Question

The area of a right angled triangle is $40$ sq.cm and its perimeter is $40$ cm.Find the length of its hypotenuse.

Answer 1

Let the hypotenuse $AC=h$ and the other two sides be $AB=a$ and $BC=b$

By Pythagoras theorem, we have $h^2=a^2+b^2$ .....$\left(1\right)$

Given:Area of $△ABC=40$sq.m

Hence,$\frac{1}{2}\times a\times b=40$

$\Rightarrow ab=80$ .......$\left(2\right)$

Given perimeter of $△ABC=40$ cm

That is $a+b+h=40$

$\Rightarrow a+b=40–h$

Squaring on both the sides, we get

${(a+b)}^2={(40-h)}^2$

$\Rightarrow a^2+b^2+2ab=1600-80h+h^2$

$\Rightarrow h^2+2\left(80\right)=1600-80h+h^2$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 160=1600-80h$

$\Rightarrow 80h=1600–160=1440$

$\therefore h=18$ cm

Thus the length of hypotenuse is $18$ cm.