Find the percentage increased in the …

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% ================================================================ Hope this will help you.....

400 percent