) = -(5/2)
from (iii) h = 3k + 11 = -(15/2) + 11 = 7/2
Put the value of h and k in (2 – h) 2 + (3 – k) 2 = r2
(2 – 7/2)2 + (3 + 5/2)2 = r2
=> r2 = 9/4 + 121/4 130/4 = 65/2
∴ Equation of the circle
(x – 7/2)2 + (y + 5/2)2 = 65/2
=> x2 + y2 – 7x + 5y + 49/4 + 25/4 – 65/2 = 0
=> x2 + y2 – 7x + 5y – 14 = 0

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Answer

Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 … (i)
r = 5 ∴ r 2 = 25
Centre lies on x -axis is k = 0
Equation (i) becomes (x – h) 2 + y 2 = 25 (2, 3) lies on it
∴ (2 – h) 2 + 9 = 25 => (2 – h) 2 = 16,
∴ 2 – h = ±4 => h = –2, 6
When h = –2, equation of circle
(x + 2)2 + y 2 = 25
=> x2 + y2 + 4x – 21 = 0
When h = 6, (x – 6)2 + y 2 = 25,
x 2 + y 2 – 12x + 11 = 0
Thus, required circles are x 2 + y 2 + 4x – 21 = 0
and x2 + y 2 – 12x + 11 = 0

13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Answer

a, b are the intercepts made by the circle on the co-ordinate axes at A and B, C the mid point of AB is the centre of the circle
∴ centre (a/2 , b/2);
radius = OC =
=
∴ Equation of the circle is
(x – a/2)2 + (y – b/2)2 = 

x2 + y2 – ax – by + a2/4 + b2/4 = (a2 + b2)4
=> x2 + y2 – ax – by = 0

14. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Answer

Let the centre of the circle C(2, 2), and P(4, 5) is a point on the circle
∴ radius CP = √(4 - 2)2 + (5 - 2)2
= √4 + 9 = √13
∴ Equation of the circle is
(x – 2)2 + (y – 2)2 = 13
=> x2 + y2 – 4x – 4y = 5

15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25?

Answer


↧
Search

NCERT Solutions for Class 8th: Ch 12 Exponents and Powers Geometry

February 4, 2020, 3:09 am
$
0
0

NCERT Solutions for Chapter 12 Exponents and Powers Class 8 Mathematics

Page No: 197

Exercise 12.1

1. Evaluate:
(i) 3-2
(ii) (-4)-2
(iii) (1/2)-5 

Answer

(i) 3-2
= 1/32 [ ∵ a-m = 1/am ]
= 1/9

(ii) (-4)-2
= 1/(-4)2 [∵ a-m = 1/am ]
= 1/16

(iii) (1/2)-5
= (2/1)5  [ ∵ a-m = 1/am ]
 = (2)5 = 32

2. Simplify and express the result in power notation with positive exponent:
(i) ( -4)5 ÷ (-4)8 
(ii) (1/23)2 
(iii) (-3)4 × (5/3)4 
(iv) (3-5 ÷ 3-10) × 3-5
(v) 2-3 × (-7)-3 

Answer

(i) ( -4)5 ÷ (-4)8
= (-4)5-8 [ ∵ am÷ an = am-n]
= (-4)-3
= 1/(-4)3 [∵ a-m = 1/am ]

(ii) (1/23 )2
= 12/(23)2 [∵ (a/b)m = am/an]
= 1/23×2
= 1/26 [(am)n = am×n]

(iii) (-3)4 × (5/3)4
= (-3)4 × 54/34 [∵ (a/b)m = am/an]
= {(-1)4 × 34} × 54/34 [ ∵ (ab)m = ambm]
= 34-4 × 54 [ ∵ am ÷ an = am-n]
= 30 × 54
= 54 [a0 = 1]

(iv) (3-7 ÷ 3-10) × 3-5
= 3-7-(-10) × 3-5 [∵ am ÷ an = am-n]
= 3-7+10 × 3-5
= 33 × 3-5 = 33+(-5) [∵ am × an = am+n]
=3-2
= 1/32 [a-m = 1/am]

(v) 2-3 × (-7)-3
= 1/23 ×1/(-7)3 [∵ a-m = 1/am]
= 1/{2 × (-7)}3
= 1/(-14)3 [∵ (ab)m = ambm]

3. Find the value of:
(i) (30 + 4-1) × 22
(ii) (2-1 × 4-1) ÷ 2-2 
(iii) (1/2)-2 + (1/3)-2 + (1/4)-2 
(iv) (3-1 + 4-1 + 5-1)0 
(v) {(-2/3)-2}2 

Answer 

(i) ( 30 + 4-1) × 22
= (1 + 1/4) × 22 [∵ a-m = 1/am]
= {(4 + 1)/4} × 22
= 5/4 × 22
= 5/22 × 22
= 5 × 22-2 [∵ am ÷ an = am-n]
= 5 × 20
= 5 × 1
= 5 [∵ a0 = 1]
(ii) ( 2-1 × 4-1) ÷ 2-2 =[∵ a-m = 1/am]
= [∵ am ×an = am+n]
=[∵ am ÷ an = am-n]
= 1/2 [∵ a-m = 1/am]

(iii) (1/2)-2 + (1/3)-3 + (1/4)-2
= (2-1)-2 + (3-1)-2 + (4-1)-2  [∵ a-m = 1/am]
= 2-1×(-2) + (3)-1×(-2) + (4)-1×(-2) [∵ (am)n = am×n]
= 22 + 32 + 42
= 4 + 9 + 16
= 29

(iv) (3 + 4 + 5)0
= (1/3 + 1/4 + 1/5)0 [∵ a-m = 1/am]
= {(20 + 15 + 12)/60}0
= (47/60)0
= 1 [∵ a0 = 1]

(v) [∵ (am)n = am×n]
=  [∵ a-m = 1/am]
= 81/16

4. Evaluate:
(i) (8-1 × 53)/2-4 
(ii) ( 5-1 × 2-1) × 6-1 

Answer

(i) (8-1 × 53)/2-4
= (8-1 × 53)/2-4
= {(23)-1 × 53}/2-4
= (2-3 × 53)/2-4  [∵ (am)n = am×n]
= 2-3(-4) × 53
= 2-3+4 × 53 [∵ am ÷ an = am-n]
= 2×125
= 250

(ii)( 5-1 × 2-1) 6-1
= (1/5 × 1/2) × 1/6 [∵ a-m = 1/am]
= 1/10 × 1/6 = 1/60

5. Find the value of m for which 5m ÷ 5-3 = 55 

Answer

5m ÷ 5-3 = 55
= 5m-(-3) = 55 [∵ am ÷ an = am-n]
⇒ 5m+3 = 55
Comparing exponents both sides, we get
⇒ m + 3 = 5
⇒ m = 5 - 3
⇒ m = 2

6. Evaluate:
(i) 
(ii) 

Answer

(i) 
= [∵ a-m = 1/am]
={3 -4} = -1
(ii) 
= 
=  [∵ am ÷ an = am-n]
=  [∵ a-m = 1/am]
= 512/125

7. Simplify:
(i) 
(ii) 

Answer

(i)
= 
=  
= 

(ii)
=
=
= 
= 
= 
= 1×1×3125 [∵ a0 = 1]
= 3125

Page No. 200

Exercise 12.2

1. Express the following numbers in standard form:
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

Answer

(i) 0.0000000000085
= (0.0000000000085×1012)/1012
= 8.5 × 10-12

(ii) 0.00000000000942
= (0.00000000000942×1012)/1012
= 9.42 ×10-12

(iii) 6020000000000000
= (6020000000000000×1015)/1015
= 6.02 × 1015

(iv) 0.00000000837
= (0.00000000837×109)/109
= 8.37 × 10-9

(v) 31860000000
= 31860000000 × 1010/1010 = 3.186 ×1010

2. Express the following numbers in usual form:
(i) 3.02 × 10-6 
(ii) 4.5 × 104 
(iii) 3 × 10-8 
(iv) 1.0001 × 109 
(v) 5.8 × 1012 
(vi) 3.61492 × 106 

Answer

(i) 3.02 × 10-6
= 3.02/106
= 0.00000302

(ii) 4.5 × 104
= 4.5 × 10000
= 45000

(iii) 3 × 10-8
= 3/108
= 0.00000003

(iv) 1.0001 × 109
= 1000100000

(v) 5.8 × 1012
= 5.8 × 1000000000000
= 5800000000000

(vi) 3.61492 × 106
= 3.61492 × 1000000
= 3614920

3. Express the number appearing in the following statements in standard form:
(i)1 micron is equal to 1/1000000 m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m.
(iv) Size of a plant cell is 0.00001275 m.
(v) Thickness if a thick paper is 0.07 mm.

Answer

(i) 1 micron
= 1/1000000
= 1/106 = 1 × 10-6 m

(ii)Charge of an electron is
0.00000000000000000016 coulombs.
= (0.00000000000000000016×1019)/1019
= 1.6 × 10-19 coulomb

(iii) Size of bacteria = 0.0000005
5/10000000
=5×107
=5×10−7

(iv) Size of a plant cell is 0.00001275 m
= (0.00001275 × 105)/105
= 1.275 × 10-5 m

(v) Thickness of a thick paper = 0.07 mm
= 7/100 mm = 7/102 = 7 × 10-2 mm

4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Answer

Thickness of one book = 20 mm
Thickness of 5 books = 20 × 5 = 100 mm
Thickness of one paper = 0.016 mm
Thickness of 5 papers = 0.016 × 5 = 0.08 mm
Total thickness of a stack = 100 + 0.08
=100.08 mm
= 100.08 × 102/102
= 1.0008 × 102 mm

↧
↧

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.2

February 4, 2020, 4:18 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.2

Chapter 11 Conic Sections Exercise 11.2 NCERT Solutions for Class 11 Maths will prove useful guide in completing your homework and improving your problem solving skills. Class 11 Maths NCERT Solutions are prepared by Studyrankers subject matter experts who have given in depth concepts in every question so students can easily understand them.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.2

1. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum
y2 = 12x.

Answer

y 2 = 12x is the equation of parabola => 4a = 12
a = 3

Focus F (3, 0)
Axis is x-axis, i.e., y = 0
Directrix is x = –3
Length of Latus Rectum = 4a = 12

2. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum
x 2 = 6y.

Answer

x 2 = 6y is the equation of parabola
4a = 6,

∴ a = 6/4 = 3/2

Focus (0, 3/2)
Axis is y-axis i.e., x = 0
Directrix y = -(3/2)

3. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
y 2 = –8x.

Answer

The given equation involves y 2, so the axis of symmetry is along the x-axis.
The coefficient of x is negative, so the parabola opens to the left. Comparing with the given equation y 2 = 4ax, we find that a = 2. Thus the focus of the parabola is (–2, 0) and the equation of the directrix of the parabola is x = 2. Length of the Latus rectum LL' is 4a = 4 × 2 = 8


4. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
x 2 = –16y

Answer

The given equation involves x 2, so the axis of symmetry is along the y-axis.
The coefficient of y is negative, so the parabola opens downward. Comparing with the given equation x 2 = – 4ay, we find that a = 4. Thus the focus of the parabola is (0, –4) and the equation of the directrix of the parabola is y = 4. Length of the Latus rectum LL' is 4a = 4 × 4 = 16.

5. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum.
y 2 = 10x

Answer

y 2 = 10x is the equation of parabola
=> 4a = 10, a = 5/2

Focus (5/2 , 0)
Axis is x-axis, i.e., y = 0
Directrix is x = -(5/2)
Length of latus rectum = 10

6. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for given curve
x 2 = –9y

Answer

Equation of the parabola is x 2 = –9y, 4a = 9, a = 9/4

Focus, (0, -(9/4))
Axis of parabola is Y-axis i.e x = 0, Directrix is y = 9/4
and length of latus rectum = 9.

7. Find the equation of the parabola that satisfies the given conditions:
Focus (6, 0); directrix x = –6.

Answer

Since the focus (6, 0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form of either y 2 = 4ax or y 2 = – 4ax. Since the directrix is x = – 6 and the focus is (6, 0) the parabola is to be of the form of y 2 = 4ax with a = 6. Hence, the required equation is y 2 = 4(6)x = 24x.

8. Find the equation of parabola that satisfies the given focus (0, –3) and directrix y = 3.

Answer

Focus (0, –3) and directrix y = 3


vertex is the mid point of (0, –3), (0, 3)
i.e. vertex is (0, 0)
and a = 3, ∴ 4a = 12
∴ Equation of parabola x 2 = –12y

9. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), Focus (3, 0)

Answer

Vertex (0, 0), Focus is (3, 0)
a = 3
∴ 4a = 12
∴ Equation of parabola is y 2 = 12x.

10. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), Focus (–2, 0).

Answer

Vertex (0, 0), Focus is (–2, 0)
a = 2, focus being (–2, 0) it is directed towards OX'.
∴ Equation of parabola is y 2 = –8x.

11. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.

Answer

Since the parabola is symmetric about x-axis and has its vertex at the origin, the equation is of the form y 2 = 4ax or y 2 = – 4ax, where the sign depends on whether the parabola opens to the right or left. But the parabola passes through (2, 3) which lies in the first quadrant, it must open to the right. Thus the equation is of the form y 2 = 4ax.
Since the parabola passes through (2, 3) we have
32 = 4a(2) => 9 = 8a => a = 9/8
Therefore, the equation of the parabola is
y2 = 4(9/8)x = 9/2 x => 2y2 = 9x.

12. Find the equation of the parabola whose vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Answer

Vertex (0, 0), parabola passes through (5, 2) symmetric about y-axis
Let the equation of parabola be x 2 = 4ay (5, 2) lies on it
∴ 25 = 4a. 2 ∴ a = 25/8
∴ Equation of parabola is
x2 = 25/2 y or 2x2 = 25y
↧

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.3

February 4, 2020, 6:58 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.3

If you want NCERT Solutions of Chapter 11 Conic Sections Exercise 11.3 then you can find them here. These Class 11 Maths NCERT Solutions will guide you in covering all the important points and concepts of the chapter easily. These NCERT Solutions are updated as per the latest pattern released by CBSE.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.3

1. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Answer

Since denominator of x 2 is larger than the denominator of y2, the major axis is along the x-axis. Comparing the given equation with
x2/a2 + y2/b2 = 1, we get a = 6 and b = 4.
Also √a2 - b2 = √36 - 16 = √20
Therefore, the coordinates of the foci, (–c, 0) and (c, 0) are (-√20, 0) and (√20, 0). - Vertices, (–a, 0) and (a, 0) are (–6, 0) and (6, 0). Length of the major axis, 2a is 12 units.
The length of the minor axis, 2b is 8 units and eccentricity, e = c/a is √20/6 = √5/3
The length of latus rectum = 2b2/a = 2(4)2/6
= 16/3 units.

2. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x2/4 + y2/25 = 1

Answer

Since the denominator of y 2 is larger than the denominator of x 2, the major axis is along the y-axis. Comparing the given equation with the standard equation
x2/b2 + y2/a2 = 1
We have b = 2 and a = 5
Also c =√a2 - b2 = √25 - 4 = √21
and e = c/a = √21/5
Hence, the foci, (0, c) and (0, –c) are (0, √21) and (0, – √21); the vertices are (0, 5) and (0, –5); length of the major axis, 2a is 10 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is √21/5 . The length of latus rectum = 2b2/a = 8/5 units.

3. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x2/16 + y2/9 = 1

Answer

Equation of ellipse is x2/16 + y2/9 = 1
Here, a 2 = 16, ∴ a = 4 and b 2 = 9, ∴ b = 3 Major axis is along x-axis,
c2 = a2 – b2 = 16 – 9 = 7 ∴ c = √7
Co-ordinates of foci (± c, 0) i.e., (±√7, 0)
Vertices are (± a, 0) i.e., (± 4, 0)
Length of major axis = 2a = 2 × 4 = 8
Length of minor axis = 2b = 2 × 3 = 6
e = c/a = √7/4
Latus rectum = 2b2/a = (2 × 9)/4 = 9/2

4. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x2/25 + y2/100 = 1

Answer

Equation of ellipse is x2/25 + y2/100 = 1
Major axis is along y-axis
a 2 = 100, ∴ a = 10, b 2 = 25 ∴ b = 5
c2 = a2 – b2 = 100 – 25 = 75 ∴ c = 5√3
Foci are (0, ± c) i.e., (0, ± 5√3)
Vertices are (0, ± a) i.e., (0, ± 10)
Length of major axis = 2a = 2 × 10 = 20
Length of minor axis = 2b = 2 × 5 = 10
e = c/a = 5√3/10 = √3/2
Length of Latus rectum = 2b2/a = (2 × 25)/10 = 5

5. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x2/49 + y2/36 = 1

Answer

x2/49 + y2/36 = 1 is the equation of ellipse.
a 2 = 49, major axis is along x-axis
a 2 = 49 ∴ a = 7, b 2 = 36, b = 6
∴ c2 = a2 – b2 = 49 – 36 = 13 c = √13
Foci are (± c, 0) i.e., (± √13, 0)
Vertices are (± a, 0) i.e., (± 7, 0)
Length of major axis = 2b = 2 × 7 = 14
Length of minor axis = 2b = 2 × 6 = 12
Length of Latus rectum = 2b2/a = (2 × 36)/7 = 72/7
Ecentricity, e = c/a = √13/7

6. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x2/100 + y2/400 = 1

Answer

x2/ 100 + y2/400 = 1 is the equation of ellipse.
Major axis is along y-axis
a 2 = 400, ∴ a = 20, b 2 = 100 ∴ b = 10
c2 = a2 – b2 = 400 – 100 = 300 ∴ c = 10√3
Vertices are (0, ± a) i.e., (0, ± 20)
∴ Foci are (0, ± c) i.e., (0, ±10√3)
Length of major axis = 2a = 2 × 20 = 40
Length of minor axis = 2b = 2 × 10 = 20
Eccentricity, e = c/a = 10√3/20 = √3/2
Length of Latus rectum = 2b2/a = (2 × 100)/20 = 10

7. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
36x2 + 4y2 = 144

Answer

The given equation of the ellipse can be written if standard form as
36x2/144 + 4y2/144 = 1
Or x2/4 + y2/36 = 1
Since the denominator of x2 < denominator of y 2, the major axis is along the y-axis. Comparing the given equation with the standard equation
x2/b2 + y2/a2 = 1,
we have b = 2 and a = 6.
Also, c =√a - b =√36 - 4 =√32 = 4√2
e = c/a = √34/6 = 4√2/6 = 2√2/3
Hence, the foci, are (0, 4√2) and (0, – 4√2). Vertices, are (0, 6) and (0, –6); the length of the major axis, 2a is 12 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is 2√2/3.
The length of latus rectum,
2b2/a is (2(22))/6 = 4/3 units.

8. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x 2 + y 2 = 16.

Answer

The ellipse is 16x 2 + y 2 = 16
Dividing by 16 we get x2/1 + y2/ 16 = 1
Major axis is along y-axis a 2 = 16,
=> a = 4, b 2 = 1,
=> b = 1
and c2 = a2 – b2 = 16 – 1 = 15, ∴ c = 15
foci are (0, ± c) i.e. (0, ± 15)
vertices are (0, ± a) i.e. (0, ±4)
Length of major axis = 2a = 2 × 4 = 8;
Length of minor axis = 2b = 2 × 1 = 2
Eccentricity = e = c/a = 15/4 ;
Length of the latus rectum = 2b2/a = (2 × 1)/4 = 1/2

9. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
4x2 + 9y2 = 36

Answer

The given equation of the ellipse can be written in standard form as
4x2/36 + 9y2/36 = 36/36 => x2/9 + y2/4 = 1
Since the denominator of x 2 > denominator of y 2, the major axis is along the x-axis. Comparing the given equation with the standard equation
x2/a2 + y2/b2 = 1
We have a = 3 and b = 2.
Also, c = √a2 - b2 = √9 - 4 = √5
and e = c/a = √5/3
Hence, the foci, are (√5, 0) and (– √5, 0). vertices, are (3, 0) and (–3, 0); the length of the major axis, 2a is 6 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is √5/ 3
The length of latus rectum
= 2b2/a = 2(22)/3 = 8/3 units.

10. Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 5, 0), Foci (± 4, 0)

Answer

Vertices (± 5, 0), Foci (± 4, 0)
=> (± a, 0) = (±5, 0) and (± ae, 0) = (± 4, 0)
∴ a = 5 and ae = 4
 => e = 4/a = 4/5
Also b 2 = a 2 (1 – e 2) (gives)
b2 = 25(1 – 16/25) = 9
b = 3
∴ The equation of ellipse x2/a2 + y2/b2 = 1 becomes
x2/25 + y2/9 = 1 => 9x2 + 25y2 = 225 which is the equation of the required ellipse.

11. Find the equation for the ellipse that satisfies the given conditions:
Vertices (0, ± 13), Foci (0, ± 5)

Answer

Foci (0, ± 5), vertices (0, ± 13)
(0, ± ae) = (0, ± 5) and (0, ± a) = (0, ± 13)
=> ae = 5 and a = 13 ∴ ae/a = 5/13
b2 = a2 – a2e2 = 132 – 52 = 169 – 25 = 144
∴ b= 12
∴ equation of the required ellipse
= x2/144 + y2/169 = 1

12. Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 6, 0), foci (± 4, 0)

Answer

Since the vertices are on x-axis, the equation will be of the form.
x2/a2 + y2/b2 = 1
Where a is the semi-major axis.
Given that a = 6, c = ± 4
Therefore from the relation
c2 = a2 – b2 , we get
16 = 36 – b 2
=> b 2 = 20 or b = 2√5
Hence, the equation of the ellipse is
x2/36 + y2/20 = 1

13. Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Answer

Since the vertices are on x-axis, the equation will be of the form
x2/a2 + y2/b2 = 1
Where a is the semi-major axis.
Given a = 3, b = 2
Hence, the equation of the ellipse is
x2/9 + y2/4 = 1

14. Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (0, ± √5), ends of minor axis (± 1, 0)

Answer

Ends of major axis (0, ± √5).
Major axis is the y-axis and a = √5
Ends of minor axis are (± 1, 0)
b = 1
Equation of ellipse is x2/1 + y2/5 = 1

15. Find the equation for the ellipse that satisfies the given conditions:
Length of major axis 26, Foci (± 5, 0)

Answer

Length of major axis = 2a = 26 ∴ a = 13
Foci are (± 5, 0), c = 5,
∴b 2 = a 2 – c 2
= 169 – 25 = 144
Major axis is x-axis.
Equation of ellipse is x2/169 + y2/144 = 1

16. Find the equation for the ellipse that satisfies the given conditions:
Length of minor axis 16, foci (0, + 6)

Answer

Since the foci are on the y-axis, the major axis will be on y-axis. The equation will be of the form
x2/b2 + y2/a2 = 1
Given 2b = 16 or b = 8 and c = 16
We know that c 2 = a 2 – b 2 or 36 = a 2 – 64
or a2 = 100 or a = 10
Hence, the equation of the ellipse is
x2/64 + y2/100 = 1

17. Find the equation for the ellipse that satisfies the given conditions:
Foci (± 3, 0), a = 4

Answer

Since the foci are on the x-axis, the major axis will be on x-axis. The equation will be of the form
x2/a2 + y2/b2 = 1
Given c = ± 3; a = 4
Therefore, c 2 = a 2 – b 2 => 9 = 16 – b 2
and b 2 = 7 => b = √7
Hence, the equation of the ellipse is x2/16 + y2/ 7 = 1

18. Find the equation for the ellipse that satisfies the given conditions:
b = 3, c = 4, centre at the origin; foci on x -axis.

