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) = x^8-3 = x⁵

(ii) (m⁸)/(m³) = m^8-3 = m⁵

(iii) (p¹⁴)/(p⁸) = p^14-8 = p⁶

(iv) (30k⁹)/(6k²) = 5k^9-2 = 5k⁷

(v) (-15x⁸)/(5x⁵) = -3x^8-5 = -3x³

(vi) (-9a¹⁰)/(-3a⁹) = 3a

(vii) (-9y⁶)/(9y²) = -y^6-2 = -y⁴

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

(ix) (-18h⁴k⁵)/(-6k²h²) = 3h^4-2k^5-2 = 3h²k³

(x) [(1/8)x⁶]/[(1/4)x³] = (4x⁶)/(8x³) = x^6-3/₄ = x³/₄

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

Solution:

Length = 10x

Breadth = ^area/_length = 800x²/10x = 80x

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3. 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

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4. 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

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5. 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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