Question

The perimeter of an isosceles triangle is $42cm$ and its base is $1\frac{1}{2}$ times its congruent sides, then the area of the triangle is approximately equal to

Answer 1

Let the equal side of the isosceles triangle be $x$.

Base of isosceles triangle is $\frac{3}{2}x$.

Given perimeter of isosceles triangle $=42$ cm and we know that perimeter is sum of all sides, therefore,

$x+x+\frac{3}{2}x=42\Rightarrow \frac{7}{2}x=42\Rightarrow 7x=84\Rightarrow x=12$

Thus, the equal sides are $12$ cm each and base is $18$ cm.

Since we have found all the three sides of triangle, therefore we find the area of triangle by Herons formula:

$S=\frac{18+12+12}{2}=\frac{42}{2}=21$ cm

$A=\sqrt{21\times (21-18)\times (21-12)\times (21-12)}=\sqrt{21\times 3\times 9\times 9}=\sqrt{5103}=71.435=72$

Hence, the area of the triangle is $72$ cm${}^2$(approx).