Answer

Since the foci are on the x-axis, the equation will be of the form
x2/a2 + y2/b2 = 1
Given that b = 3; c = 4
Therefore, c 2 = a 2 – b 2
or 16 = a2 – 9 => a 2 = 25 => a = 5
Hence, the equation of the ellipse is
x2/25 + y2/9 = 1

19. Find the equation of the ellipse whose centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)

Answer

Major axis is y-axis
Let the ellipse be x2/b2 + y2/a2 = 1
since (3, 2) and (1, 6) lies on it
∴ 9/b2 + 4/a2 = 1 …..(i);
1/b2 + 36/a2 = 1 ……(ii)
Subtracting (9 – 1)/b2 + (4 – 36)/a2 = 0
=> 8/b2 – 32/a2 = 0 => 8a2 = 32b2∴ a 2 = 4b 2
Putting the value in (i), it become 9/b2 + 4/4b2 = 1
∴ 10/b2 = 1
∴ b 2 = 10
Now, a 2 = 4b 2 = 4 × 10 = 40
∴ Equation of the ellipse x2/10 + y2/40 = 1

20. If major axis on the x-axis and passes through the points (4, 3) and (6, 2), then find the equation for the ellipse that satisfies the given condition.

Answer

Major axis is x-axis
Let the equation of the ellipse be x2/a2 + y2/b2 = 1
(4, 3) and (6, 2) lies on it 16/a2 + 9/b2 = 1… (i)
36/a2 + 4/b2 = 1 …(ii)
Subtracting -20/a2 + 5/b2 = 0 => 5a 2 = 20 b 2
a 2 = 4b 2
Putting the value of a 2 in eqn. (i) 16/4b2 + 9/b2 = 1
∴ b2 = 13 and a 2 = 4b 2 = 4 × 13 = 52
∴ Equation of ellipse is x2/52 + y2/13 = 1
↧

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

February 4, 2020, 8:57 am
$
0
0

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

Chapter 11 Conic Sections Exercise 11.4 Class 11 Maths NCERT Solutions is very helpful guide in boosting your marks in examinations. NCERT Solutions for Class 11 Maths will help in covering the important concepts related to syllabus through which you can solve difficult questions easily.


1. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola x2/16 – y2/9 = 1.

Answer

Comparing the equation x2/16 – y2/9 = 1 with the standard equation x2/a2 – y2/b2 = 1,
we have, a = 4, b = 3
and c = √(a2 + b2) = √(16 + 9) = √25 = 5
Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 4, 0).
Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b2/a = 2(3)2/4 = 9/2 units.

2. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
y2/9 – x2/27 = 1

Answer

Comparing the equation y2/9 – x2/27 = 1 with the standard equation.
we have, a = 3, b = 3√3
and c = √(a2 + b2) = √(9 + 27) = √36 = √6
Therefore, the coordinates of the foci are (0, ± 6) and that of vertices are (0, ± 3). Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b2/a = 2(3√3)2/3 = 54/3 = 18 units.

3. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola.
9y 2 – 4x 2 = 36

Answer

9y 2 – 4x 2 = 36 is the equation of hyperbola
i.e., y2/4 – x2/9 = 1
∴ a 2 = 4, b 2 = 9, ∴ c 2 = a 2 + b 2 = 4 + 9 = 13,
a = 2, b = 3, c = √3
Axis is y-axis
Foci (0, ± √13), vertices = (0, ± 2)
Eccentricity = e = c/a = √13/2 ,
Latus rectum = 2b2/a = (2 × 9)/2 = 9

4. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola
16x 2 – 9y2 = 576

Answer

Equation of hyperbola is
16x 2 – 9y 2 = 576
or x2/36 – y2/64 = 1
∴ a 2 = 36, b 2 = 64, c 2 = 36 + 64 = 100
∴ a = 6, b = 8, c = 10
Axis is x-axis
Foci are (± 10, 0)
Vertices are (± 6, 0)
Eccentricity = c/a = 10/6 = 5/3
Latus rectum = 2b2/a = (2 × 64)/6 = 64/3

5. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas
5y 2 – 9x 2 = 36

Answer

Equation of hyperbolas is 5y 2 – 9x 2 = 36
=> y2/36/5 - x2/4 = 1 => -x2/4 + y2/36/5 = 1
∴ b2 = 36/5 , a 2 = 4, c 2 = b 2 + a 2
= 4 + 36/5 = 56/5
∴ b = 6/√5 , a = 2, c = 2√14/√5
Axis is along y-axis ∴Foci are (0, ± 2√14/√5) ,
vertices are (0, ± 6/√5)
Eccentricity, e = c/b = 2√14/√5 × √5/6 = √14/3;
Latus rectum = 2a2/b = 2 × 4/6/√5 = 4√5/3

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
49y 2 – 16x 2 = 784

Answer

Dividing the equation by 784 on both sides, we have
49y2/a – 16x2/784 = 1 => y2/16 – x2/49 = 1
Comparing the equation with the standard equation y2/a2 – x2/b2 = 1 we find that a = 4, b = 7 and
c = √(a2 + b2 ) = √( 16 + 49) = √65
Therefore the coordinates of the foci are (0, ± √65) and that of vertices are (0, ± 4). Also the eccentricity e = c/a = √65/4 and the length of latus rectum is 2b2/a = 2(7)2/4 = 49/2 units.

7. Find the equation of the hyperbola satisfying the given conditions:
Vertices (± 2, 0), foci (± 3, 0)

Answer

Since the foci are on x-axis, the equation of the hyperbola is of the form
x2/a2 – y2/b2 = 1
Given: vertices are (± 2, 0), a = 2
Also, since foci are (± 3, 0), c = 3 and
b 2 = c 2 – a 2 = 9 – 4 = 5
Therefore, the equation of the hyperbola is
x2/4 – y2/5 = 1

8. Find the equation of the hyperbola satisfying the given conditions: Vertices (0, ± 5), foci (0, ± 8)

Answer

Since the foci are on y-axis, the equation of the hyperbola is of the form
y2/a2 – x2/b2 = 1
Given: vertices are (0, ± 5), a = 5
Also, since foci are (0, ± 8), c = 8 and
b 2 = c 2 – a 2 = 64 – 25 = 39
Therefore, the equation of the hyperbola is
x2/25 – y2/39 = 1

9. Find the equation of the hyperbola satisfying the given conditions:
Vertices (0, ± 3), foci (0, ± 5)

Answer

Vertices are (0, ± 3) ∴ a = 3
Foci are (0, ±5)
∴ c = 5, b 2 = c 2 – a 2 = 25 – 9 = 16
∴ b = 4
∴ Equation of hyperbola is (axis being y-axis)
y2/9 – x2/16 = 1

10. Find the equation of the hyperbola satisfying the given conditions:
Foci (± 5, 0), the transverse axis is of length 8.

Answer

Foci are (± 5, 0) ∴ c = 5
Transverse axis = 8 ∴ a = 4
b 2 = c 2 – a 2 = 25 – 16 = 9 ∴ b = 3
Axis of hyperbola is x-axis since (± 5, 0) lies on it.
x2/16 – y2/9 = 1

11. Find the equation of the hyperbola satisfying the given conditions:
Foci (0, ± 13), the conjugate axis is of length 24.

Answer

Since the foci are on y-axis, the equation of the hyperbola is of the form
y2/a2 – x2/b2 = 1
Given: foci are (0, ± 13), c = 13
Length of conjugate axis = 2b = 24 => b = 12
or b 2 = c 2 – a 2 or 144 = 169 – a 2
or a 2 = 25 or a = 5
The equation of the hyperbola is
y2/25 – x2/144 = 1
or 144 y 2 – 25x 2 = 3600

12. Find the equation of the hyperbola satisfying the given conditions:
Foci (± 3√5, 0), the latus rectum is of length 8.

Answer

Since the foci are on x-axis, the equation of the hyperbola is of the form
x2/a2 – y2/b2 = 1
Given: foci are (± 3√5, 0), c = 3√5
and length of latus rectum = 2b2/a = 8
As b 2 = 4a
We have c 2 = a 2 + b 2
or 45 = a 2 + 4 a
or a 2 + 4 a – 45 = 0
or a 2 + 9a – 5a – 45 = 0
or a(a + 9) – 5 (a – 9) = 0
or (a + 9)(a – 5) = 0
or a = –9 or a = 5
Since a cannot be negative, we take a = 5 and so b 2 = 20
Therefore, the equation of the required hyperbola is x2/25 – y2/20 = 1 => 4x2 – 5y2 = 100

13. Find the equation of hyperbola satisfying the given condition:
Foci (± 4, 0), the latus rectum is of length 12.

Answer

Foci are (± 4, 0) ∴ c = 4
or c2 = a 2 + b 2∴ 16 = a 2 + b 2 … (i)
Latus rectum = 2b2/a = 12
∴ b 2 = 6a … (ii)
Eliminating b 2 from (i) and (ii)
∴ 16 = a 2 + 6a or a 2 + 6a – 16 = 0
or (a + 8) (a – 2) = 0
a2– 8, ∴ a = 2 ∴ b 2 = 6a = 6 × 2 = 12
a 2 = 4, b 2 = 12, Axis is x-axis
x2/4 – y2/12 = 1

14. Find the equation of hyperbola satisfying the given condition:
Vertices (± 4, 0), e = 4/3

Answer

Veritices are (± 7, 0) ∴ a = 7
e = c/a = 4/3 ∴ c = 4/3 a = 4/3 × 7 = 28/3
b2 = c2 – a2 = (28/3)2 – 49 = 343/9
Axis is along x-axis
x2/49 - y2/343/9 = 1 or 7x2 – 9y2 = 343

15. Find the equation of the hyperbola satisfying the given condition:
Foci (0, ± √10), passing through (2, 3)

Answer

Let the equation of hyperbola be
y2/b2 - x2/a2 = 1 …(i)
∴ be = √10
Also, a 2 = b 2 (e 2 – 1) = b 2 e 2 – b 2 = 10 – b 2 …(ii)
Thus, the equation of the hyperbola is
y2/b2 - x2/(10 – b2) = 1
As, it passes through the point (2, 3)
∴ 9/b2 - 4/(10 – b2) = 1 => b 4 – 23b 2 + 90 = 0
=> (b 2 – 5)(b 2 – 18) = 0 => b 2 = 18, 5
When b 2 = 18, then from (ii) a 2 = –8 which is not possible and when b 2 = 5, then from (ii) a 2 = 5
Hence, the required equation of the hyperbola is y2/5 - x2/5 = 1 => y2 - x2 = 5
↧

NCERT Solutions for Class 8th: Ch 13 Direct and Inverse Proportions Geometry

February 4, 2020, 9:44 am
$
0
0

NCERT Solutions for Chapter 13 Direct and Inverse Proportions Class 8 Mathematics

Page No: 208

Exercise 13.1

1. Following are the car parking charges near a railway station upto:
4 hours Rs.60
8 hours Rs.100
12 hours Rs.140
24 hours Rs.180
Check if the parking charges are in direct proportion to the parking time.

