Class 8th
Maths
Sol.
Vol. of Packet
No. of Match Box = —————————
Vol. of Matchbox
8 × 8 × 8
No. of Match Box = ——————
2 × 2 × 2
8 × 8 × 8
No. of Match Box = ——————
8
No. of Match Box = 8 × 8
No. of Match Box = 64
(i) 9(y+1)(x+2)
(ii) m(m+2)(m+1)
Sol. (i)
9(y+1)(x+2)
= 9[(y+1)(x+2)]
= 9[(y(x+2)+1(x+2)]
= 9[xy+2y+1x+2]
= 9xy+18y+9x+18
Sol. (ii)
m(m+2)(m+1)
= m[(m+2)(m+1)]
= m[(m(m+2)+1(m+2)]
= m[m²+2m+1m+2]
= m[m²+3m+2]
= m³+3m²+2m
Q.8 find the value by applying the suitable identities
(i) (5x –4y)²
(ii) (7z + 3y)(7z + 3y)
Sol.
(i) (5x –4y)²
By identity
(a – b)² = (a)² – 2.a.b + (b)²
(5x – 4y)²
= (5x)² – 2.5x.4y + (4y)²
= 25x² – 40xy + 16y²
(ii) (7z + 3y)(7z + 3y)
By identity
(a + b)(a + b) = (a + b)²
(a + b)² = (a)² – 2.a.b + (b)²
(7z + 3y)(7z + 3y)
= (7z + 3y)²
= (7z)² + 2.7z.3y + (3y)²
= 49x² + 42yz + 9y²
Q.9 By using the suitable identities find the value of
(956)² –(44)²
Sol.
(956)² –(44)²
By identity
(a)² –(b)² = (a+b)(a–b)
(956)² –(44)²
= (956+44)(956–44)
= (1000)(912)
= 912000
Q. 10 The width of the cuboid is 3 cm less than its length and height is 5 cm less than its length, find the volume of the cuboid?
Sol.
Let the Length of the cuboid (l)= m
then Breadth (b) = (m–3)
Height. (h) = (m–5)
volume of the cuboid = l×b×h
= m(m – 3)(m – 5)
= m[(m – 3)(m – 5)]
= m[(m(m – 5) –1(m – 5)]
= m[m² – 5m –1m + 5]
= m[m² – 6m + 5]
= m³ – 6m² + 5m
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