Question

The percentage increase in the area of triangle, if its each side is quadrupled is equal to (a) 1500% (b) 1200% (c) 900% (d) 800%

Answer 1

Let a,b,c be the sides of the original ∆ & s be its semi perimeter.

S= (a+b+c)/2

2s= a+b+c.................(1)

The sides of a new ∆ are 2a,2b,2c

[ given: Side is doubled]

Let s' be the new semi perimeter.

s'= (2a+2b+2c)/2

s'= 2(a+b+c) /2

s'= a+b+c

S'= 2s. ( From eq 1)......(2)


Let ∆= area of original triangle

∆= √s(s-a)(s-b)(s-c).........(3)


&

∆'= area of new Triangle

∆' = √s'(s'-2a)(s'-2b)(s'-2c)

∆'= √ 2s(2s-2a)(2s-2b)(2s-2c)

[From eq. 2]


∆'= √ 2s×2(s-a)×2(s-b)×2(s-c)

= √16s(s-a)(s-b)(s-c)

∆'= 4 √s(s-a)(s-b)(s-c)

∆'= 4∆. (From eq (3))

Increase in the area of the triangle= ∆'- ∆= 4∆ - 1∆= 3∆

%increase in area= (increase in the area of the triangle/ original area of the triangle)× 100

% increase in area= (3∆/∆)×100


% increase in area= 3×100=300 %

Hence, the percentage increase in the area of a triangle is 300%

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Hope this will help you...