Answer

Charges per hour:
 C1= 60/4 = Rs.15
C2= 100/8 = Rs.12.50
C3= 140/12 = Rs.11.67
C4= 180/24= Rs.7.50
Here, the charges per hour are not same, i.e., C1 ≠ C2 ≠ C3 ≠ C4
Therefore, the parking charges are not in direct proportion to the parking time.

2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Parts of red pigment1471220
Parts of base8…………..……..…….

Answer

Let the ratio of parts of red pigment and parts of base be a/b
Here a1 = 1, b1 = 8
⇒ a1/b1 = 1/8 = k (say)
When a2 = 4, b2 = ?
 k = a2/b2 ⇒ b2 = a2/k = 4/1/8 = 4 × 8 = 32
When a = 7, b = ?
k = a3/b3⇒ b3 = a3/k = 7/1/8 = 7 × 8 = 56
When a = 12, b = ?
k = a4/b4 ⇒ b4 = a4/k = 12/1/8 = 12 × 8 = 96
When a5 = 20, b5 =?
k = a5/b5⇒ b5 = a5/k = 20/1/8 = 20 × 8 = 160

Parts of red pigment1471220
Parts of base8325696160


3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Answer

Let the parts of red pigment mix with 1800 mL base be x.
Parts of red pigment1x
Parts of base751800
Since it is in direct proportion.
∴ 1/75 = x/1800
⇒ 75 × x = 1 × 1800
⇒ x = (1 × 1800)/75 = 24 parts
Hence with base 1800 mL, 24 parts red pigment should be mixed.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Answer

Let the number of bottles filled in five hours be x.
Hours65
Bottles840x
Here ratio of hours and bottles are in direct proportion.
6/840 = 5/x
⇒ 6 × x = 5 × 840
⇒ x = (5 × 840)/6 = 700 bottles
Hence machine will fill 700 bottles in five hours.

5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Answer

Let Actual length of bacteria be 'a'
It is enlarged 50,000 times so 50000 × a = 5 cm
Actual length of bacteria
= 5/50000 = 1/10000 cm = 10-4 cm
Let enlarged length of bacteria be x
Length5x
Enlarged length50,00020,000
Here length and enlarged length of bacteria are in direct proportion.
∴ 5/50000 = x/20000
⇒ x × 50000 = 5 × 20000
⇒ x = (5 × 20000)/50000 = 2 cm
Hence the enlarged length of bacteria is 2 cm.

6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Answer

Let the length of model ship be x.
Length of actual ship (in m)1228
Length of model ship (in cm)9x
Here length of mast and actual length of ship are in direct proportion.
∴ 12/9 = 28/x
⇒ x × 12 = 28 ×9
⇒ x = (28 × 9)/12= 21 cm
Hence length of the model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9× 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?

Answer

(i) Let sugar crystals be x.
Weight of sugar (in kg)25
No. of crystals9 × 106x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/(9 × 106) = 5/x
⇒ x × 2 = 5 × 9 × 106
⇒ x = (5 × 9× 106)/2
= 22.5 × 106 = 2.25 × 107
Hence the number of sugar crystals is 2.25 × 107
Weight of sugar (in kg)21.2
No. of crystals9× 106x

(ii) Let sugar crystals be x.
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/(9 × 106) = 1.2/x
⇒ x × 2 = 1.2 ×9 × 106
⇒ x = (1.2 × 9 × 106)/2
= 0.6 × 9 × 106 = 5.4 × 106
Hence the number of sugar crystals is 5.4 × 106

8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Answer

Let distance covered in the map be x.
Actual distance ( in km)1872
Distance covered in map(in cm)1x
Here actual distance and distance covered in the map are in direct proportion.
∴ 18/1 = 72/x
⇒ x × 18 = 72 × 1
⇒ x = (72 × 1)/18 = 4 cm
Hence distance covered in the map is 4 cm.

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.

Answer

Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5 × 100 + 60 = 560 cm
3 m 20 cm = 3 × 100 + 20 = 320 cm
10 m 50 cm = 10 × 100 + 50 = 1050 cm
5 m = 5 × 100 = 500 cm

(i) Let the length of the shadow of another pole be x.
Height of pole ( in cm)5601050
Length of shadow (in cm)320x
∴ 560/320 = 1050/x
⇒ x ×560 = 1050 × 320
⇒ x = (1050 × 320)/ 560 = 600 cm = 6 m
Hence length of the shadow of another pole is 6 m.

(ii) Let the height of the pole be x.
Height of pole ( in cm)560x
Length of shadow (in cm)320500
∴ 560/320 = x/500
⇒ x × 320 = 560 × 500
⇒ x = (560 × 500)/320
= 875 cm = 8 m 75 cm
Hence height of the pole is 8 m 75 cm.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Answer

Let distance covered in 5 hours be x km.
∵ 1 hour = 60 minutes
∴ 5 hours = 5 × 60 = 300 minutes
Distance (in Km)14x
Time ( in minutes)25300
Here distance covered and time are in direct proportion.
∴ 14/25 = x/300
⇒ x × 25 = 14 × 300
⇒ x = (14 × 300)/25 = 168 km

Page No. 213

Exercise 13.2

1. Which of the following are in inverse proportion:
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.

Answer

(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work.
(ii) Time and distance covered in direct proportion.
(iii) It is a direct proportion because more area of cultivated land will yield more crops.
(iv) Time and speed are inverse proportion because if time is less, speed is more.
(v) It is a inverse proportion. If the population of a country increases, the area of land per person decreases.

2. In a Television game show, the prize money of Rs.1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners:
No. of winners124581020
Prize for each winner (in’)1,00,00050,000---------------

Answer

Here number of winners and prize money are in inverse proportion because winners are increasing, prize money is decreasing.
When the number of winners are 4, each winner will get = 100000/4 = Rs. 25,000
When the number of winners are 5, each winner will get = 100000/5 = Rs. 20,000
When the number of winners are 8, each winner will get = 100000/8 = Rs. 12,500
When the number of winners are 10, each winner will get = 100000/10 = Rs. 10,000
When the number of winners are 20, each winner will get = 100000/20 = Rs. 5,000

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table:
(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
No. of spokes4681012
Angle between a pair of consecutive spokes90°60°------
---

Answer

Here the number of spokes are increasing and the angle between a pair of consecutive spokes is decreasing. So, it is a inverse proportion and angle at the centre of a circle is 360°.
When the number of spokes is 8, then angle between a pair of consecutive spokes = 360°/8 = 45°
When the number of spokes is 10, then angle between a pair of consecutive spokes= 360°/10 = 36°
When the number of spokes is 12, then angle between a pair of consecutive spokes= 360°/12 = 30°


No. of spokes4681012
Angle between a pair of consecutive spokes90°60°45°36°30°

(i) Yes, the number of spokes and the angles formed between a pair of consecutive spokes is in inverse proportion.
(ii) When the number of spokes is 15, then angle between a pair of consecutive spokes= 360°/15 = 24°
(iii) The number of spokes would be needed = 360°/40 = 9°

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Answer

∵ Each child gets = 5 sweets
∴ 24 children will get 24 × 5 = 120 sweets
Total number of sweets = 120
If the number of children is reduced by 4, then children left = 24 – 4 = 20
Now each child will get sweets = 120/20
= 6 sweets

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?

Answer

Let the number of days be x
Total number of animals = 20 + 10 = 30
Animals2030
Days6x
Here the number of animals and the number of days are in inverse proportion.
∴ 20/30 = x/6
⇒ 30 × x = 20 × 6
⇒ x = (20 × 6)/30 = 4
Hence the food will last for four days.

6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?

Answer

Let time taken to complete the job be x
Persons34
Days4x
Here the number of persons and the number of days are in inverse proportion.
∴ 3/4 = x/4
4 × x = 3 × 4
⇒ x = (3 × 4)/4 = 3 days
Hence they will complete the job in 3 days.

7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Answer

Let the number of boxes be x.
No. of bottles in each box1220
Boxes25x
Here the number of bottles and the number of boxes are in inverse proportion.
∴ 12/20 = x/25
⇒ x × 20 = 12 × 25
⇒ x = (12 × 25)/20 = 15
Hence 15 boxes would be filled.

8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Answer

Let the number of machines required be x.
Days6354
Machines42x
Here the number of machines and the number of days are in inverse proportion.
∴ 63/54 = x/42
x × 54 = 63 × 42
x = (63 × 42)/54 = 49
Hence 49 machines would be required.

9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 80 km/hr?

Answer

Let the number of hours be x.
Speed (in km/hr)6080
Time (in hours)2x
Here the speed of car and time are in inverse proportion.
∴ 60/80 = x/2
⇒ x 80 = 60 × 2
⇒ x = (60×2)/80
=3/2
= 1.1/2 hrs.
Hence the car will take 1.1/2 hours to reach its destination.

10. Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?

Answer

(i) Let the number of days be x
Persons21
Days3x
Here the number of persons and the number of days are in inverse proportion.
∴ 2/1 = x/3
⇒ x × 1 = 2 × 3
⇒ x = (2 × 3)/1 = 6 days

(ii) Let the number of persons be x
Persons2X
Days13
Here the number of persons and the number of days are in inverse proportion.
∴ 2/x = 1/3
⇒ x×1 = 2×3
⇒ x = (2×3)/1
= 6 persons

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Answer

Let the duration of each period be x.
Period89
Duration of period(in minutes)45x
Here the number of periods and the duration of periods are in inverse proportion.
∴ 8/9 = x/45
⇒ x×9 = 8×45
⇒ x = (8×45)/9 = 40 minutes
Hence, duration of each period would be 40 minutes.

↧
↧

Study Materials for Class 8th CBSE

February 4, 2020, 12:32 pm
$
0
0

Study Materials for Class 8th CBSE| NCERT Solutions, Revision Notes, Important Questions, Sample Papers and Previous Year Questions Papers

Students of Class 8th need to prepare well as in this class you will build fundamentals that will help you in upper classes. The syllabus is not vast but it is very important as it covers basic things which will help you in competitive exams also. Thus, our experts at StudyRankers have prepared detailed study materials which will in grasping the concepts of the chapter easily. These preparation contains all the important things - NCERT Solutions, Revision Notes, Important Questions, Sample Papers and Previous Year Question Papers - that will help in scoring more marks in the examination.

NCERT Solutions for Class 8

The answers given at the end of each chapter known as NCERT Solutions will help you in checking understanding the chapter. You need to solve these questions after reading the chapter to check your progress. We have prepared NCERT Solutions chapterwise as per the marking scheme. While solving any questions, you're getting problem then you can check these solutions.

