RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3

Other Exercises

Factorize each of the following algebraic expressions.

Question 1.
6x (2x – y) + 7y (2x – y)
Solution:
6x (2x – y) + 7y (2x – y)
= (2x – y) (6x + 7y)
[∵ (2x – y) is common]

Question 2.
2r (y – x) + s (x – y)
Solution:
2r (y – x) + s (x – y)
-2r (x – y) +s (x – y)
= (x – y) (-2r + s)                   [(x – y) is common]
= (x-y) (s-2r)

Question 3.
7a (2x – 3) + 2b (2x – 3)
Solution:
7a (2x – 3) + 3b (2x – 3)
= (2x – 3) (7a + 3b)               [(2x – 3) is common]

Question 4.
9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
9a (6a – 5b) – 12a2 (6a – 5b)
HCF of 9 and 12 = 3
∴ 3a (6a – 5b) (3 – 4a)
{(6a – 5b) is common}

Question 5.
5 (x – 2y)2 + 3 (x – 2y)
Solution:
5 (x – 2y)2 + 3 (x – 2y)
= 5 (x – 2y) (x – 2y) + 3 (x – 2y)
= (x – 2y) {5 (x – 2y) + 3}
{(x – 2y) is common}
= (x – 2y) (5x – 10y + 3)

Question 6.
16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
16 (2l – 3m)2 – 12 (3m-2l)
= 16 (2l – 3m) (2l – 3m) + 12 (2l – 3m)
HCF of 16, 12 = 4 4 (2l-3m) {4 (2l- 3m) + 3}
{(2l – 3m) is common}
= 4 (2l -3m) (8l- 12m+ 3)

Question 7.
3a (x – 2y) – b (x – 2y)
Solution:
3a (x – 2y) – b (x – 2y)
= (x – 2y) (3a – b)
{(x – 2y) is common}

Question 8.
a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
a2 (x + y) + b2 (x + y) + c2 (x + 3’)
= (x + y) (a2 + b2 + c2)
{(x + y) is common}

Question 9.
(x-y)2 + (x -y)
Solution:
(x – y)2 + (x- y) = (x – y) (x – y) + (x – y)
= (x – y) (x – y + 1)                          {(a – y) is common}

Question 10.
6 (a + 2b) – 4 (a + 2b)2
Solution:
6 (a + 2b) – 4 (a + 2b)2
= 6 (a + 2b) – 4 (a + 2b) (a + 2b)
HCF of 6, 4 = 2
= 2 {a + 2b) {3 – 2 {a + 2b)
{2 (a + b) is common}
= 2 (a + 2b) (3-2 a- 4b)

Question 11.
a (x -y) + 2b (y – x) + c (x -y)2
Solution:
a (x -y) + 2b (y – x) + c (x -y)2
= a (x – y) – 2b (x – y) + c (x – y) {x – y)
= (x – y) {x – 2b + c (x – y)}
{(a – y) is common}
= (a – y) (a – 2b + cx – cy)

Question 12.
– 4 (a – 2y)2 + 8 (a – 2y)
Solution:
– 4 (x – 2y)2 + 8 (x – 2y)
= – 4 (x – 2y) (x – 2y) + 8 (x – 2y)
{- 4 (x – 2y) is common}
= – 4 (x – 2y) (x – 2y – 2)
= 4 (x – 2y) (2 – x + 2y)

Question 13.
x3 (a – 2b) + a2 (a – 2b)
Solution:
x3 (a – 2b) + x2 (a – 2b)
HCF of x3, x2 = x2
∴ 
x2 (a – 2b) (x + 1)
{x2 (x – 2b) is common}
= x2 (x – 2b) (x + 1)

Question 14.
(2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
(2x – 3y) (a + b) + (3x – 2y) (a + b)
= (a + b) {2x – 3y + 3x – 2y}
{(x + b) is common}
= (a + b) (5x – 5y)
= 5 (a + b) (x – y)

Question 15.
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
Solution:
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
= 4 (x + y) (3a – b) – 6 (x + y) (3a – 2b)
HCF of 4, 6 = 2
= 2 (x + y) {2 (3a – b) – 3 (3a – 2b)}
= 2 (x + 3) {6a – 2b – 9a + 6b}
= 2 (x +y) {-3a + 4b}
= 2 (x + y) (4b – 3a)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Other Exercises

Factorize each of the following quadratic polynomials by using the method of completing the square.
Question 1.
p2 + 6p + 8
Solution:
p2 + 6p + 8
= p2 + 2 x p x 3 + 32 – 32 + 8   (completing the square)
= (p2 + 6p + 32) – 1
= (p + 3)2 – 12
= (P + 3)2 – (1)2       { ∵ a2 + b2  = (a+b) (a-b)}
= (p +3+1) (p + 3 -1)
= (p+4) (p+ 2)

Question 2.
q2 – 10q + 21
Solution:
q2 – 10q + 21
= (q)2 – 2 x q x 5 + (5)2 – (5)2 + 21   (completing the square)
= (q)2 – 2 x q x 5 + (5)2 -25+21
= (q)2-2 x q x 5 + (5)2 – 25 +21
= (q)2-2 x q x 5 + (5)2 – 4
= (q – 5)2 – (2)     {∵ a2 – b2 = (a + b) (a – b)}
= (q- 5 + 2) (q-5-2)
=(q- 3) (q-7)

Question 3.
4y2 + 12y + 5
Solution:
4y+12y + 5
= (2y)2 + 2 x 2y x 3 + (3)2 – (3)2 + 5    (completing the square)
= (2y + 3)2 – 9 + 5
= (2y + 3)2 – 4
= (2y + 3)2-(2)2   {∵ a2 – b2 = (a + b) (a – b)}
= (2y + 3 + 2) (2y + 3 – 2)
= (2y + 5) (2y+ 1)

