Question

The hypotenuse of a right angled triangle is $6$ meters more than twice the shortest side. If the third side is $2$ meters less than the hypotenuse. find the sides of the triangle.

Answer 1

Step 1: Finding the quadratic equation in $x$

Let the shortest side of the right-angled triangle be $x$ meters.

Since it is given that, the hypotenuse of a right-angled triangle is $6$ meters more than twice the shortest side and the third side of triangle is $2$ meters less than the hypotenuse.

So, we have hypotenuse$=\left(2x+6\right)$ meters

And the third side $=\left(2x+6\right)-2$

$=\left(2x+4\right)$ meters

By Pythagoras theorem,

$$\begin{gathered} \Rightarrow {(2x+6)}^2=x^2+{(2x+4)}^2 \\ \Rightarrow 4x^2+24x+36=x^2+4x^2+16x+16 \\ \Rightarrow x^2-8x-20=0 \end{gathered}$$

Step 2: Finding the sides of the triangle

By factorization method

$$\begin{gathered} x^2-8x-20=0 \\ \Rightarrow x^2-10x+2x-20=0 \\ \Rightarrow (x-10)(x+2)=0 \\ \Rightarrow x=-2,10 \end{gathered}$$

The side of the triangle cannot have a negative value.

So, the shortest side $\left(x\right)=10$ meters

Hypotenuse $\left(2x+6\right)=2\times 10+6$

$=26$ meters

And the third side $\left(2x+4\right)=2\times 10+4$

$=24$ meters

Hence, the sides of the right-angled triangle are $10$ meters, $24$ meters, and $26$ meters.