Revision Notes for Class 8 NCERT Textbook

Revision Notes are an easy way to look on what you read. The notes for Class 8th prepared by us will come to rescue if have lack of time and prepare yourself quickly. These notes cover all the important points and concepts so you don't have to worry that you have missed an important topic.
  • Notes for Class 8 Math
  • Notes for Class 8 Science
  • Notes for Class 8 English
  • Notes for Class 8 Hindi
  • Notes for Class 8 Social Science

Important Questions for Class 8

In the examination, most of the questions which are asked not present in the NCERT textbook. For solving those questions, you must prepare questions from the insides of the chapter. We have extracted all the important questions from the chapters so you can practice every portion of the chapter.
  • Important Questions for Class 8 Math
  • Important Questions for Class 8 Science
  • Important Questions for Class 8 English
  • Important Questions for Class 8 Hindi
  • Important Questions for Class 8 Social Science

Sample Papers for Class 8

We prepared many sample papers for Class 8 which includes important questions with answers that could be asked in the examinations. These papers can be utilized by you to check your understanding about specific subject so that you can know if you have prepared yourself for that thoroughly or not.

  • Sample Papers for Class 8 Math
  • Sample Papers for Class 8 Science
  • Sample Papers for Class 8 English
  • Sample Papers for Class 8 Hindi
  • Sample Papers for Class 8 Social Science

Previous Year Question Papers

Previous Year Question Papers will help you a lot if you want to find which questions are repeating and very important. You can practice these questions before examination to get more marks. We have provided PYQs along with answers so that you can use that if you find any difficulty while solving.
  • Previous Year Question Papers for Class 8 Math
  • Previous Year Question Papers for Class 8 Science
  • Previous Year Question Papers for Class 8 English
  • Previous Year Question Papers for Class 8 Hindi
  • Previous Year Question Papers for Class 8 Social Science
        ↧

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1

        February 5, 2020, 12:06 am
        $
        0
        0

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1

        If you're looking for NCERT Solutions of Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1 then you're at right place. Class 11 Maths NCERT Solutions will be useful guide that will increase your efficiency so you can easily solve the questions that come in the examinations. It will surely help you to improve your marks and boost your confidence.

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1

        1. A point is on the x-axis. What are its y-coordinate and z-coordinate?

        Answer

        Any point on x-axis is (x 1, 0, 0) i.e y-coordinate and z-coordinate of this point are 0 and 0 respectively.

        2. A point is in the XZ-plane. What can you say about its y-coordinate?

        Answer

        Any point lying on XZ-plane is (x1, 0, z1) i.e. its y-coordinate is zero.


        3. Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5), (–3, –1, 6), (2, –4, –7) 

        Answer

        The octant in which the given point lie:
        (i) (1, 2, 3), XOYZ
        (ii) (4, – 2, 3), XOY’Z
        (iii) (4, – 2, – 5), XOY’Z’
        (iv) (4, 2, – 5), XOYZ’
        (v) (– 4, 2, – 5), X’OYZ’
        (vi) (– 4, 2, 5), X’OYZ
        (vii) (– 3, – 1, 6), X’OY’Z
        (viii) (2, – 4, –7), XOY’Z’

        4. Fill in the blanks:
        (i) The x-axis and y-axis taken together determine a plane known as ...........
        (ii) Coordinates of points in XY-plane are of the form ......
        (iii) Coordinate planes divide the space into ......... octants.

        Answer

        (i) The x-axis and y-axis together determine a plane known as XY-plane.
        (ii) Coordinates of points in XY-plane are of the form (x1, y 1, 0).
        (iii) Coordinate planes divides the space into 8 octants.
        ↧
        Search

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

        February 5, 2020, 1:52 am
        $
        0
        0

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

        NCERT Solutions of Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2 is very helpful if you want to crack examination with good marks. The subject matters experts of Studyrankers have prepared best accurate and detailed NCERT Solutions for Class 11 Maths which will be useful for you in understanding the concepts embedded in a question.

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

        1. Find the distance between the following pairs of points:
        (i) (2, 3, 5), (4, 3, 1)
        (ii) (–3, 7, 2), (2, 4, –1)
        (iii) (–1, 3, –4), (1, –3, 4)
        (iv) (2, –1, 3), (–2, 1, 3)

        Answer

        (i) The given points are (2, 3, 5) and (4, 3, 1).
        Required distance
        = √((4 – 2)2 + (3 – 3)2 + (1– 5)2)
        = √(4 + 0 + 16) = √20 = 2√5
        (ii) The given points are (–3, 7, 2) and (2, 4, –1).
        Required distance
        = √( (2 + 3)2 + (4 – 7)2 + (–1– 2)2)
        = √(25 + 9 + 9) = √43
        (iii) The given points are (–1, 3, – 4) and (1, –3, 4).
        Required distance
        = √( (1 + 1)2 + (–3 – 3)2 + (4 + 4)2)
        = √(4 + 36 + 64) = √104 = 2√26
        (iv) The given points are (2, –1, 3) and (–2, 1, 3).
        Required distance
        = √((2 + 2)2 + (-1 - 1)2 + (3 - 3)2)
        = √(16 + 4 + 0) = √20 = 2√5

        2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

        Answer

        Let the given points be A(–2, 3, 5), B(1, 2, 3) and C(7, 0, –1).

        Thus AB + BC = AC
        Hence, the points A, B, C are collinear.

        3. Verify the following
        (i) (0, 7, – 10), (1, 6, – 6) and (4, 9, –6) are the verticles of an isosceles triangle.
        (ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
        (iii) Verify the given coordinates (–1, 2, 1), (1, –2, 5) (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

        Answer

        (i) Let A(0, 7, – 10), B(1, 6, –6) and C(4, 9, –6) are the given vertices.
        Now

        and

        Thus, AB = BC. Hence A, B and C are the vertices of an isosceles triangle.

        (ii) Let the vertices be A(0, 7, 10), B(–1, 6, 6) and C(–4, 9, 6).

        Hence A, B and C are the vertices of a rightangled triangle.

        (iii) The given points are A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) and D(2, –3, 4), then by distance formula


        Now, since AB = CD and BC = DA i.e., opposite sides are equal and AC ≠ BD  i.e. the diagonals are not equal. So, points are the vertices of parallelogram.

        4. Find the equation of the set of points P, which are equidistant from (1, 2, 3) and (3, 2, –1).

        Answer

        Let A(1, 2, 3) and B(3, 2, – 1) be the given points and the coordinates of points P be (x, y, z) We are given AP = BP or AP2 = BP2



        5. Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(– 4, 0, 0) is equal to 10.

        Answer

        The given points are A(4, 0, 0) and (– 4, 0, 0). Let the point P be (x, y, z)

        Squaring both sides
        Squaring again, we get
        ↧

        KCET 2020 application process begins at kea.kar.nic.in

        February 5, 2020, 3:02 am
        $
        0
        0

        KCET 2020 application process begins at kea.kar.nic.in

        Karnataka Common Entrance Test (KCET) application process will starts from today that is 5th February on the official website- kea.kar.nic.in. As per the official website, the application process will start from 11 am.

        KCET 2020 application process begins at kea.kar.nic.in

        The candidates who want appear for the KCET 2020 can now register and fill up their qualification details. The online application process that starts from today will end on March 2, 2020. The last day to pay application fee online is March 6, 2020. Also, the successful applicants will be given chance to correct their application form, if required, from March 19 to March 25, 2020. KCET 2020 admit cards will be available of download on or after April 10.

        Application Link of KCET 2020

        Click here to apply for KCET 2020 exam.

        The Karnataka Common Entrance Test (CET) 2020 will begin on April 22 and end on April 24. Kannada Language Test for Horanadu and Gadinadu Kannadiga candidates will be held on April 24. The Karnataka CET 2020 exams will be carried out in two shifts.

        On April 22nd (first day of exam), KEA will hold exam for Biology and Mathematics paper. On April 23rd, exams for Physics and Chemistry papers will be held. Each paper will carry 60 marks in total. The Language test will carry 50 marks.

        Eligibility Criteria for KCET 2020


        Eligibility criteria is different for different course in KCET 2020.

        CoursesEligibility Criteria
        B.Tech/BEPassed 12th/2nd PUC with English as one of the language subject and obtaining 45% aggregate marks in Physics & Mathematics along with Chemistry/Bio-technology/Biology/Electronic/Computer (40% for reserve category candidates)
        MBBS/BDS/Ayush CoursesPassed 12th/2nd PUC with English as one of the language subject and obtaining 50% aggregate marks in PCB (40% for reserve category candidates)
        B.ArchPassed 12th/2nd PUC with English as one of the Languages and physics, chemistry & mathematics subjects obtaining 50% aggregate marks in all subjects (45% for reserve category candidates)
        B.PharmaPassed in 2nd PUC/12th with PCB or PCM and obtained a minimum of 45% of marks in aggregate in Physics, Chemistry, Biology subjects (40% for reserve category candidates)
        OR
         Passed in Diploma in Pharmacy from an institution approved by PCI and secured minimum of 45% marks (40% for category candidates)
        B.V. Sc. & A.H.Passed 12th/2nd PUC with obtaining 50% aggregate marks in PCB & English (40% for reserve category candidates)

        Candidates who clear KCET 2020 will be eligible for admission to first year or first semester courses in engineering, technology, farm science, pharmacy, AYUSH (Ayurveda, Yoga and Naturopathy, Unani and Homoeopathy), architecture, medicine and dentistry.

        The medical and dental candidates, have to also clear NEET 2020 held by the Central Board Secondary Education (CBSE) while architecture applicants need to appear for the National Aptitude Test in Architecture (NATA) held by the Council of Architecture or for the Joint Entrance Examination (JEE) paper 2.

        Schedule of KCET 2020

        DateExam
        22nd AprilBiology (10:30 am- 11:50 am)
        Maths (2:30 pm- 3:50 pm)
        23rd AprilPhysics (10:30 am- 11:50 am)
        Chemistry (2:30 pm- 3:50 pm)
        24th AprilKannada language (11:30 am- 12:30 pm).
        ↧
        ↧

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

        February 5, 2020, 11:23 pm
        $
        0
        0

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

        Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3 NCERT Solutions for Class 11 Maths are helpful for the purpose of examinations. You can clear your basic doubts by practicing these questions. Class 11 Maths NCERT Solutions prepared by Studyrankers experts that are correct and detailed so you can use them whenever you find a problem difficult.

        NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

        1. Find the co ordinates of the point which divides the line segment joining the points (–2, 3, 5) and (1, – 4, 6) in the ratio (i) 2: 3 internally (ii) 2: 3 externally.

        Answer

         (i) Let P(x, y, z) be the point which divides the line segment joining the points (–2, 3, 5) and (1, –4, 6) in the ratio 2: 3 internally. Coordinates of point P are given by ratio (l, m)

        (ii) If P divides AB externally in the ratio l: m then the point P is
        ∴The point (–8, 17, 3) divides AB externally in the ratio 2: 3.

        2. Given that P(3, 2, –4), Q(5, 4, –6) and R(9, 8, –10) are collinear. Find the ratio in which Q divides PR.

        Answer

        Let Q be divide PR in the ratio k: 1
        This shows that the point Q lies on it and it divides PR in the ratio 1: 2.

        3. Find the ratio in which the YZ-plane divide the line segment formed by joining the points (– 2, 4, 7) and (3, – 5, 8).

        Answer

        Let the points be A(–2, 4, 7) and B(3, –5, 8) on yz-plane, x-coordinate = 0.
        Let the ratio be K: 1.
        The coordinates of C are
        Hence required ratio is 2: 3.

