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NCERT solution class 8 chapter 9 Algebraic Expression and Identities

Exercise 9.1  Exercise 9.2  Exercise 9.3  Exercise 9.4  Exercise 9.5

Exercise 9.2

Question 1

Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) − 4p, 7p

(iii) − 4p, 7pq

(iv) 4p3, − 3p

(v) 4p, 0

Sol :

The product will be as follows.

(i)

= 4 × 7p

= 4 × 7 × p

= 28p

 

(ii)

= − 4p × 7p

= − 4 × p × 7 × p

= (− 4 × 7) × (p × p)

= − 28 p2

 

(iii)

= − 4p × 7pq

= − 4 × p × 7 × p × q

= (− 4 × 7) × (p × p × q)

= − 28p2q

 

(iv)

= 4p3 × − 3p

= 4 × (− 3) × p × p × p × p

= − 12 p4

 

(v)

= 4p × 0

= 4 × p × 0

= 0

 


Question 2

Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

Sol :

We know that,

⇒ Area of rectangle = Length × Breadth

⇒Area of 1st rectangle = p × q

= pq

⇒Area of 2nd rectangle = 10m × 5n

= 10 × 5 × m × n

= 50 mn

⇒Area of 3rd rectangle = 20x2 × 5y2

= 20 × 5 × x2 × y2

= 100 x2y2

⇒Area of 4th rectangle = 4x × 3x2

= 4 × 3 × x × x2

= 12x3

⇒Area of 5th rectangle = 3mn × 4np

= 3 × 4 × m × n × n × p

= 12mn2p

 


Question 3

Complete the table of products.

\dfrac{\text{First monomial}}{\text{Second monomial}}

2x

−5y

3x2

−4xy

7x2y

−9x2y2

2x

4x2

−5y

−15x2y

3x2

−4xy

7x2y

−9x2y2

Sol :

 

The table can be completed as follows.

\dfrac{\text{First monomial}}{\text{Second monomial}}

2x

−5y

3x2

−4xy

7x2y

−9x2y2

2x

4x2

−10xy

6x3

−8x2y

14x3y

−18x3y2

−5y

−10xy

25y2

−15x2y

20xy2

−35x2y2

45x2y3

3x2

6x3

−15x2y

9x4

−12x3y

21x4y

−27x4y2

−4xy

−8x2y

20xy2

−12x3y

16x2y2

−28x3y2

36x3y3

7x2y

14x3y

−35x2y2

21x4y

−28x3y2

49x4y2

−63x4y3

−9x2y2

−18x3y2

45x2y3

−27x4y2

36x3y3

−63x4y3

81x4y4

 


Question 4

Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

 

Sol :

We know that,

Volume = Length × Breadth × Height

(i) Volume = (5a)×(3a2)×(7a4)

= (5×3×7)×(a×a2×a4)

= 105 a7

 

(ii) Volume = (2p)×(4q)×(8r)

= (2×4×8)×(p×q×r)

= 64pqr

 

(iii) Volume = (xy)×(2x2y)×(2xy2)

= (2×2)×(xy × x2y × xy2)

= 4×(x.x2.x)×(y.y.y2)

= 4 x(1+2+1) y(1+1+2)

= 4x4y4

 

(iv) Volume = (a)×(2b)×(3c)

= (2×3)×(a×b×c)

= 6abc

 


Question 5

Obtain the product of 

(i) xy , yz , zx

Sol:

⇒(xy)×(yz)×(zx)

⇒ xy.yz.zx

⇒ xx.yy.zz

⇒ x2y2z2

 


(ii) a , -a2 , a3

Sol :

⇒(a)×(-a2)×(a3)

⇒ -a(1+2+3)

⇒ -a6

 


(iii) 2 , 4y , 8y2 , 16y3

Sol:

⇒ (2)×(4y)×(8y2)×(16y3)

⇒ (2×4×8×16)×(y1×y2×y3)

⇒ 1024 y(1+2+3)

⇒ 1024 y6

 


(iv) a , 2b , 3c , 6abc

Sol:

⇒ (a)×(2b)×(3c)×(6abc)

⇒ (2×3×6)×(a×a)×(b×b)×(c×c)

⇒ 36 a2b2c2

 


(v) m , -mn , mnp

Sol:

⇒ -(m)×(mn)×(mnp)

⇒ -(m×m×m)×(n×n)×(p)

⇒ -m3n2p


 

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