Question 4.
p2 + 6p- 16
Solution:
p2 + 6p – 16
= (p)2 + 2 x  p x 3 + (3)2 – (3)2 – 16    (completing the square)
= (p)2 + 2 x p x 3 + (3)2 – 9 – 16
= (p + 3)2 – 25
= (p + 3)2 – (5)2     {∵ a2 -b2 = {a + b) (a – b)}
= (p + 3 + 5)(p + 3-5)
= (p + 8) (p – 2)

Question 5.
x2 + 12x + 20
Solution:
x2 + 12x + 20
= (x)2 + 2 x x x 6 + (6)2 – (6)2 + 20   (completing the square)
= (x)2 + 2 x x x6 + (6)2 -36 + 20
= (x + 6)2 -16
= (x + 6)2 – (4)2   {∵ a2 – b2 = (a + b) (a – b)}
= (x + 6 + 4) (x + 6 – 4)
= (x + 10) (x + 2)

Question 6.
a2 – 14a – 51
Solution:
a2 – 14a-51
= (a)2 – 2 x x 7 + (7)2 – (7)2 – 51       (completing the square)
= (a)2 – 2 x a x 7 + (7)2 – 49 – 51
= (a – 7)2 – 100
= (a – 7)2 – (10)2    {∵  a2 – b2 = (a + b) (a – b)}
= (a – 7 + 10) (a – 7 – 10)
= (a + 3) (a – 17)

Question 7.
a2 + 2a – 3
Solution:
a2 + 2a – 3
= (a)2 + 2 x a x 1 + (1)2 – (1)2 – 3   (completing the square)
= (a)2 + 2 x a x 1 + (1)2 – 1 – 3
= (a + 1)2 – 4
= (a + 1)2 – (2){∵ a2 – b2 = (a + b) (a – b)}
= (a + 1 + 2) (a + 1 – 2)
= (a + 3) (a – 1)

Question 8.
4x2 – 12x + 5
Solution:
4x2 – 12x + 5
= (2x)2 – 2 x 2x x 3 + (3)2 – (3)2 + 5  (completing the square)
= (2x)2 – 2 x 2x x 3 + (3)2 -9 + 5
= (2x – 3)2 – 4
= (2x – 3)2 – (2)2      
{∵ a2b2 = (a + b) (a – b)}
=
(2x – 3 + 2) (2x – 3 – 2)
= (2x – 1) (2x – 5)

Question 9.
y2 – 7y + 12
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 1
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 2

Question 10.
z2-4z-12
Solution:
z2 – 4z – 12
= (z)2 – 2 z x 2 + (2)2 – (2)2 – 12  (completing the square)
= (z)2 – 2 x z x 2 + (2)2 – 4 – 12
= (z-2)2-16
= (z-2)2-(4)2   {∵ a2 – b2 = (a + b) (a – b)}
= (z – 2 + 4) (z – 2 – 4)
= (z + 2)(z-6)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

Other Exercises

Factorize each of the following expressions :
Question 1.
qr-pr + qs – ps
Solution:
qr- pr + qs-ps
Arranging in suitable groups = r(q-p) +s (q-p)    {(q – p) is common}
= (q-p) (r + s)

Question 2.
p2q -pr2-pq + r2
Solution:
p2q -pr2-pq + r2
= p2q -pq-pr2 + r2 (Arranging in group)
= pq(p- 1)-r2(p-1) {(p – 1) is common}
= (p – 1) (pq – r2)

Question 3.
1 + x + xy + x2y
Solution:
1 + x + xy + x 2y
= 1 (1 + x) +xy (1 +x)
= (1 + x) (1 + xy) {(1 + x) is common}

Question 4.
ax + ay – bx – by
Solution:
ax + ay – bx – by
= a (x + y) – b (x + y)   {(x + y) is coinmon}
= (x+y) (a- b)

Question 5.
xa2 + xb2 -ya2 – yb2
Solution:
xa2 + xb2 – ya2 – yb2
= x (a2 + b2) -y (a2 + b2)   {(a2 + b2) is common}
= {a2 + b2) (x -y)

Question 6.
x2 + xy + xz + yz
Solution:
x2 + xy + xz + yz
= x (x + y) + z(x + y) {(x + y) is common}
= (x + y) (x + z)

Question 7.
2ax + bx + 2ay + by
Solution:
2ax + bx + 2ay + by
= x {2a + b) + y (2a + b)      {(2a + b) is common}
= (2a + b) (x + y)

Question 8.
ab- by- ay +y2
Solution:
ab – by – ay + y2
= b(a-y)-y(a-y)    {(a -y) is common}
= (a-y) (b – y)

Question 9.
axy + bcxy -az- bcz
Solution:
axy + bcxy – az – bcz
= xy (a + bc) – z (a + bc)       {(a + bc) is common}
= (a + bc) (xy – z)

Question 10.
lm2 – mn2 – lm + n2
Solution:
lm2 – mn2 – lm + n2
= m (lm – n2)- 1 (lm – n2)  {(lm – n2) is common}
= (lm – n2) (m – 1)

Question 11.
x– y+ x – x2y2
Solution:
x3 -y2 + x – x2y2
⇒ x3 + x – x2y2 – y2
= x(x2+ 1)-y2(x2+ 1)        {(x2 + 1) is common}
= (x2 + 1) (x -y2)

Question 12.
6xy + 6 – 9y – 4x
Solution:
6xy + 6 – 9y – 4x
= 6 xy – 4x – 9y + 6
2x (3y – 2) – 3 (3y – 2)    {(3y – 2) is common}
= (3y-2) (2x – 3)

Question 13.
x22ax – 2ab + bx
Solution:
x2 – 2ax – 2ab + bx
⇒ x2 – 2ax + bx – 2ab
= x (x – 2a) + b (x – 2a)   {(x – 2a) is common}
= (x – 2a) (x + b)

Question 14.
x3 – 2x2y + 3xy2 – 6y3
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)     {(x – 2y) is common}
= (x – 2y) (x2 + 3y2)