        4. Using Section Formula, show that the points A(2, – 3, 4), B(–1, 2, 1) and C(0, 1/3 , 2) are collinear.

        Answer

        Given A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) let C divides AB in the ratio K: 1.
        Now coordinates of C are
        ((-k + 2)/(k + 1)) , ((2k – 3)/(k + 1)) , ((k + 4)/(k + 1))
        Let (-k + 2)/(k + 1) = 0 => k = 2
        Again (2k – 3)/(k + 1) = (2 × 2 – 3)/(2 + 1) = 1/3
        And (k + 4)/(k + 1) = (2 + 4)/(2 + 1) = 2
        Thus the point C(0, 1/3 , 2) divides AB in the ratio 2: 1 and is the same as C.

        5. Find the coordinates of the points which trisect the line segment joining the points P(4, 2, – 6) and Q(10, – 16, 6).

        Answer

        Let A(x 1, y 1, z 1) and B(x 2, y 2, z 2) trisect the line segment PQ.
        => A divides PQ in the ratio 1: 2
        ∴ the coordinates of A(x 1, y 1, z 1) are

        {(10 + 8)/(1 + 2) , (-16 + 4)/(1 + 2), ( 6 – 12)/(1 + 2)} or (6, -4, -2)
        Further B divides PQ in the ratio 2: 1

        ∴ The coordinates of B(x2, y2, z2) are
        {(2 × 10 + 4)/(2 + 1) , (-32 + 2)/(2 + 1), (12 – 6)/ ( 2+ 1)} i.e., (8, -10, 2)
        Thus points A(6, -4, -2) and B(8, -10, 2) trisect the line segment PQ.
        ↧

        NCERT Solutions for Class 8th: Ch 14 Factorisation Geometry

        February 5, 2020, 11:54 pm
        $
        0
        0

        NCERT Solutions for Chapter 14 Factorisation Class 8 Mathematics

        Page No: 220

        Exercise 14.1

        1. Find the common factors of the given terms.
        (i) 12x, 36
        (ii) 2y, 22xy
        (iii) 14pq, 28p2q2
        (iv) 2x, 3x2, 4 
        (v) 6abc, 24ab2, 12a2b 
        (vi) 16x3, -4x2, 32x 
        (vii) 10 pq, 20qr, 30rp
        (viii) 3x2y3, 10x3y2, 6x2y2z 

        Answer

        (i) 12x = 2×2×3×x
        36 = 2×2×3×3
        Hence, the common factors are 2, 2 and 3 = 2×2×3 = 12

        (ii) 2y = 2×y
        22xy = 2×11×x×y
        Hence, the common factors are 2 and y = 2×y = 2y

        (iii) 14pq = 2×7×p×q
        28p2q2 = 2×2×7×p×p×q×q
        Hence, the common factors are 2×7×p×q = 14pq

        (iv) 2x = 2×x×1
        3x2 = 3×x×x×1
        4 = 2×2×1
        Hence, the common factor is 1.

        (v) 6abc = 2×3×a×b×c
        24ab2 = 2×2×2×3×a×b×b
        12a2b = 2×2×3×a×a×b
        Hence, the common factors are 2×3×a×b = 6ab

        (vi) 16x3 = 2×2×2×x×x×x
        -4x2 = (-1)×2×2×x×x
        32x = 2×2×2×2×2×x
        Hence the common factors are 2×2×x = 4x

        (vii) 10pq = 2×5×p×q
        20qr = 2×2×5×q×r
        30rp = 2×3×5×r×p
        Hence the common factors are 2×5 = 10

        (viii) 3x2y3 = 3×x×x×y×y×y
        10x3y2 = 2×5×x×x×x×y×y
        6x2y2z = 2×3×x×x×y×y×z
        Hence the common factors are x×x×y×y = x2y2

        2. Factorize the following expressions.
        (i) 7x - 42
        (ii) 6p - 12q
        (iii) 7a2 + 14a 
        (iv) -16z + 20z3 
        (v)20l2m + 30alm 
        (vi) 5x2y - 15xy2 
        (vii) 10a2 - 15b2 + 20c2 
        (viii) -4a2 + 4ab - 4ca 
        (ix) x2yz + xy2z + xyz2 
        (x) ax2y + bxy2 + cxyz 

        Answer

        (i) 7x - 42 = 7×x - 2×3×7
        Taking common factors from each term,
        = 7(x - 2×3)
        = 7(x - 6)

        (ii) 6p - 12q = 2×3×p - 2×2×3×q
        Taking common factors from each term,
        = 2×3(p - 2q)
        = 6(p - 2q)

        (iii) 7a2 + 14a = 7×a×a + 2×7×a
        Taking common factors from each term,
        = 7×a(a + 2)
        = 7a(a + 2)

        (iv) -16z + 20z3
        = (-1)×2×2×2×2×z + 2×2×5×z×z× z
        Taking common factors from each term,
        = 2×2×z (-2×2 + 5×z×z)
        = 4z (-4 + 5z2)

        (v) 20l2m + 30alm
        = 2×2×5×l×l×m + 2×3×5×a×l×m
        Taking common factors from each term,
        = 2×5×l×m(2×l + 3×a)
        = 10 lm(2l +3a)

        (vi) 5x2y - 15xy2
        = 5×x×x×y - 3×5×x×y×y (Taking common factors from each term)
        = 5×x×y(x - 3y)
        = 5xy(x - 3y)

        (vii) 10a2 - 15b2 + 20c2
        = 2×5×a×a - 3×5×b×b + 2×2×5×c×c
        Taking common factors from each term,
        = 5(2×a×a - 3×b×b + 2×2×c×c)
        = 5(2a2 - 3b2 + 4c2)

        (viii) -4a2 + 4ab - 4ca
        = (-1)×2×2×a×a + 2×2×a×b - 2×2×c×a
        Taking common factors from each term,
        = 2×2×a(-a + b -c)
        = 4a (-a + b - c)

        (ix) x2yz + xy2z + xyz2
        = x×x×y×z + x×y×y×z + z×y×z×z
        Taking common factors from each term,
        = x×y×z( x + y + z)
        = xyz(x + y +z)

        (x) ax2y + bxy2 + cxyz
        = a×x×x×y + b×x×y×y + c×x×y×z
        Taking common factors from each term,
        = x×y(a×x + b×y + c×z)
        = xy(ax + by +cz)

        3. Factorize:
        (i) x2 + xy + 8x + 8y 
        (ii) 15xy - 6x + 5y -2 
        (iii) ax + bx - ay - by 
        (iv) 15pq + 15 + 9q + 25p 
        (v) z - 7 + 7xy -xyz 

        Answer

        (i) x2 + xy + 8x + 8y
        = x(x + y) + 8(x + y)
        = (x + y)(x + 8)

        (ii) 15xy - 6x + 5y - 2
        = 3x(5y - 2) + 1(5y - 2)
        = (5y -2)(3x + 1)

        (iii) ax + bx - ay - by
        = (ax + bx) - (ay + by)
        = x(a + b) - y(a + b)
        = (a + b)(x - y)

        (iv) 15pq + 15 + 9q + 25p
        = 15pq + 25p + 9q + 15
        = 5p(3q + 5) + 3(3q + 5)
        = (3q + 5)(5p + 3)

        (v) z -7 + 7xy - xyz = 7xy - 7 - xyz + z
        = 7(xy - 1) - z(xy - 1)
        = (xy -1)(7 - z) = (-1)(1 - xy)(-1)(z - 7)
        = (1 - xy)(z - 7)

        Page No. 223

        Exercise 14.2

        1. Factorize the following expressions:(i) a2 + 8a + 16 
        (ii) p2 - 10p + 25 
        (iii) 25m2 + 30m + 9 
        (iv) 49y2 + 84yz + 36z2
        (v) 4x2 - 8x + 4
        (vi) 121b2 - 88bc + 16c2 
        (vii) (l + m)2 - 4lm [Hint: Expand (l + m)2 first]
        (viii) a4 + 2a2b2 + b4 

        Answer

        (i) a2 + 8a + 16 = a2 + (4 + 4)a + 4 × 4
        Using identity x2 + (a + b)x + ab = (x + a)(x + b),
        Here x = a, a = 4 and b = 4
        a2 + 8a + 16 = (a + 4)(a + 4) = (a + 4)2

        (ii) p2 - 10p + 25 = p2 +(-5-5)p + (-5)(-5)
        Using identity x2 + (a +b)x + ab = ( x + a)(x + b),
        Here x = p, a = -5 and b = -5
        p2 - 10p + 25 = (p -5)(p- 5) = (p - 5)2


        (iii) 25m2 + 30m + 9 = (5m)2 + 2 × 5m × 3 + (3)2
        Using identity a2 + 2ab + b2 = (a + b)2 , here a= 5m, b = 3
         25m2 + 30m + 9 = (5m + 3)2

        (iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 × 7y × 6z + (6z)2
        Using identity a2 + 2ab + b2 = (a + b)2 , here a = 7y, b = 6z
        49y2 + 84yz + 36z2 = (7y + 6z)2

        (v) 4x2 - 8x + 4 = (2x)2 - 2 × 2x ×2 + (2)2
        Using identity a2 - 2ab + b2 = (a - b)2 , here a = 2x, b = 2
        4x2 - 8x + 4 = (2x - 2)2
        = (2)2 (x - 1)2 = 4( x - 1)2

        (vi) 121b2 - 88bc + 16c2 = (11b)2 - 2 × 11b × 4c + (4c)2
        Using identity a2 - 2ab + b2 = (a - b)2 , here a = 11b, b = 4c
        121b2 - 88bc + 16c2 = (11b - 4c)2

        (vii) (l + m)2 - 4lm
        = l2 + 2 × l ×m + m2 - 4lm [ ∵ (a + b)2 = a2 + 2ab + b2 ]
        = l2 + 2lm + m2 - 4lm
        = l2 - 2lm + m2
        = (l - m)2 [ ∵ (a- b)2 = a2 - 2ab + b2 ]


        (viii) a4 + 2a2b2 + b4 = (a2)2 + 2 × a2 × b2 + (b2)2
        = (a2 + b2)2 [∵ (a + b)2 = a2 + 2ab + b ]

        2. Factorize:
        (i) 4p2 - 9q2         (ii) 63a2 - 112b2 
        (iii) 49x2 - 36
        (iv) 16x5 - 144x2 
        (v) (l + m)2 - (l -m)2 
        (vi) 9x2y2 - 16 
        (vii) (x2 - 2xy + y2) - z2 
        (viii) 25a2 - 4b2 + 28bc - 49c2 

        Answer

        (i) 4p2 - 9q2 = (2p)2 - (3q)2
        = (2p -3q)(2p + 3q) [∵ a2 - b2 = (a - b)(a +b)]


        (ii) 63a2 - 112b2 = 7(9a2 - 16b2)
        = 7 [ (3a)2 - (4b)2]
        = 7(3a - 4b)(3a + 4b)  [∵ a2 - b2 = (a - b)(a +b)]


        (iii) 49x2 - 36 = (7x)2 - (6)2
        = (7x - 6)(7x + 6)  [∵ a2 - b2 = (a - b)(a +b)]


        (iv) 16x5 - 144x3 = 16x3(x2 - 9)
        = 16x3 [(x)2 - (3)2]
        = 16x3 (x - 3)(x + 3) [∵ a2 - b2 = (a - b)(a +b)]


        (v) (l + m)2 - (l - m)2
        = [(l + m) + ( l - m)][(l + m)- (l - m)] [∵ a2 - b2 = (a - b)(a +b)]
        = (l + m + l - m)(l + m - l +m)
        = (2l) (2m) = 4lm

