Question

The length of the hypotenuse of the right triangle exceeds the length of the base by $2cm$ and exceeds twice the length of the altitude by $1cm$. Find the length of each side of the $△$.

Answer 1

Let hypotenuse be $a$

Given, $base=a-2$ ---------------- (1)

Given, $altitude=\frac{(a-1)}{2}$ ---------------- (2)

By Pythagoras theorem

$a^2=(a-2{)}^2+{\left(\frac{a-1}{2}\right)}^2$

$a^2=\frac{a^2-2a+1}{4}+a^2-4a+4$

$4a^2=a^2-2a+1+4(a^2-4a+4)$

$4a^2=a^2-2a+1+4a^2-16a+16$

$a^2-18a+17=0$

$(a-17)(a-1)=0$

$a=17$ ------------ $a$ cannot be $1$ as the base will be $-1$ in that case

Substitute $a=17$ in (2) we get

Length of each side is

$altitude=\frac{17-1}{2}=8cm$

$hypotenuse=17cm$

$base=17-2=15cm$