Question 15.
abx2 + (ay – b) x-y
Solution:
abx2 + (ay – b) x-y
= abx2 + ayx – bx -y 
= ax (bx + y) – 1 (bx + y)               {(bx +y) is common}
= (bx + y) (ax – 1)

Question 16.
(ax + by)2 + (bx – ay)2
Solution:
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + a2y2 + by2
= x2 (a2 + b2) + y2 (a2 + b2)         {(a2 + b2) is common}
= (a2 + b2) (x2 + y2)

Question 17.
16 (ab)3 -24 (a- b)2
Solution:
16 (a – b)3 -24 (a- b)2
HCF of 16, 24 = 8
and HCF of (a – b)3, (a – b)2 = (a – b)2
∴16 (a – b)3 – 24 (a – b)2
= 8 (a-b)2 {2 (a-b)- 3}
{8 (a – b)2 is common}
= 8 (a – b)2 (2a – 2b – 3)

Question 18.
ab (x2 + 1) + x (a2 + b2)
Solution:
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + b2x + a2x + ab
= bx (ax + b) + a (ax + b)  {(ax + b) is common}
= (ax + b) (bx + a)

Question 19.
a2x2 + (ax2 + 1) x + a
Solution:
a2x2 + (ax2 + 1) x + a
= a2x2 + ax3 + x + a
= ax3 + a2x2 + x + a
= ax2 (x + a) + 1 (x + a) {(x + a) is common}
= (x + a) (ax2 + 1)

Question 20.
a(a- 2b -c) + 2bc
Solution:
a(a- 2b -c) + 2bc
= a2– 2ab -ac +2bc
= a (a – 2b) – c (a – 2b) {(a – 2b) is common}
= (a – 2b) (a – c)

Question 21.
a (a + b – c)- bc
Solution:
a (a + b – c) – bc
= a2 + ab – ac – bc
= a (a + b) – c (a + b)   {(a + b) is common}
= (a + b) (a – c)

Question 22.
x2 – 11xy – x +11y
Solution:
x2 – 11xy-x + 11y
= x2 -x – 11 xy + 11 y
= x (x – 1) – 11y (x – 1)   {(x – 1) is common}
= (x- 1) (x- 11y)

Question 23.
ab – a – b + 1
Solution:
ab – a-b + 1
= a (b – 1) – 1 (b – 1)    {(b – 1) is common}
= (b – 1) (a – 1)

Question 24.
x2 + y – xy – x
Solution:
x2 + y – xy – x
= x2 – x- xy + y
= x (x – 1) – y (x – 1)   {(x – 1) is common}
= (x- 1) (x-y)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2

Other Exercises

Solve each of the following equations and also check your result in each case :
Question 1.
\(\frac { 2x + 5 }{ 3 }\) = 3x – 10
Solution:
\(\frac { 2x + 5 }{ 3 }\) = \(\frac { 3x – 10 }{ 1 }\)
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 1

Question 2.
\(\frac { a – 8 }{ 3 }\) = \(\frac { a – 3 }{ 2 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 2

Question 3.
\(\frac { 7y + 2 }{ 5 }\) = \(\frac { 6y – 5 }{ 11 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 3
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 4

Question 4.
x – 2x + 2 – \(\frac { 16 }{ 3 }\) x + 5 = 3 – \(\frac { 7 }{ 2 }\) x.
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 5
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 6

Question 5.
\(\frac { 1 }{ 2 }\) x + 7x – 6 = 7x + \(\frac { 1 }{ 4 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 7

Question 6.
\(\frac { 3 }{ 4 }\) x + 4x = \(\frac { 7 }{ 8 }\) + 6x – 6
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 8
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 9

Question 7.
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10
Solution:
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 10

Question 8.
\(\frac { 6x + 1 }{ 2 }\) + 1 = \(\frac { 7x – 3 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 11
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 12

Question 9.
\(\frac { 3a – 2 }{ 3 }\) + \(\frac { 2a + 3 }{ 2 }\) = a + \(\frac { 7 }{ 6 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 13
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 14

Question 10.
x – \(\frac { x – 1 }{ 2 }\) = 1 – \(\frac { x – 2 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 15
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 16

Question 11.
\(\frac { 3x }{ 4 }\) – \(\frac { x – 1 }{ 2 }\) = \(\frac { x – 2 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 17

Question 12.
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)
Solution:
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 18
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 19

Question 13.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 20
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 21
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 22

Question 14.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 23
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 24
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 25

Question 15.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 26
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 27
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 28

Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 29

Question 17.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 30
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 31
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 32

Question 18.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 33
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 34
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 35

Question 19.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 36
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 37
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 38

Question 20.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 39
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 40
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 41

Question 21.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 42
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 43
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 44
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 45

Question 22.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 46
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 47
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 48
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 49

Question 23.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 50
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 51
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 52

Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.

Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 53
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 54
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 55

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RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2

Other Exercises

Factorize the following :

Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3)        (HCF of 3, 9 = 3)

Question 2.
5x – 15x2
Solution:
5x- 15x2 = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x2 = x}

Question 3.
20a12b2 – 15a8b4
Solution:
20a12b2 – 15a8b4
{HCF of 20, 15 = 5, a12, a8 = a8, b2, b4 = b2}
= 5ab2(4a4 – 3b2)

Question 4.
72xy – 96x7y6
Solution:
72xy – 96x7y6
HCF of 72, 96 = 24 of x6x7 = x6, y7,y6 = y6
∴ 72x7y6 – 96x7y6 = 24x6y6 (3y – 4x)

Question 5.
20X3 – 40x2 + 80x
Solution:
20x3 – 40x2 + 80x
HCF of 20, 40,80 = 20
HCF of x3, x2, x = x
∴ 20x3 – 40x2 + 80x = 20x (x2 – 2x + 4)

Question 6.
2x3y2 – 4x2y3 + 8xy4
Solution:
2x3y2 – 4x2y3 + 8xy4
HCF of 2, 4, 8 = 2
HCF of x3, x2, x = 1
and HCF of y2, y3, y4 = y2
∴ 2x3y2 – 4x2y3 + 8xy4
= 2xy2 (x2 – 2xy + 4y2)

