Question

FIND THE PERCENTAGE INCREASE IN AN AREA OF TRIANGLE IF ITS EACH SIDE IS DOUBLE....?

Answer 1

let a,b,c are the sides of triangle and s be its  semiperimeter.
then s=1/2(a+b+c)
the sides of the triangle after doubling are 2a,2b,2c and let the new semi perimeter be z
z=1/2(2a+2b+2c)=2*1/2(a+b+c)=2s
area before doubling=√s(s-a)(s-b)(s-c)
area after doubling=√z(z-2a)(z-2b)(z-2c)=√2s(2s-2a)(2s-2b)(2s-2c)
=√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)
increase in area=4√s(s-a)(s-b)(s-c) - √s(s-a)(s-b)(s-c) = 3√s(s-a)(s-b)(s-c)
% increase in area =3√s(s-a)(s-b)(s-c)*100/√s(s-a)(s-b)(s-c)
=300%