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

**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 a^{2}.

Let the side of an equilateral triangle be X.

Then initial area = sqrt(3)/4 x X^{2}.

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 (121X^{2} / 400)

= 121 x X^{2} x sqrt(3)/1600

Then change in area = sqrt(3)/4 x X^{2} - 121 x X^{2} x sqrt(3)/1600

= sqrt(3)/4 x X^{2} [1 - 121/400]

= sqrt(3)/4 x X^{2} [279/400]

Therefore, decrease % = [change in area /original area] x 100

= {{sqrt(3)/4 x X^{2} [ 279/400]}/sqrt(3)/4 x X^{2}} 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)(a^{2})

Given that, the side of another triangle = 2a

Then area = A1 = (sqrt 3/4)(2a)^{2} = (sqrt3/4)[4(a)^{2}] = 4[(sqrt3/4)(a^{2})]

From A and A1, we have

4A = A1.

A/A1 = 1/4

Hence the answer is A:A1 = 1:4.