NCERT solution class 8 chapter 9 Algebraic Expression and Identities exercise 9.2

Exercise 9.2

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Q 1 - Ex 9.2 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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

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Q 2 - Ex 9.2 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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 1^st rectangle = p × q
= pq
⇒Area of 2^nd rectangle = 10m × 5n
= 10 × 5 × m × n
= 50 mn
⇒Area of 3^rd rectangle = 20x² × 5y²
= 20 × 5 × x² × y²
= 100 x²y²
⇒Area of 4^th rectangle = 4x × 3x²
= 4 × 3 × x × x²
= 12x³
⇒Area of 5^th rectangle = 3mn × 4np
= 3 × 4 × m × n × n × p
= 12mn²p

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Q 3 - Ex 9.2 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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

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Q 4 - Ex 9.2 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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

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Q 5 - Ex 9.2 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 5
Obtain the product of 
(i) xy , yz , zx
Sol:
⇒(xy)×(yz)×(zx)
⇒ xy.yz.zx
⇒ xx.yy.zz
⇒ x²y²z²

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(ii) a , -a² , a³
Sol :
⇒(a)×(-a²)×(a³)
⇒ -a^(1+2+3)
⇒ -a⁶

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(iii) 2 , 4y , 8y² , 16y³
Sol:
⇒ (2)×(4y)×(8y²)×(16y³)
⇒ (2×4×8×16)×(y¹×y²×y³)
⇒ 1024 y^(1+2+3)
⇒ 1024 y⁶

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(iv) a , 2b , 3c , 6abc
Sol:
⇒ (a)×(2b)×(3c)×(6abc)
⇒ (2×3×6)×(a×a)×(b×b)×(c×c)
⇒ 36 a²b²c²

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(v) m , -mn , mnp
Sol:
⇒ -(m)×(mn)×(mnp)
⇒ -(m×m×m)×(n×n)×(p)
⇒ -m³n²p

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