Question

In a right-angled triangle, the length of its base is one less than twice the perpendicular, and the hypotenuse is one more than twice the perpendicular. Find the length of the base, perpendicular and hypotenuse if the perimeter is$40\mathrm{cm}$.

Answer 1

Find the length of the base, perpendicular, and hypotenuse of the right-angled triangle.

Step1:Form the equation.

Let the perpendicular length be$x$, the base is$2x-1$ and the hypotenuse be$2x+1$.

$$\begin{gathered} \mathrm{Perimeter}\mathrm{of}\mathrm{right}\mathrm{angled}\mathrm{triangle}=\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{sides}\mathrm{of}\mathrm{triangle} \\ 40=\mathrm{x}+\left(2\mathrm{x}-1\right)+\left(2\mathrm{x}+1\right) \end{gathered}$$

Step2:Find the value of$x$

$$\begin{gathered} 40=\mathrm{x}+\left(2\mathrm{x}-1\right)+\left(2\mathrm{x}+1\right) \\ 40=\mathrm{x}+2\mathrm{x}-1+2\mathrm{x}+1 \\ 40=5\mathrm{x} \\ \mathrm{x}=\frac{40}{5} \\ \therefore \mathrm{x}=8 \end{gathered}$$

$\therefore$The length of the perpendicular side is $x=8\mathrm{cm}$

and the length of the base is $2x-1=2\times 8-1=16-1=15\mathrm{cm}$

and the length of the hypotenuse is$2x+1=2\times 8+1=16+1=17\mathrm{cm}$.

$\therefore$The sides of the right-angled triangle are$8\mathrm{cm},15\mathrm{cm},17\mathrm{cm}$.