Question 7.
10m3n2 + 15m4n – 20m2n3
Solution:
10m3n2 + 15m4n – 20m2n3
HCF of 10, 15, 20 = 5
HCF of m3, m4, m2 = m2
HCF of n2, n, n3 = n
10m3n2 + 15m4n – 20m2n3
5m2n(2mn + 3m2– 4n2)

Question 8.
2a4b4 – 3a3b5 + 4a2b5
Solution:
2a4b4 – 3a3b5 + 4a2b5
HCF of 2, 3, 4= 1
HCF of a4, a3,
a2 = a2
HCF of b4, b5 b5 = b4
∴ 2a4b4 – 3a3b5 + 4a2b5 = a2b4
(2a2 – 3ab
+ 4b)

Question 9.
28a2 + 14a2b2 – 21a4
Solution:
28a2 + 14a2b2 – 21a4
HCF of 28, 14,21 =7
HCF of a2, a2, a4 = a2
HCF of 1, b2, 1 = 1
28a2 + 14a2b2-21a4 = 7a2
(4 + 2b2 – 3a2)

Question 10.
a4b – 3a2b2 – 6ab3
Solution:
a4b – 3a2b2 – 6ab3
HCF of 1,3,6 = 1
HCF of a4, a2, a = a
HCF of b, b2, b3 = b
∴ a4b – 3a2b2 – 6ab3 = ab (a3 – 3ab – 6b2)

Question 11.
2l2mn – 3lm2n + 4lmn2
Solution:
2l2mn – 3lm2n + 4lmn2
HCF 2, 3,4 = 1,
HCF of l2,l,l = l
HCF of m, m2, m = m
HCF of n, n, n2 = n
∴ 2lmn – 3lm2n + 4lmn2
= lmn (21 -3m + 4n)

Question 12.
x4y2 – x2y4 – x4y4
Solution:
x4y2 – x2y4 – x4y4
HCF of x4, x2, x4 = x2
HCF of y2, y4, y4 =y2
∴ x4y2 – x2y4 – x4y4 = x2y2 (x-y2 -x2y2)

Question 13.
9 x2y + 3 axy
Solution:
9 x2y + 3 axy
HCF of 9, 3 = 3
HCF of x2, x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x2y + 3axy = 3xy (3x + a)

Question 14.
16m – 4m2
Solution:
16m – 4m2
HCF of 16, 4 = 4
HCF of m, m2 = m
∴ 16m – 4m2 = 4m (4 – m)

Question 15.
-4a2 + 4ab – 4ca
Solution:
-4a2 + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a2, a, a = a
∴ -4a2 + 4ab – 4ca = -4a (a – b + c)

Question 16.
x2yz + xy2z + xyz2
Solution:
x2yz + xy2z + xyz2
HCF of x2, x, x = x
HCF of y,y2,y=y
HCF of z, z,z2 = z
∴ x2yz + xy2z + xyz2 = xyz (x + y + z)

Question 17.
ax2y + bxy2 + cxyz
Solution:
ax2y + bxy2 + cxyz
HCF of x2, x, x = x,
HCF of y,y2,y = y
ax2y + bxy2 + cxyz = xy (ax + by + cz)

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RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1

Other Exercises

Find the greatest common factors (GCF / HCF) of the following polynomials : (1 – 14)

Question 1.
2x2 and 12x2
Solution:
2x2 and 12x2
HCF of 2 and 12 =2
HCF of x2,x2=x2
∴ HCF = 2x2

Question 2.
(6xy3 and 18x2y3
Solution:
6x3y and 18xy
HCF of 6, 18 = 6
HCF of x3 and x2 = x2
HCF of y and y3 -y
∴ HCF = 6x2y

Question 3.
7x, 21x2 and 14xy2
Solution:
7x, 21x2 and 14xy2
HCF of 7, 21 and 14 = 7
HCF of x, x2, x = x
∴ HCF = 7x

Question 4.
42x2yz and 63x3y2z3
Solution:
42x2yz and 63x3y2z3
HCF of 42 and 63 = 21
HCF of x2, x3 = x2
HCF of y,y2=y
HCF of z,z3 = z
∴ HCF = 21 x2yz

Question 5.
12ax2,6a2x3 and 2ax5
Solution:
12ax2, 6a2x3 and 2a3x5
HCF of 12, 6,2 = 2
HCF of a, a2, a3 = a
HCF of x2, x3, x5 = x2
∴ HCF = 2ax2

Question 6.
9x2, 15x2y3, 6xy2 and 21x2y2
Solution:
9x2, 15xV, 6xy2 and 21x2y2
HCF of 9, 15, 6,21 = 3
HCF of x2, x2, x, x2 = x
HCF of 1, y3, y2, y2 =2
∴ HCF = 3x

Question 7.
4a2b3 -12a3b, 18a4b3
Solution:
4a2b3, -12a3b, 18a4b3
HCF of 4, 12, 18 = 2
HCF of a2, a3, a4 = a2
HCF of b3,b, b3 = b
∴ HCF = 2a2b

Question 8.
6x2y2, 9xy3, 3x3y2
Solution:
6x2y2, 9xy3, 3x3y2
HCF of 6, 9, 3 = 3
HCF of x2, x, x3 = x
HCF of y2,y3,y2=y2
∴ HCF = 3xy2

Question 9.
a2b3, a3b2
Solution:
a2b3, a3b2
HCF of a2, a3 = a2
HCF of b3, b2 = b2
∴ HCF = a2b2

Question 10.
36a2b2c4, 54a5c2,90a4b2c2
Solution:
36a2b2c4, 54a5c2,90a4b2c2
HCF of 36, 54, 90 = 18
HCF of a2, a5, a4 = a2
HCF of b2, 1,b2= 1
HCF of c4,c2,c2 = c2
∴ HCF = 18a2 x 1 x c2 = 18a2c2

