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

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 4a,4b,4c [ given: Sides are quadrupled]

Let s’ be the new semi perimeter.

s’= (4a+4b+4c)/2

s’= 4(a+b+c) /2

s’= 2(a+b+c)

s’= 4s ( From eq 1)……(2)

Let ∆= area of original triangle

∆= √s(s-a)(s-b)(s-c)………(3)

&

∆’= area of new Triangle

∆’ = √s'(s’-4a)(s’-4b)(s’-4c)

∆’= √ 4s(4s-4a)(4s-4b)(4s-4c)

[From eq. 2]

∆’= √ 4s×4(s-a)×4(s-b)×4(s-c)

= √256s(s-a)(s-b)(s-c)

∆’= 16 √s(s-a)(s-b)(s-c)

∆’= 16∆. (From eq (3))

Increase in the area of the triangle= ∆’- ∆= 16∆ – 1∆= 15∆

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

% increase in area= (15∆/∆)×100

% increase in area= 15×100=1500 %

Hence, the percentage increase in the area of a triangle is 1500%