Page 96 - Start Up Mathematics_8 (Non CCE)

2. Find the area of the rectangles with the following length and breadth respectively:
3 4 2 3 2 1 2 3 2 4
23
2
2
(a) ab , ab (b) 3.2x , 4.1x (c) 1 xyz, 3 xyz 2 (d) l m n, 4lm n
2 9 3 2
3. Multiply:
3 3 3 14
2
2
2 2 2
2
(a) 4xy , –3x y and 6x yz (b) 6 a b c , a b c and –5
7 3
-8 -3 9 2 6 2 Ê -1 ˆ
3
3
2
(c) x , x y , xyz and –yz (d) 5x , (–10xy ), Á xy 3 ˜ and –15
4
2
15 4 16 Ë 25 ¯
2 3
2
2
2
3
(e) 0.25x, (–0.1x ), (2.5x ) and (–1.2x) (f) 2 1 p q, -4 pr , 3 1 q r and –7
2 7 4
4. Find the volume of the rectangular boxes whose length, breadth and height are given as follows:
Length Breadth Height
(a) 3.5x y 1.3x 3 0.5xy 2
2
2 2
(b) 1 p q r 3 pqr 2 pqr
2 4
(c) l m 4 m n l n
2 5
3
6 2
(d) 3a b c 5a b c 2a b c
3 5 2
4 2 5
2 3 2
ˆ
2 2
4
2
3
5. Add the product of 2p q and Ê -3 pq to the product of (–5p q) and (pq ).
Á
˜
¯
Ë 4
Ê 4 ˆ 2 1 2 Ê 7 2 ˆ Ê 2 2 2 ˆ
2
6. From the product of Á 3 lmn , (–5lmn ) and l mn, subtract the product of Á 3 lmn and Á Ë 3 lm n .
3
2
˜
˜
˜
Ë
Ë
¯
¯
¯
3
7. Express the following products as monomials and verify the result for x = 2, y = –1 and z = 1.
3
3
2
2
(a) (x y) ¥ (–3xy ) ¥ Ê Á Ë 1 yz 2 ˆ (b) (2.1x y) ¥ (–3xy ) ¥ (–0.5)
˜
¯
6
-1
4 3
9
3 5
(c) (y z ) ¥ (–2y z ) ¥ (3yz) (d) Ê Á Ë 7 z 2 ˆ ˜ ¯ ¥ Ê 7 z 7 ˆ ˜ ¯ ¥ (8z )
Á
Ë 8
8. Find two monomials with positive integer coefficients, whose product is the given monomial.
2
2
(a) 3lmn (b) 7xy (c) pq (d) 2a bc
9. Evaluate the following:
2
2
2
2
2
(a) (2a b)(3ab )(0.5ab) for a = –2, b = 1 (b) (–2l ) ¥ (–4lm ) ¥ (–8mn ) for l = 3, m = 2, n = 1.5
-16
ˆ
2
2 3
6
3
¥
(c) (5x ) ¥ (–1.5x y ) ¥ (–12xy ) for x = 1, y = 0.5 (d) Ê Á Ë 15 xyz 2 ˆ Ê -25 xz for x = 1, y = –4, z = 2
˜
˜ Á
¯
¯ Ë 24
Multiplication of a monomial by a binomial
You have already learnt that multiplication of literals is distributive over their addition. Thus,
x ¥ (y + z) = (x ¥ y) + (x ¥ z)
Utilizing this property in multiplication of a monomial and a binomial, we can see that if A, B and C are three
monomials, then
A ¥ (B + C) = (A ¥ B) + (A ¥ C) and (B + C) ¥ A = (B ¥ A) + (C ¥ A)
A ¥ (B – C) = (A ¥ B) – (A ¥ C) and (B – C) ¥ A = (B ¥ A) – (C ¥ A)
88