Division – Exercise 3.4.2 – Class IX

1. Find the quotient in each case:

(i) (x8)/(x3)

(ii) (m8)/(m3)

(iii) (p14)/(p8)

(iv) (30k9)/(6k2)

(v) (-15x6)/(5x5)

(vi) (-9a10)/(-3a9)

(vii) (-9y6)/(9y2)

(viii) (2x2y)/(xy/2)

(ix) (-18h4h5)/(-62h2)

(x) [(1/8)x6]/[(1/4)x5]

Solution:

(i) (x8)/(x3) = x8-3 = x5

(ii) (m8)/(m3) = m8-3 = m5

(iii) (p14)/(p8) = p14-8 = p6

(iv) (30k9)/(6k2) = 5k9-2 = 5k7

(v) (-15x8)/(5x5) = -3x8-5 = -3x3

(vi) (-9a10)/(-3a9) = 3a

(vii) (-9y6)/(9y2) = -y6-2 = -y4

(viii) (2x2y)/(xy/2) = (4x2y)/(xy) = 4x

(ix) (-18h4k5)/(-6k2h2) = 3h4-2k5-2 = 3h2k3

(x) [(1/8)x6]/[(1/4)x3] = (4x6)/(8x3) = x6-3/4 = x3/4

1. Area of a rectangle = length x breadth = 800x2

Solution:

Length = 10x

Breadth = area/length = 800x2/10x = 80x

1. An isoscless right triangle has its area 20x2. What is the area of the equilateral triangle constructed on one of its side?

(hint: area of an equilateral triangle with side length a is √3a2/4)

Solution:

In an isoscless right tringale ABC

AB = BC = a cm

Area of ∆ABc = 1/2 x AB x BC

20x2 = 1/2 x a x a

20x2 = a2/2

40x2 = a2 ————-(1)

The area of an equilateral triangle with AB or BC as side

Area of an equilateral triangle = √3a2/4 = √3 x 40 x2/4 = 10√3x2

1. The area of a rectangle is 36x4 and one of its sides is 4x. A square is constructed on the other side. What is ratio of the area of the square to that of the rectangle?

Solution:

Area of a rectangle A1 = 36x4

Length L1 = 4x

area of the square of length 4x,

A2 = (4x)2 = 16x2

A1/A2 = (36x4)/(16x2) = 9/4x2

1. The area of a square is 64x2 an equilateral triangle is constructed on one of its sides. What is the altitude of this triangle?

Solution:

Area of the square = 64x2

Length of the sides = √(64x2) = 8x

Altitude of the equilateral ∆with side 8x is

(8x)2 = (4x)2 + h2

64x2 = 16x2 + h2

h2 = 64x2 – 16x2 = 48x2

h = √(48x2) = √(16x3x) = 4√(3x)

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