NCERT Solution For Class 8 Maths Chapter – 9 Algebraic Expression Exercise – 9.2
Q1. Find the product of the following pairs of monomials.
- 4 , 7p
- -4p , 7p
- -4p , 7pq
- 4p3 , -3p
- 4p , 0
Solution:-
- 4 , 7p = 4 x 7p = 28p
- -4p x 7p = (-4 x 7) x (p x q) = -28p2
- -4p x 7pq = (-4 x 7) x (p x pq) = -28p2q
- 4p3 x -3p = (4 x -3) (p3 x p ) = -12p4
- 4p x 0 = 0
Q2. Find the area of rectangles with the following pairs of monomials as their lengths and breadths , respectively.
(p,q) ; (10m , 5n) ; (20x2 , 5y2) ; (4x , 3x2) ; (3mn , 4np)
Solution:-
Area of rectangle = Length x Breadth . So , it is multiplication of two monomials.
The results can be written square units
- P x q = pq
- 10m x 5n = 50 mn
- 20x2 x 5y2 = 100x2y2
- 4x x 3x2 == 12x3
- 3mn x 4np = 12mn2p
Q3. Complete the following table of products:
Solution:-
Q4. Obtain the volume of rectangular boxes with the following length , breadth and height respectively.
- 5a , 3a2 , 7a4
- 2p , 4q , 8r
- Xy , 2x2y , 2xy2
Solution:-
Volume of rectangular = Length x Breadth x Height . To evaluate volume of rectangular boxes, multiply all the monomials.
- 5a x 3a2 x 7a4 = (5 x 3 x 7) (a x a2 x a4) = 105a7
- 2p x 4q x 8r = (2 x 4 x 8) (p x q x r) = 64 pqr
- Y x 2x2y x 2xy2 = ( 1 x 2 x 2) (x x x2 x x x y x y x y2) = 4x4y4
- A x 2b x 3c = (1 x 2 x 3) (a x b x c) = 6abc
Q5. Obtain the product of
- Xy , yz , zx
- A , -a2 , a3
- 2 , 4y , 8y2 , 16y3
- A , 2b , 3c , 6abc
- M , -mn , mnp
Solution:-
- Xy x yz x zx = x2y2z2
- A x -a2 x a3 = -a6
- 2 x 4y x 8y2 x 16y3 = 1024y6
- A x 2b x 3c x 6abc = 36a2 b2 c2
- M x -mn x mnp = -m3 n2 p