Question 11.
x3, – yx2
Solution:
x3, – yx2
HCF of x3, x2 = x2
HCF of 1, y= 1
∴ HCF = x2

Question 12.
15a3, -45a2, -150a
Solution:
15a3,-45a2,-150a
HCF of 15,45, 150 = 15
HCF of a3, a2, a = a
∴ HCF = 15a

Question 13.
2x3y2, 10x2y3, 14xy
Solution:
2x3y2, 10x2y3, 14xy
HCF of 2, 10, 14 = 2
HCF of x3, x2, x = x
HCF of y2,y3,y=y
∴ HCF = 2xy

Question 14.
14x3y5, 10x5y3, 2x2y2
Solution:
14x3y5, 10x5y3, 2x2y2
HCF of 14, 10, 2, = 2
HCF of x3, x5, x2 = x2
HCF of y5,y3,y2=y2
∴ HCF = 2xy

Find the greatest common factor of the terms in each of the following expressions:

Question 15.
5a4 + 10a3 – 15a2
Solution:
5a4 + 10a3– 15a2
HCF of 5, 10, 15 = 5
HCF of a4, a3, a2 = a2
∴ HCF = 5a2

Question 16.
2xyz + 3x2y + 4y2
Solution:
2xyz + 3x2y + 4y2
HCF of 2, 3,4 = 1
HCF of x, x2, 1 = 1
HCF of y,y,y2 =y
HCF of z, 1, 1 = 1
∴ HCF = y

Question 17.
3a2b2 + 4b2c2 + 12a2b2c2
Solution:
3a2b2 + 4b2c2 + 12a2b2c2
HCF of 3, 4, 12 = 1
HCF of a2, 1, a2 = 1
HCF of b2, b2, b2 = b2
HCF of 1, c2, c2 = 1
∴ HCF = b2

 

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RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

Other Exercises

Question 1.
Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 1
Solution:
In the figure, here are three triangles and one trapezium.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 2
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 3
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 4

Question 2.
Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 5
Solution:
(i) In the figure ABCDEF,
Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 6
Area of the square = 18 x 18 cm² = 324 cm²
Area of trapezium FCDE = \(\frac { 1 }{ 2 }\) (CF + ED) x 8 cm²
= \(\frac { 1 }{ 2 }\) (18 + 7) x 8
= \(\frac { 1 }{ 2 }\) x 25 x 8 cm²
= 100 cm²
Total area of fig. ABCDEF = 324 + 100 = 424 cm²
(ii) In the figure ABCDEF,
Join BE.
The figure consists of one rectangle BCDE and one trapezium ABEF
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 7
Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm²
Area of trapezium ABEF,
= \(\frac { 1 }{ 2 }\) (BE + AF) x height
= \(\frac { 1 }{ 2 }\) (15 + 6) x 8 cm²
= \(\frac { 1 }{ 2 }\) x 21 x 8 cm²
= 84 cm²
Area of the figure ABCDEF = 300 + 84 = 384 cm²
(iii) In the figure ABCDEFGH,
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 8
HC= AB = EF = 6 cm
AH = BC = 4 cm
DE = GF = 5 cm
Join HC.
In right ∆CDE,
ED² = CD² + CE²
⇒ (5)² = (4)² + (CE)²
⇒ 25 = 16 + (CE)²
⇒ (CE)² = 25 – 16 = 9 = (3)²
CE = 3 cm
The figure consist a rectangle and a trapezium
Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm²
Area of trapezium GDEF,
= \(\frac { 1 }{ 2 }\) (GD + EF) x CE 1
= \(\frac { 1 }{ 2 }\) (GH + HC + CD + EF) x CE
= \(\frac { 1 }{ 2 }\) (4 + 6 + 4 + 6) x 3 cm²
= \(\frac { 1 }{ 2 }\) x 20 x 3 cm²
= 30 cm²
Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²

Question 3.
There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 9
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 10
Find the area of this park using both ways. Can you suggest some another way of finding its area ?
Solution:
In first case, the figure ABCDE is divided into 2 trapezium of equal area.
Now area of trapezium DFBC
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 11
Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m²
In second case, the figure ABCDE is divided into two parts, namely one square and other triangle.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 12
Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²

Question 4.
Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 13
Solution:
In the figure ABCDEF,
AD = 90 cm
BL = 30 cm
AO = 60 cm
CN = 40 cm
AN = 50 cm
EO = 60 cm
AM = 20 cm
FM = 20 cm
AL = 10 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 14
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 15
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 16
Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²

Question 5.
Find the area of the following regular hexagon:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 17
Solution:
In the regular hexagon MNOPQR There are two triangles and one rectangle.
Join MQ, MO and RP
NQ = 23 cm,
NA = BQ = \(\frac { 10 }{ 2 }\) = 5 cm
MR = OP = 13 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 18
In right ∆BDQ,
PQ² = BQ² + BP²
⇒ (13)² = (5)² + BP²
⇒ 169 = 25 + BP²
⇒ BP² = 169 – 25 = 144 = (12)²
BP = 12 cm
PR = MO = 2 x 12 = 24 cm
Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm²
Area of ∆PRQ = \(\frac { 1 }{ 2 }\) x PR x BQ
= \(\frac { 1 }{ 2 }\) x 24 x 5 = 60 cm²
Similarly area ∆MON = 60 cm²
Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²

Hope given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2

Other Exercises

Question 1.
Find the area, in square metres, of the trapezium whose bases and altitude are as under:
(i) bases = 12 dm and 20 dm, altitude =10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 1
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 2
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 3

Question 2.
Find the area of trapezium with base 15 cm and height 8 cm. If the side parallel to the given base is 9 cm long.
Solution:
In the trapezium ABCD,
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 4
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 5

Question 3.
Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.
Solution:
Length of parallel sides of a trapezium are 16 dm and 22 dm i.e.
b1 = 16 dm, b2 = 22 dm
and height (h) = 12 dm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 6