        (vi) 9x2y2 - 16 = (3xy)2 - (4)2
        = (3xy - 4)(3xy + 4) [ ∵ a2 - b2 = (a - b)(a +b)]

        (vii) ( x2 - 2xy + y2) - z2 = ( x - y)2 - z2   [∵ (a -b)2 = a2 -2ab + b2]
         = ( x - y - z)( x - y + z) [ ∵ a2 - b2 = (a - b)(a +b)]
        (viii) 25a2 - 4b2 + 28bs - 49c2
        = 25a2 - (4b2 - 28bc + 49c2)
        = 25a2 - [ (2b)2 - 2 × 2b × 7c + (7c)2]
        = 25a2 - (2b - 7c)2 [ ∵ (a -b)2 = a2 -2ab + b2]
        = (5a)2 - (2b - 7c)2
        = [5a - (2b - 7c)][5a + (2b - 7c)] [ ∵ a2 - b2 = (a - b)(a +b)]
        = (5a - 2b + 7c)(5a + 2b - 7c)

        3. Factorize the expressions:
        (i) ax2 + bx(ii) 7p2 + 21q2 
        (iii) 2x3 + 2xy2 + 2xz2
        (iv) am2 + bm2 + bn2 + an2 
        (v) (lm + l ) + m + 1
        (vi) y( y + z) + 9 ( y + z) 
        (vii) 5y2 - 20y - 8z + 2yz
        (viii) 10ab + 4a + 5b + 2 
        (ix) 6xy - 4y + 6 - 9x

        Answer

        (i) ax2 + bx = x(ax + b)

        (ii) 7p2 + 21q2 = 7(p2 + 3q2)

        (iii) 2x3 + 2xy2 + 2xz2 = 2x( x2 + y2 + z2)

        (iv) am2 + bm2 + bn2 + an2
        = m2( a + b) + n2(a + b)
        = (a + b )(m2 + n2)

        (v) (lm + l) + m + 1
        = l(m + 1) + 1(m + 1)
        = (m + 1)( l + 1)

        (vi) y(y + z) + 9(y + z)
        = (y + z)(y + 9)

        (vii) 5y2 - 20y - 8z + 2yz
        = 5y2 - 20y + 2yz - 8z
        = 5y(y - 4) + 2z(y - 4)
        = (y - 4)(5y + 2z)

        (viii) 10ab + 4a + 5b + 2
        = 2a(5b + 2) + 1 (5b + 2)
        = (5b + 2)(2a + 1)

        (ix) 6xy - 4y + 6 - 9x
        = 6xy - 9x - 4y + 6
        = 3x(2y - 3) - 2(2y - 3)
        = (2y - 3) (3x - 2)

        4. Factorize:
        (i) a4 - b4
        (ii) p4 - 81 
        (iii) x4 - (y + z)4
        (iv)x4 - (x -z)4 
        (v) a4 - 2a2b2 + b4 

        Answer

        (i) a4 - b4 = (a2)2 - (b2)2
        = (a2 - b2)( a2 + b2) [ ∵ a2 - b2 = (a - b)(a +b)]
        = (a - b)(a + b)(a2 + b2) [ ∵ a2 - b2 = (a - b)(a +b)]

        (ii) p4 - 81 = (p2)2 - (9)2
        = (p2 - 9)(p2 + 9) [∵ a2 - b2 = (a - b)(a +b)]
        = (p2 - 32)(p2 + 9)
        = ( p - 3)(p + 3)(p2 + 9) [∵ a2 - b2 = (a - b)(a +b)]

        (iii) x4 - (y + z)4 = (x2)2 - [(y + z)2]2
        = [x2 - (y + z)2][ x2 + (y + z)2] [∵ a2 - b2 = (a - b)(a +b)]
        = [x -(y +z)][x + (y + z)][x2 + (y + z)2] [∵ a2 - b2 = (a - b)(a +b)]
        = (x - y - z) (x + y + z) [x2 + (y + z)2]

        (iv) x4 - (x - z)4 = (x2)2 - [(x - z)2]2
        = [x2 -(x - z)2][x2 + (x - z)2] [ ∵ a2 - b2 = (a - b)(a +b)]
        = [x - (x - z)][x + (x - z)] [x2 + (x - z)2] [∵ a2 - b2 = (a - b)(a +b)]
        = [x - x + z] [x + x - z] [x2 + x2 - 2xz + z2] [∵ (a -b)2 = a2 -2ab + b2]
        = z(2x - z) (2x2 - 2xz + z2)

        (v) a4 - 2a2b2 + b4 = (a2)2 - 2a2b2 + (b2)2
        = (a2 - b2)2 [∵ a2 - b2 = (a - b)(a +b)]
        = [(a - b)(a + b)]2 [∵ a2 - b2 = (a - b)(a +b)]
        = (a -b)2 (a + b)2 [ (xy)m = xmym]

        5. Factorize the following expressions:
        (i) p2 + 6p + 8
        (ii) q2 - 10q + 21 
        (iii) p2 + 6p - 16 

        Answer

        (i) p2 + 6p + 8 = p2 + ( 4 + 2)p + 4 × 2
        = p2 + 4p + 2p + 4 ×2
        = p(p + 4) + 2 ( p + 4)
        = (p + 4)(p + 2)

        (ii) q2 - 10q + 21 = q2 - ( 7 + 3)q + 7 × 3
        = q2 - 7q - 3q + 7 × 3
        = q(q - 7) - 3(q - 7)
        = (q - 7)( q - 3)

        (iii) p2 + 6p - 16
        = p2 + (8 - 2)p - 8×2
        = p2 + 8p - 2p - 8×2
        = p(p + 8) - 2(p + 8)
        = ( p + 8)(p -2)

        Page No. 227

        Exercise 14.3

        1. Carry out the following divisions:
        (i) 2x4 ÷ 56x
        (ii) -36y3 ÷ 9y2
        (iii) 66pq2r3 ÷ 11 qr2
        (iv) 34x3y3x3 ÷ 51xy2z3
        (v) 12a8b8 ÷ (-6a6b4)

        Answer

         (i) 2x4 ÷ 56x
        = 28x4/56x
        = 28/56 × x4/x
        = 1/2 x3 [xm ÷ xn = xm-n]

        (ii) -36y3 ÷ 9y2 = -36y3/9y2
        = -36/9 × y3/y2
        = -4y [xm ÷ xn = xm-n]

        (iii) 66pq2r3 ÷ 11qr2
        = 66pq2r3/11qr2
        = 66/11 × pq2r3/qr2
        = 6pqr [xm ÷ xn = xm-n]

        (iv) 34x3y3z3 ÷ 51xy2z3
        = 34x3y3z3/51xy2z3
        = 34/51 ×x3y3z3/xy2z3
        = 2/3x2y [xm ÷ xn = xm-n]

        (v) 12a8b8 ÷ (- 6a6b4)
        = 12a8b8/- 6a6b4
        = 12/-6 × a8b8/a6b4
        = -2a2b4 [xm ÷ xn = xm-n]

        2. Divide the given polynomial by the given monomial:
        (i) (5x2 - 6x) ÷ 3x
        (ii) (3y8 - 4y6 + 5y4) ÷ y4
        (iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
        (iv) (x3 + 2x2 + 3x) ÷2x
        (v) (p3q6 - p6q3) ÷ p3q3

        Answer

        (i) (5x2 - 6x) ÷3x
        = (5x2 - 6x)/3x
        = 5x2/3x - 6x/3x = (5/3)x - 2 = 1/3 (5x - 6)

        (ii) (3y8 - 4y6 + 5y4) ÷ y4
        = (3y8 - 4y6 + 5y4)/ y4
        = 3y8/y4 - 4y6/y4 + 5y4/y4 = 3y4 - 4y2 + 5

        (iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
        = {8(x3y2z2 + x2y3z2 + x2y2z3)}/4 x2y2z2
        = 8 x3y2z2/4 x2y2z2 + 8 x2y3z2/4x2y2z2 + 8 x2y2z3/4x2y2z2
        = 2x + 2y + 2z
        = 2(x + y + z)

        (iv) (x3 + 2x2 + 3x) ÷ 2x
        = (x3 + 2x2 + 3x)/2x
        = x3/2x + 2x2/2x + 3x/2x = x2/2 + 2x/2 + 3/2
        = 1/2( x2 + 2x + 3)
        (v) (p3q6 - p6q3) ÷ p3q3
        = (p3q6 - p6q3)/p3q3
        = p3q6/p3q3 - p6q3/p3q3 = q3 - p3

        3. Work out the following divisions:
        (i) (10x - 25) ÷ 5
        (ii) (10x - 25) ÷ (2x - 5)
        (iii) 10y (6y + 21) ÷ 5(2y + 7)
        (iv) 9x2y2(3z - 24) ÷ 27xy(z - 8)
        (v) 96abc(3a - 12)(5b - 30) ÷ 144(a -4)(b - 6)

        Answer

        (i) (10x - 25) ÷ 5
        = (10x - 25)/5
        = {5(2x - 5)}/5
        = 2x -5

        (ii) (10x - 25) ÷ (2x - 5)
        = (10x - 25)/(2x - 5)
        = {5(2x - 5)/(2x - 5)
        = 5

        (iii) 10y(6y + 21) ÷ 5(2y + 7)
        = {10y(6y + 21)}/5(2y + 7)
        = {2×5×y× 3(2y + 7)}/5(2y + 7)
        = 2×y×3
        = 6y

        (iv) 9x2y2(3z - 24) ÷ 27xy(z - 8)
        = {9x2y2(3z - 24)}/27xy(z - 8)
        = 9/27 × {xy × xy × 3(z - 8)}/xy(z - 8)
        = xy

        (v) 96abc(3a - 12)(5b - 30) ÷ 144(a- 4)(b - 6)
        = {96abc(3a - 12)(5b - 30)}/144(a - 4)(b - 6)
        = {12×4×2×abc× 3(a-4) × 5(b-6)}/{12×4×3 (a - 4)(b - 6)
        = 10abc

        4. Divide as directed:
        (i) 5(2x + 1)(3x + 5) ÷ (2x + 1)
        (ii) 26xy(x + 5)(y - 4) ÷ 13x(y - 4)
        (iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
        (iv) 20(y + 4)(y2 + 5y + 3) ÷ 5(y + 4)
        (v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)

        Answer

        (i) 5(2x + 1)(3x + 5) ÷ (2x + 1)
        = {5(2x + 1)(3x +5)}/(2x + 1)
        = 5(3x + 5)

        (ii) 26xy( x + 5)(y - 4) ÷ 13x(y - 4)
        26xy( x + 5)(y -4) ÷ 13x(y - 4)
        = {26xy(x + 5)(y - 4)}/13x(y - 4)
        = {13×2×xy(x + 5)(y - 4)}/13x(y - 4)
        = 2y(x + 5)

        (iii) 52pqr( p + q)(q + r)( r + p) ÷ 104pq(q + r)(r + p)
        = {52pqr(p + q)(q + r)( r + p)}/{52 × 2 × pq(q + r)(r + p)}
        = (1/2)r (p + q)

        (iv) 20( y + 4)(y2 + 5y + 3) ÷ 5(y + 4)
        = {20(y + 4)(y2 + 5y + 3)}/5(y + 4)
        = 4(y2 + 5y + 3)

        (v) x( x + 1)(x + 2)(x + 3) ÷ x(x + 1)
        = {x(x + 1)(x + 2)(x + 3)}/x(x + 1)
        = (x + 2)(x + 3)

