Question

The hypotenuse of a right triangle is $1$ metres more than twice the shortest side. If the third side is $7$ metres less than the hypotenuse, find the sides of the triangle.

Answer 1

consider a given right angle triangle

let whose shortest side is base(B) and third side is length(L) and H be a hypotenuse.

According to the given question,

$H=2B+6......\left(1\right)$

$L=H-2........\left(2\right)$

By equation (1),

$B=\frac{H-6}{2}.......\left(3\right)$

We know that,

By Pythagoras theorem,

$H^2=B^2+L^2$

$\Rightarrow H^2={\left(\frac{H-6}{2}\right)}^2+{(H-2)}^2$

$\Rightarrow H^2=\frac{H^2+36-12H}{4}+H^2+4-4H$

$\Rightarrow 4H^2=H^2+36-12H+4H^2+16-16H$

$\Rightarrow H^2+36-12H+16-16H=0$

$\Rightarrow H^2+52-28H=0$

$\Rightarrow H^2-28H+52=0$

$\Rightarrow H^2-(26+2)H+52=0$

$\Rightarrow H^2-26H-2H+52=0$

$\Rightarrow H(H-26)-2(H-26)=0$

$\Rightarrow (H-26)(H-2)=0$

$\Rightarrow H=2,26$

$H=2$ is very small then $H=26$ exists.

Put the value of $H$ in equation (2) and (3) to, we get

$L=H-2$

$L=26-2=24$

$B=\frac{H-6}{2}$

$B=\frac{26-6}{2}$

$B=\frac{20}{2}$

$B=10$

Hence, $B=10m$, $L=24m$ and $H=26m$