Question 4.
Find the height of a trapezium, the sum of lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm²
Solution:
Sum of parallel sides (b1 + b2) = 60 cm
Area of trapezium = 600 cm²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 7

Question 5.
Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm.
Solution:
Area of a trapezium = 65 cm²
Bases are 13 cm and 26 cm
i.e. b1 = 13 cm, b2 = 26 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 8

Question 6.
Find the sum of the lengths of the bases of trapezium whose area is 4.2 m² and whose height is 280 cm.
Solution:
Area of trapezium = 4.2 m²
Height (h) = 280 cm = 2.8 m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 9

Question 7.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate the area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle id the sum of the areas of two triangles.
Solution:
In trapezium ABCD, parallel sides or bases are 10 cm and 15 cm and height = 6 cm
Area of trapezium
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 10
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 11
Area of trapezium = 90 – 15 = 75 cm²
= area of rectangle – areas of two triangles.

Question 8.
The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them:
Solution:
Area of trapezium = 960 cm²
Parallel sides are 34 cm and 46 cm
b1 + b2 = 34 + 46 = 80 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 12
Distance between parallel sides = 24 cm

Question 9.
Find the area of the figure as the sum of the areas of two trapezium and a rectangle.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 13
Solution:
In the figure,
One rectangle is ABCD whose sides are 50 cm and 10 cm.
Two trapezium of equal size in which parallel sides are 30 cm and 10 cm and height
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 14

Question 10.
Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 15
Solution:
Top of a table is of trapezium in shape whose parallel sides are 1 m and 1.2 m and distance between them (h) = 0.8 m
Area of trapezium = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x height
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 16

Question 11.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom, and the area of the cross-section is 72 m², determine its depth.
Solution:
Area of cross-section = 72 m²
Parallel sides of the trapezium = 10 m and 6 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 17

Question 12.
The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Solution:
Area of trapezium = 91 cm²
Height (h) = 7 cm.
Sum of parallel sides
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 18
One parallel side = 9 cm
and second side = 9 + 8 = 17 cm
Hence parallel sides are 17 cm, 9 cm

Question 13.
The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.
Solution:
Area of trapezium = 384 cm²
Perpendicular distance (h) = 12 cm
Sum of parallel sides
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 19
First parallel side = 8 x 3 = 24 cm
Second side = 8 x 5 = 40 cm

Question 14.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along 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.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 20
Solution:
Area of the trapezium shaped field = 10500 m²
and perpendicular distance between them (h) = 100 m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 21

Question 15.
The area of trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.
Solution:
Area of a trapezium = 1586 cm²
and distance between the parallel sides (h) = 26
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 22

Question 16.
The parallel sides of a trapezium are 25 cm and 13 cm ; Its nonparallel sides are equal each being 10 cm, find the area of the trapezium.
Solution:
Parallel sides of a trapezium ABCD are 25 cm and 13 cm
i.e. AB = 25 cm, CD = 13 cm
and each non-parallel side = 10 cm
i.e., AD = BC = 10 cm
From C, draw CE || DA and draw CL ⊥ AB
CE = DA = CB = 10 cm
and EB = AB – AE = AB – DC = 25 – 13 = 12 cm
Perpendicular CL bisects base EB of an isosceles ∆CED
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 23

Question 17.
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm each.
Solution:
In trapezium ABCD, parallel sides are AB and DC.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 24
AB = 25 cm, CD = 13 cm
and other sides are 15 cm each i.e. AD = CB = 15 cm
From C, draw CE || DA and CL ⊥ AB
AE = DC = 13 cm
and EB = AB – AE = 25 – 13 = 12 cm
Perpendicular CL bisects the base EB of the isosceles triangle CEB
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 25

Question 18.
If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.
Solution:
Area of trapezium = 28 cm²
Altitude (h) = 4 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 26
One of the parallel side = 6 cm
Second parallel side = 14 – 6 = 8 cm

Question 19.
In the figure, a parallelogram is drawn in a trapezium the area of the parallelogram is 80 cm², find the area of the trapezium.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 27
Solution:
Area of parallelogram (AECD) = 80 cm²
Side AE (b) = 10 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 28

Question 20.
Find the area of the field shown in the figure by dividing it into a square rectangle and a trapezium.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 29
Solution:
Produce EF to H to meet AB at H and draw DK || EH
HF = 4 cm, KD = HE = 4 + 4 = 8 cm
HK = ED = 4 cm,
KB = 12 – (8) = 4 cm
Now, area of square AGFH = 4 x 4 = 16 cm²
area of rectangle KDEH = l x b = 8 x 4 = 32 cm²
and area of trapezium BCDK.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 30
Total area of the figure = 16 + 32 + 22 = 70 cm²

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

Other Exercises

Question 1.
Find the following products :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 4

Question 2.
Evaluate the following :
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) 102 x 106 = (100 + 2) (100 + 6)
= (100)2 + (2 + 6) x 100 + 2 x 6
= 10000 + 800 + 12 = 10812

(ii) 109 x 107 = (100 + 9) (100 + 7)
= (100)2 + (9 + 7) x 100 + 9 x 7
=10000 + 1600 + 63 = 11663

(iii) 35 x 37 = (30 + 5) (30 + 7)
= (30)2 + (5 + 7) x 30 + 5 x 7
= 900 + 12 x 30 + 35
= 900 + 360 + 35 = 1295

(iv) 53 x 55 = (50 + 3) (50 + 5)
= (50)2 + (3 + 5) x 50 + 3 x 5
= 2500 + 8 x 50 + 15
= 2500 + 400+ 15 = 2915

(v)103 x 96 = (100 + 3) (100-4)
= (100)2 + (3 – 4) x 100 + 3 x (-4)
= 10000+ (-1) x 100-12
= 10000 – 100 – 12 = 10000 – 112 = 9888

(vi) 34 x 36 = (30 + 4) (30 + 6)
= (30)2 + (4 + 6) x 30 + 4 x 6
= 900 + 10 x 30 + 24
= 900 + 300 + 24 = 1224

