Question

find the percentage increased in area when sides of triangle is tripled

Answer 1

Let x,y,z be the sides of the original ∆ & s be its semi perimeter.

S= $\frac{x+y+z}{2}$

3s= x+y+z.................(i)

The sides of a new ∆ are 3x,3y,3z

[ given: Side is tripled]

Let s' be the new semi perimeter.

$s'= \frac{(3x+3y+3z)}{2} \\ s'= \frac{3(x+y+z)}{2} \\ s'= x+y+z$

S'= 3s. ( From eq 1)......(2)

Let ∆= area of original triangle

∆= √s(s-x)(s-y)(s-z).........(3)

&

∆'= area of new Triangle

∆' = √s'(s'-3x)(s'-3y)(s'-3z)

∆'= √ 3s(3s-3x)(3s-3y)(3s-3z)

[From eq. 2]

∆'= √ 3s×3(s-x)×3(s-y)×3(s-z)

= √81s(s-x)(s-y)(s-z)

∆'= 9√s(s-x)(s-y)(s-z)

∆'= 9∆. (From eq (3))

Increase in the area of the triangle= ∆'- ∆= 9∆ - 1∆= 8∆

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

% increase in area= (9∆/∆)×100

% increase in area= 9×100=900 %

Hence, the percentage increase in the area of a triangle is 900%