Question

The altitude of a right triangle is $7cm$ less than its base. If the hypotenuse is $13cm$, find the other two sides.

Answer 1

Apply Pythagoras Theorem to find the other two sides

Given

The hypotenuse $=13cm$

Let the base of the right triangle be $xcm$

The altitude of the right triangle $=(x-7)cm$

From Pythagoras' theorem, we know,

${\mathrm{Base}}^2+{\mathrm{Altitude}}^2={\mathrm{Hypotenuse}}^2$

$\Rightarrow x^2+{(x-7)}^2={13}^2$

$$\begin{gathered} \Rightarrow x^2+x^2+49-14x=169\because {(a-b)}^2=a^2-2ab+b^2 \\ \Rightarrow 2^{}{x}^2-14x+49=169 \\ \Rightarrow 2x^2-14x+49-169=0 \\ \Rightarrow 2x^2-14x-120=0 \\ \Rightarrow x^2-7x-60=0(dividingtheequationby2) \\ \Rightarrow x^2-12x+5x-60=0 \\ \Rightarrow x(x-12)+5(x-12)=0 \\ \Rightarrow (x-12)+(x+5)=0 \end{gathered}$$

Thus, either $x=12orx=-5$

Since sides cannot be negative, $x$ can only be $12$

Hence, the base of the given triangle is $12cm$

The altitude of the triangle $=\left(x-7\right)cm$

$=\left(12-7\right)cm$

$=5cm$
Hence the remaining two sides of triangle are $12cm$ and $5cm$