(vii) 994 x 1006 = (1000 – 6) (1000 + 6)
= (1000)2 + (-6 + 6) x 1000 + (-6) x 6
= 1000000 + 0-36
= 1000000-36 = 999964

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1

Other Exercises

Question 1.
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 ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 1

Question 2.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 2
Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = \(\frac { b }{ 2 }\) = \(\frac { 28 }{ 2 }\) = 14 m
Area = \(\frac { 1 }{ 2 }\) πr²
= \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²

Question 3.
A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = \(\frac { 22 }{ 7 }\)).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 3
= \(\frac { 22 }{ 7 }\) x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
breadth (b) = 15 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 4
Area of rectangular piece = l x b = 20 x 15 = 300 m²
Radius of each quadrant (r) = 3.5 m
Total area of 4 quadrants
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 5
Area of the remaining portion = 300 – 38.5 m² = 261.5 m²

Question 5.
The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 6
Solution:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 7

Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = \(\frac { 22 }{ 7 }\))
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 8
Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 9

Question 7.
The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = \(\frac { 22 }{ 7 }\))
Solution:
Diameter of the wheel (d) = 90 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 10

Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d1) = 16 cm
Second diagonal (d2)
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 11

Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d1) = 7.5 cm
and diagonal (d2) = 12 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 12

Question 10.
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.
Solution:
In quadrilateral shaped field ABCD,
diagonal AC = 24 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 13
and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = \(\frac { 1 }{ 2 }\) x AC x (BL + DM)
= \(\frac { 1 }{ 2 }\) x 24 x (13 + 8) m²
= 12 x 21 = 252 m²

Question 11.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 14

Question 12.
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 Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 15
Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810

Question 13.
A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 16
Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50

Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d1) = 24 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 17
Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = \(\frac { 1 }{ 2 }\) AC = \(\frac { 1 }{ 2 }\) x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 18

Question 15.
The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 19
Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 20

Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²

Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 21

Question 18.
In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.
Solution:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 22

Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 23

Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960

Question 21.
A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ?
Solution:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m

Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x bh= \(\frac { 1 }{ 2 }\) x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d1 = 22 cm
Second diagonal (d2)
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 24

 

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6

Other Exercises

Question 1.
Write the following squares of bionomials as trinomials :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 1
Solution:
Using the formulas
(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(i) (a + 2)2 = (a)2 + 2 x a x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= a2 + 4a + 4
(ii) (8a + 3b)2 = (8a)2 + 2 x 8a * 3b + (3b)2 = 642 + 48ab + 9 b2
(iii) (2m+ 1)2 = (2m)2 + 2 x 2m x1 + (1)2
= 4m2 + 4m + 1
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 3

Question 2.
Find the product of the following binomials :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 4
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 5
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 6
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 7

Question 3.
Using the formula for squaring a binomial, evaluate the following :
(i) (102)2
(ii) (99)2
(iii) (1001)2

(iv) (999)2
(v) (703)
2
Solution:
(i) (102)2 = (100 + 2)2
= (100)2 + 2 x 100 x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= 10000 + 400 + 4 = 10404
(ii) (99)2 = (100 – 1)2
= (100)2 – 2 x 100 X 1 +(1)2
{(a – b)2 = a2 – 2ab + b2}
= 10000 -200+1
= 10001 -200 =9801
(iii) (1001 )2 = (1000 + 1)2
{(a + b)2 = a2 + 2ab + b2}
= (1000)2 + 2 x 1000 x 1 + (1)2
= 1000000 + 2000 + 1 = 1002001
(iv) (999)2 = (1000 – 1)2
{(a – b)2 = a2 – 2ab + b2}
= (1000)2 – 2 x 1000 x 1 + (1)2
= 1000000 – 2000 + 1
= 1000001 -2000 = 998001

Question 4.
Simplify the following using the formula:
(a – b) (a + b) = a2 – b2 :
(i) (82)2 (18)2
(ii) (467)2 (33)2
(iii) (79)2 (69)2
(iv) 197 x 203
(v) 113 x 87
(vi) 95 x 105
(vii) 1.8 x 2.2
(viii) 9.8 x 10.2
Solution:
(i) (82)2 – (18)2 = (82 + 18) (82 – 18)
{(a + b)(a- b) = a2 – b2} = 100 x 64 = 6400
(ii) (467)2 – (33)2 = (467 + 33) (467 – 33)
= 500 x 434 = 217000
(ii) (79)2 – (69)2 = (79 + 69) (79 – 69)
148 x 10= 1480
(iv) 197 x 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000-9 = 39991
(v) 113 x 87 = (100 + 13) (100- 13)
= (100)2 – (13)2
= 10000- 169 = 9831
(vi) 95 x 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25 = 9975
(vii) 8 x 2.2 = (2.0 – 0.2) (2.0 + 0.2)
= (2.0)2 – (0.2)2
= 4.00 – 0.04 = 3.96
(viii)9.8 x 10.2 = (10.0 – 0.2) (10.0 + 0.2)
(10.0)2 – (0.2)2
= 100.00 – 0.04 = 99.96

Question 5.
Simplify the following using the identities :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 8
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 9
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 10

Question 6.
Find the value of x, if
(i)  4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii)  5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
4x = (52 + 48) (52 – 48)
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 11

Question 7.
If x + \(\frac { 1 }{ x }\)= 20, find the value of x2+ \(\frac { 1 }{ { x }^{ 2 } }\)

Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 12

Question 8.
If x – \(\frac { 1 }{ x }\) = 3, find the values of x2 + \(\frac { 1 }{ { x }^{ 2 } }\) and x4 + \(\frac { 1 }{ { x }^{ 4 } }\)

Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 13

Question 9.
If x2 – \(\frac { 1 }{ { x }^{ 2 } }\)= 18, find the values of x+ \(\frac { 1 }{ x }\)  and x– \(\frac { 1 }{ x }\)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 14

