**Find the quotient in each case:**

**(i) (x ^{8})/(x^{3})**

**(ii) (m ^{8})/(m^{3})**

**(iii) (p ^{14})/(p^{8})**

**(iv) (30k ^{9})/(6k^{2})**

**(v) (-15x ^{6})/(5x^{5})**

**(vi) (-9a ^{10})/(-3a^{9})**

**(vii) (-9y ^{6})/(9y^{2})**

**(viii) (2x ^{2}y)/_{(xy/2)}**

**(ix) (-18h ^{4}h^{5})/(-6^{2}h^{2})**

**(x) [(1/8)x ^{6}]/[(1/4)x^{5}]**

Solution:

(i) (x^{8})/(x^{3}) = x^{8-3} = x^{5}

(ii) (m^{8})/(m^{3}) = m^{8-3} = m^{5}

(iii) (p^{14})/(p^{8}) = p^{14-8} = p^{6}

(iv) (30k^{9})/(6k^{2}) = 5k^{9-2} = 5k^{7}

(v) (-15x^{8})/(5x^{5}) = -3x^{8-5} = -3x^{3}

(vi) (-9a^{10})/(-3a^{9}) = 3a

(vii) (-9y^{6})/(9y^{2}) = -y^{6-2} = -y^{4}

(viii) (2x^{2}y)/_{(xy/2) }= (4x^{2}y)/(xy) = 4x

(ix) (-18h^{4}k^{5})/(-6k^{2}h^{2}) = 3h^{4-2}k^{5-2} = 3h^{2}k^{3}

(x) [(1/8)x^{6}]/[(1/4)x^{3}] = (4x^{6})/(8x^{3}) = x^{6-3}/_{4} = x^{3}/_{4}

- A
**rea of a rectangle = length x breadth = 800x**^{2}

Solution:

Length = 10x

Breadth = ^{area}/_{length} = 800x^{2}/10x = 80x

**An isoscless right triangle has its area 20x**^{2}. 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 √3a ^{2}/_{4})**

Solution:

In an isoscless right tringale ABC

AB = BC = a cm

Area of ∆ABc = ^{1}/_{2} x AB x BC

20x^{2} = ^{1}/_{2} x a x a

20x^{2} = a^{2}/_{2}

40x^{2} = a^{2} ————-(1)

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

Area of an equilateral triangle = √3a^{2}/4 = √3 x 40 *x ^{2}*/

_{4}= 10√3x

^{2}

**The area of a rectangle is 36x**^{4}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 A_{1} = 36x^{4}

Length L_{1} = 4x

area of the square of length 4x,

A_{2} = (4x)^{2} = 16x^{2}

^{A1}/_{A2} = (36x^{4})/(16x^{2}) = ^{9}/_{4x}^{2}

**The area of a square is 64x**^{2}an equilateral triangle is constructed on one of its sides. What is the altitude of this triangle?

Solution:

Area of the square = 64x^{2}

Length of the sides = √(64x^{2}) = 8x

Altitude of the equilateral ∆with side 8x is

(8x)^{2} = (4x)^{2} + h^{2}

64x^{2} = 16x^{2} + h^{2}

h^{2} = 64x^{2} – 16x^{2} = 48x^{2}

h = √(48x^{2}) = √(16x3x) = 4√(3x)

## One thought on “Division – Exercise 3.4.2 – Class IX”

Comments are closed.