- 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
- Area of a rectangle = length x breadth = 800x2
Solution:
Length = 10x
Breadth = area/length = 800x2/10x = 80x
- 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
- 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
- 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)
1 thought on “Division – Exercise 3.4.2 – Class IX”
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