Question 10.
Ifx+y = 4 and xy = 2, find the value of x2+y2.
Solution:
x + y = 4
Squaring on both sides,
(x + y)2 = (4)2
⇒ x2 +y2 + 2xy = 16
⇒ x2+y2 + 2 x 2 = 16                       (∵ xy = 2)
⇒ x2 + y2 + 4 = 16
⇒ x2+y2 = 16 – 4= 12           ‘
∴ x2+y2 = 12

Question 11.
If x-y = 7 and xy = 9, find the value of x2+y2.
Solution:
x-y = 7
Squaring on both sides,
(x-y)2 = (7)2
⇒ x2+y2-2xy = 49
⇒ x2 + y2 – 2 x 9 = 49                    (∵ xy = 9)
⇒ x2 +y2 – 18 = 49
⇒ x2 + y2 = 49 + 18 = 67
∴ x2+y2 = 67

Question 12.
If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
3 x + 5y = 11, xy = 2
Squaring on both sides,
(3x + 5y)2 = (11)2
⇒ (3x)2 + (5y)2 + 2 x 3x x 5y = 121
⇒ 9x2 + 25y2 + 30 x 7 = 121
⇒ 9x2 + 25y2+ 30 x 2 = 121           (∵ xy = 2)
⇒ 9x2 + 25y2 + 60 = 121
⇒ 9x2 + 25y2 = 121 – 60 = 61
∴ 9x2 + 25y2 = 61

Question 13.
Find the values of the following expressions :
(i)16x2 + 24x + 9, when X’ = \(\frac { 7 }{ 45 }\)
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = \(\frac { 4 }{ 3 }\)
(iii) 81x2 + 16y2-72xy, whenx= \(\frac { 2 }{ 3 }\) andy= \(\frac { 3 }{ 4 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 15
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 16

Question 14.
If x + \(\frac { 1 }{ x }\) = 9, find the values of x+ \(\frac { 1 }{ { x }^{ 4 } }\).
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 17
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 18

Question 15.
If x + \(\frac { 1 }{ x }\) = 12, find the values of x–  \(\frac { 1 }{ x }\).
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 19

Question 16.
If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy.
Solution:
2x + 3y = 14, 2x – 3y= 2
We know that
(a + b)2 – (a – b)2 = 4ab
∴ (2x + 3y)2 – (2x – 3y)2 = 4 x 2x x 3y = 24xy
⇒ (14)2 – (2)2 = 24xy
⇒ 24xj= 196-4= 192
⇒ xy = \(\frac { 192 }{ 24 }\) = 8
∴  xy = 8

Question 17.
If x2 + y2 = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x4 +y4
Solution:
x2 + y2 = 29, xy = 2
(i) (x + y)2 = x2 + y2 + 2xy
= 29 + 2×2 = 29+ 4 = 33
∴  x + y= ±√33
(ii) (x – y)2 = x2 + y2 – 2xy
= 29- 2×2 = 29- 4 = 25
∴ x-y= ±√25= ±5
(iii) x2 + y2 = 29
Squaring on both sides,
(x2 + y2)2 = (29)2
⇒ (x2)2 + (y2)2 + 2x2y2 = 841
⇒ x4 +y + 2 (xy)2 = 841
⇒ x4 + y + 2 (2)2 = 841          (∵ xy = 2)
⇒ x4 + y + 2×4 = 841
⇒ x4 + y + 8 = 841
⇒ x4 + y = 841 – 8 = 833
∴ x4 +y = 833

Question 18.
What must be added to each of the following expressions to make it a whole square ?’
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7 = (2x)2 – 2x 2x x 3 + 7
In order to complete the square,
we have to add  32 – 7 = 9 – 7 = 2
∴ (2x)2 – 2 x 2x x 3 + (3)2
= (2x-3)2
∴ Number to be added = 2
(ii) 4x2 – 20x + 20
⇒ (2x)2 – 2 x 2x x 5 + 20
In order to complete the square,
we have to add (5)2 – 20 = 25 – 20 = 5
∴ (2x)2 – 2 x 2x x 5 + (5)2
= (2x – 5)2
∴ Number to be added = 5

Question 19.
Simplify :
(i) (x-y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8m)2 + (7m + 8m)2
(iv) (2.5p -5q)2 – (1.5p – 2.5q)2
(v) (m2 – n2m)2 + 2m3n2

Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 +y)
= (x2 – y2) (x2 + y) (x4 + y4)
= [(x2)2 – (y2)2] (x4+y4)
= (x4-y4) (x4+y4)
= (x4)2 – (y4)2 = x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
= [(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
= (4x2 – 1) (4x2 + 1) (16x4 + 1)
= [(4x2)2-(1)2] (16x4+ 1)
= (16x4-1) (16x4+ 1)
= (16x4)2– (1)2 = 256x8 – 1
(iii) (7m – 8m)2 + (7m + 8n)2
= (7m)2 + (8n)2 – 2 x 7m x 8n + (7m)2 + (8n)2 + 2 x 7m x 8n
= 49m2 + 64m2 – 112mn + 49m2 + 64m2 + 112mn
= 98 m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= (2.5p)2 + (1.5q)2 – 2 x 2.5p x 1.5q
= [(1.5p)2 + (1.5q)2 – 2 x 1.5 p x 2.5q]
= (6.25p2 + 2.25q2 – 7.5 pq) – (2.25p2 + 6.25q2-7.5pq)
= 6.25p2 + 2.25q2 – 7.5pq – 2.25p2 – 6.25q2 + 7.5pq
= 6.25p2 – 2.25p2 + 2.25g2 – 6.25q2
= 4.00P2 – 4.00q2
= 4p2 – 4q2 = 4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3M2
= (m2)2 + (n2m)2 -2 x m2 x n2m + 2;m3m2
= m4 + n4m2 – 2m3n2 + 2m3n2
= m4 + n4m2 = m4 + m2n4

Question 20.
Show that :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 22

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.