        5. Factorize the expressions and divide them as directed:
        (i) (y2 + 7y + 10) ÷ (y + 5)
        (ii) (m2 - 14m - 32) ÷ (m + 2)
        (iii) (5p2 - 25p + 20) ÷ (p - 1)
        (iv) 4yz(z2 + 6z - 16) ÷ 2y( z + 8)
        (v) 5pq(p2 - q2) ÷ 2p(p + q)
        (vi) 12xy(9x2 - 16y2) ÷ 4xy(3x + 4y)
        (vii) 39y3(50y2 - 98) ÷ 26y2(5y + 7)

        Answer

        (i) (y2 + 7y + 10) ÷ (y + 5)
        = (y2 + 7y + 10)/(y + 5)
        = {y2 + ( 2 + 5)y + 2 × 5}/(y +5)
        = (y2 + 2y + 5y + 2 × 5)/(y + 5)
        = {(y + 2)(y + 5)}/(y + 5) [∵ x2 + (a+b)x + ab = (x +a)(x+b)]
        = y + 2

        (ii) (m2 - 14m + 32) ÷ (m + 2)
        = (m2 - 14m + 32)/(m +2)
        = { m2 + (-16 + 2)m + (-16) × 2}/(m + 2)
        = {(m - 16)(m + 2)}/(m +2) [∵ x2 + (a+b)x + ab = (x +a)(x+b)]
        = (m - 16)

        (iii) (5p2 - 25p + 20) ÷ (p -1)
        = (5p2 - 25p + 20)/(p -1)
        = (5p2 - 20p -5p + 20)/(p -1)
        = {5p(p - 4) -5 (p - 4)}/(p -1)
        = {(5p - 5)(p - 4)}/(p -1) = {5(p -1)(p -4)}/(p - 1)
        = 5 (p - 4)

        (iv) 4yz (z2 + 6z - 16) ÷ 2y(z + 8)
        = {4yz(z2 + 6z - 16)}/2y(z + 8)
        = [4yz{z2 + (8 - 2)z + 8 × (-2)}]/2y(z + 8)
        = {4yz(z - 2)(z + 8)}/2y(z + 8) [∵ x2 + (a+b)x + ab = (x +a)(x+b)]
        = 2z ( z -2)

        (v) 5pq(p2 - q2) ÷ 2p( p + q)
        = {5pq(p2 - q2)}/2p(p + q)
        = {5pq(p - q)(p + q)}/2p( p + q) [∵ a2 - b2 = (a - b)(a + b)]
        = (5/2)q (p - q)

        (vi) 12xy(9x2 - 16y2) ÷ 4xy(3x + 4y)
        = {12xy (9x2 - 16y2)}/4xy(3x + 4y)
        = {12xy[(3x)2 - (4y)2]}/4xy(3x + 4y)
        = {12xy(3x - 4y)(3x + 4y)}/4xy(3x + 4y) [∵ a2 - b2 = (a - b)(a + b)]
        = 3(3x - 4y)

        (vii) 39y3(50y2 - 98) ÷ 26y2(5y + 7)
        = {39y3(50y2 - 98)}/26y2(5y + 7)
        = {39y3 × 2(25y2 - 49)}/26y2(5y + 7)
        = {39y2 × 2[(5y)2 - (7)2]}/26y2(5y + 7)
        = {39y2 × 2(5y - 7)(5y + 7)}/26y2(5y + 7) [∵ a2 - b2 = (a - b)(a + b)]
        = 3y(5y - 7)

        Page No. 228

        Exercise 14.4


        1. Find and correct the errors in the following mathematical statements:
        4(x-5) = 4x-5

        Answer

        L.H.S. = 4(x-5) = 4x- 20 ≠R.H.S.
        Hence, the correct mathematical statement is 4(x-5) = 4x- 20.

        2. x(3x+2) = 3x2+ 2

        Answer

        L.H.S. = x(3x+2) = 3x2+ 2 ≠ R.H.S.
        Hence, the correct mathematical statement is x(3x+2) = 3x2+ 2

        3. 2x + 3y = 5xy

        Answer

        L.H.S. = 2x + 3y ≠ R.H.S.
        Hence, the correct mathematical statement is 2x+ 3y = 2x+ 3y

        4. x+ 2x +3x = 5x

        Answer

        L.H.S. = x+ 2x + 3x = 6x ≠R.H.S.
        Hence, the correct mathematical statement is x+ 2x + 3x = 6x.

        5. 5y + 2y+ y-7y = 0

        Answer

        L.H.S. = 5y + 2y+ y - 7y = 8y-7y = y ≠ R.H.S.
        Hence, the correct mathematical statement is 5y+ 2y+y- 7y = 4

        6. 3x+2x = 5 x2

        Answer

        L.H.S. = 3x+ 2x = 5x ≠ R.H.S.
        Hence the correct mathematical statement is 3x+ 2x = 5x

        7. (2x)2+ 4(2x) + 7 = 2x2+ 8x+ 7

        Answer

        L.H.S. = (2x)2 + 4(2x) + 7 = 4x2 + 8x+ 7 ≠ R.H.S.
        Hence, the correct mathematical statement is (2x)2 + 4(2x) + 7 = 4x2 + 8x+ 7

        8. (2x)2+ 5x = 4x+ 5x = 9x

        Answer

        L.H.S. = (2x)2 + 5x = 4x2+ 5x ≠ R.H.S.
        Hence the correct mathematical statement is (2x)2 + 5x = 4x2+ 5x.

        9. (3x + 2)2= 3x2 + 6x + 4

        Answer

        L.H.S. = (3x + 2)2 = 3x2 + 2 × 3x × 2+ (2)2
        = 9x2 + 12x + 4 ≠ RHS
        Hence, the correct mathematical statementsis (3x + 2)2 = 9x2 + 12X + 4 × 3x

        10. Substituting x = -3 in:
        (a) x2 + 5X + 4 gives (-3)2 + 5(-3) + 4 = 9+ 2+4 = 15
        (b) x2 - 5X + 4 gives (-3)2 - 5(-3) + 4 = 9 - 15 + 4 = -2
        (c) x2 + 5X gives (-3)2 + 5(-3) = -9 - 15 = -24

        Answer

        (a) L.H.S. = x2 + 5x + 4
        Putting x = -3 in given expression,
         = (-3)2 + 5(-3) + 4 = 9 - 15 + 4 = -2 R.H.S.
        Hence, x2 + 5x + 4 gives (-3)2 + 5(-3) + 4 = 9 - 15 + 4 = -2


        (b) L.H.S. = x2 - 5X + 4
        Putting x = -3 cin given expression,
         = (-3)2 - 5(-3) + 4 = 9 + 15 + 4 = 28 ≠ R.H.S.
        Hence x2 – 5x + 4 gives (-3)2 - 5(-3) + 4 = 9 + 15 + 4 = 28


        (c) L.H.S. = x2 + 5X
        Putting x= -3 in given expression,
         = (-3)2 + 5(-3) = 9 - 15 = -6 ≠ R.H.S.
        Hence, x2 + 5X gives (-3)2 + 5(-3) = 9 - 15 = -6

        11. (y-3)2= y2 - 9 

        Answer

        L.H.S. = (y-3)2 = y2 - 2 × y × 3 +(3)2  [ (a-b)2 = a2 - 2ab + b2]
        = y2 - 6y + 9 ≠ R.H.S.
        Hence, the correct statement is (y-3)2 = y2 - 6y + 9

        12. (z+5)2 = z2 + 25

        Answer

        L.H.S. = (z+5)2 = z2 + 2 × z×5+ (5)2
        = z2 + 10z +25 [ (a-b)2 = a2 - 2ab + b2]
        Hence, the correct statement is (z+5)2 = z2 + 10z + 25

        13. (2a +3b)(a-b) = 2a2- 3b2

        Answer

        L.H.S. = (2a + 3b)(a-b) = 2a(a-b) + 3b(a-b)
        = 2a2 - 2ab + 3ab - 3b2
        = 2a2 + ab - 3b2≠ R.H.S.
        Hence, the correct statement is (2a +3b)(a-b) = 2a2 + ab - 3b2

        14. (a + 4) (a + 2) = a2+ 8

        Answer

        L.H.S. = (a+4)(a+2) =a(a+2) + 4(a+2)
        = a2 + 2a + 4a + 8 = a2 + 6a + 8 ≠ R.H.S.
        Hence, the correct statement is (a+4)(a+2) = a2+6a+ 8

        15. (a-4)(a-2) = a2- 8

        Answer

        L.H.S. = (a-4)(a-2) = a(a-2)-4(a-2)
         = a2 - 2a -4a+8 = a2- 6a + 8 ≠ R.H.S
        Hence, the correct statement is (a-4)(a-2) = a2- 6a + 8

        16. 3x2/3x2= 0 

        Answer

        L.H.S. = 3x2/3x2 =1/1 = 1 ≠ R.H.S.
        Hence, the correct statement is 3x2/3x2 =1

        17. 3x2 + 1 / 3x2 = 1+ 1 = 2

        Answer

        L.H.S. = 3x2 + 1 / 3x2 = 3x2/ 3x2 + 1/3x2
        = 1 + 1 / 3x2 R.H.S.
        Hence, the correct statement is 3x2 + 1 / 3x2 = 1 + 1/3x2 

        18. 3x/(3x+2) = 1/2

        Answer

        L.H.S. = 3x/(3x+2) ≠ R.H.S.
        Hence, the correct statement is 3x/(3x+2) = 3x/(3x+2)

        19. 3/(4x+3) = 1/4x

        Answer

        L.H.S. = 3/(4x+3) ≠ R.H.S.
        Hence, the correct statement is 3/(4x+3) = 3/(4x+3)

        20. (4x+5)/4x = 5

        Answer

        L.H.S. = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠R.H.S.
        Hence the correct statement is (4x+ 5)/4x = 1 + 5/4x

        21. (7x+5)/5 = 7x

        Answer

        L.H.S. = (7x+5)/5 = 7x/5 + 5/5 = 7x/5 + 1 ≠ R.H.S.
        Hence, the correct statement is (7x+5)/5 = 7x/5 +1
        ↧

        NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1

        February 6, 2020, 12:29 am
        $
        0
        0

        NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1

        Finding accurate NCERT Solutions of Chapter 13 Limits and Derivatives Exercise 13.1 can be tough. Thus, we at Studyrankers prepared detailed and accurate NCERT Solutions for Class 11 Maths of every question so you can find perfect solutions easily. These will help you in completing your homework and improve your problem solving ability.

        NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1

        1. Evaluate the following limits:

        Answer


        2. 

        Answer


        3. 

        Answer


        4.

        Answer


        5. 

        Answer


        6.

        Answer


        7.

        Answer


        8. 

        Answer


        9.

        Answer


        10.

        Answer


        11.

        Answer


        12.

        Answer


        13.

        Answer


        14.

        Answer


        15.

        Answer


        16.

        Answer


        17.

        Answer


        18.

        Answer


        19.

        Answer


        20.

        Answer


        21.

        Answer


        22.

        Answer


        23.

        Answer


        24.

        Answer


        25.

        Answer


        26.

        Answer


        27.

        Answer


        28.

        Answer


        29.

        Answer


        30.

        Answer


        31.

        Answer


        32.

        Answer


        ↧
        Viewing all 6190 articles
        Browse latest View live
        Search
        <script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>