Question

Find the increase percentage of triangle whose side are soubled

Answer 1

Let a,b,c be the sides of the original ∆ & s be its semi perimeter
.S= (a+b+c)/22s= 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+cS'= 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 areaof 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%

Hope this will help you.....