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Area Problems Solved Questions For IBPS, SBI and Other Bank Exams - Page 2

You will find 28 problems in 7 pages..

Area Problems Solved Questions (Page 2 of 7)

Question 1

Find the change in the area of an equilateral triangle after decreasing 45% from each of it's side?

a) 54.25% decreasing b) 54.25% increasing c) 69.75% decreasing d) 25.84% increasing

Answer : c)69.75% decreasing

Solution:

Area of an equilateral triangle with side a = sqrt(3)/4 x a2.
Let the side of an equilateral triangle be X.
Then initial area = sqrt(3)/4 x X2.

Now, New side (after decreasing) = X - 45X/100 = 55X/100 = 11X/20.
Therefore, new area = sqrt(3)/4 x (11X/20)2
= sqrt(3)/4 x (121X2 / 400)
= 121 x X2 x sqrt(3)/1600

Then change in area = sqrt(3)/4 x X2 - 121 x X2 x sqrt(3)/1600
= sqrt(3)/4 x X2 [1 - 121/400]
= sqrt(3)/4 x X2 [279/400]

Therefore, decrease % = [change in area /original area] x 100
= {{sqrt(3)/4 x X2 [ 279/400]}/sqrt(3)/4 x X2} x 100
= [ 279/400] x 100 = 279/4 = 69.75
Hence the answer is 69.75%.


Question 2

If the height and base of the triangle is decreased and increased by 60 %, then at what percentage the area of the triangle will increase or decrease ?

a) 36% decreasing b) 36% increasing c) 40% decreasing d) 40% increasing

Answer : a) 36% decreasing

Solution :

Let the initial base = b
Let the initial height = h
Then initial area = (1/2)bh

New base (after increasing) = b + 60b/100 = 160b/100
New height (after decreasing) = h - 60h/100 = 40h/100
Then new area = (1/2)(160b/100)(40h/100)
= (16/50)bh = (8/25)bh

Area decreased = (1/2)bh - (8/25)bh
= (1/2 - 8/25)bh
= (9/50)bh

Then decrease % = (9bh/50)/(bh/2) x 100 = 9bh/50 x 2/bh x 100
= 9/25 x 100 = 9 x 4 = 36
Hence the answer is 36% decreasing.


Question 3

If the height and base of the triangle are increased by 30%, then the area of the triangle :

a) 69% decreased b) 69% increased c) 30% increased d) 30% decreased

Answer : b)69% increased.

Solution:

Let the initial base = b
Let the initial height = h
Then initial area = (1/2)bh

New base (after increasing) = b + 30b/100 = 130b/100
New height (after increasing) = h + 30h/100 = 130h/100
Then new area = (1/2)(130b/100)(130h/100)
= (1/2)(13/10)2(bh)

Increased area = (1/2)(13/10)2(bh) - (1/2)bh
= (1/2)bh [169/100 - 1]
= (1/2)bh [69/100]

Therefore, Increased % = [(1/2)bh [69/100] /(1/2)bh] x 100
= 69/100 x 100 = 69
Hence the answer is 69% increased.


Question 4

Let a be the side of an equilateral triangle and A be its area. Then find the relationship between A and A1 where A1 is the area of another equilateral triangle with side 2a.

a) 1:3 b) 1:4 c) 1:2 d) 2:1

Answer : c)1:4

Solution :

Given, side of an equilateral triangle = a
Then area = A = (sqrt 3/4)(a2)

Given that, the side of another triangle = 2a
Then area = A1 = (sqrt 3/4)(2a)2 = (sqrt3/4)[4(a)2] = 4[(sqrt3/4)(a2)]
From A and A1, we have
4A = A1.
A/A1 = 1/4
Hence the answer is A:A1 = 1:4.


Area Problems Solved Questions (Page 2 of 7)

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