Question

The length of a hypotenuse of a right triangle exceeds the length of its base by $2\text{cm}$ and exceeds twice the length of the altitude by $1\text{cm}$. Find the length of each side of the triangle.

Answer 1

Step 1: Determining the quadratic equation in $x$

Let the hypotenuse of a right triangle be $x$

As given, the hypotenuse of a right triangle exceeds the length of its base by $2\text{cm}$ and exceeds twice the length of the altitude by $1\text{cm}$.

So, the base of a triangle $=x-2$

And the altitude of a triangle $=\frac{x-1}{2}$

By Pythagoras theorem,

$$\begin{gathered} \Rightarrow x^2={(x-2)}^2+{\left(\frac{x-1}{2}\right)}^2 \\ \Rightarrow x^2=\frac{4{(x-2)}^2+{(x-1)}^2}{4} \\ \Rightarrow 4x^2=4x^2-16x+16+x^2-2x+1 \\ \Rightarrow x^2-18x+17=0 \end{gathered}$$

Step 2: Finding the length of each side of the triangle

By Factorization method

$$\begin{gathered} x^2-18x+17=0 \\ \Rightarrow x^2-17x-x+17=0 \\ \Rightarrow \left(x-1\right)\left(x-17\right)=0 \\ \Rightarrow x=1,17 \end{gathered}$$

Now if take $x=1$, then the base of a triangle has the negative value $\left(1-2=-1\right)$ which is not possible.

So, the length of hypotenuse $\left(x\right)=17\text{cm}$

The length of base $\left(x-2\right)=15\text{cm}$

And the length of altitude $\left(\frac{x-1}{2}\right)=8\text{cm}$

Hence, the length of hypotenuse, base, and altitude of the right-angled triangle are $17\text{cm},15\text{cm}$ and $\text{8 cm}$ respectively.