Question

If the area of an isosceles right-angled triangle is 72 cm², then its perimeter is _________.

Answer 1

Given: Area of an isosceles right-angled triangle is 72 cm²

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base}\left(\because \mathrm{Base}=\mathrm{Height}\right) \\ =\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 72=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 144={\left(\mathrm{Base}\right)}^2 \\ \Rightarrow \mathrm{Base}=12\mathrm{cm}=\mathrm{Perpendicular} \\ \mathrm{In}\mathrm{a}\mathrm{right}-\mathrm{angled}\mathrm{triangle},\mathrm{using}\mathrm{pythagoras}\mathrm{theorem} \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2 \\ ={\left(12\right)}^2+{\left(12\right)}^2 \\ =144+144 \\ =288 \\ \Rightarrow \mathrm{Hypotenuse}=\sqrt{288}\mathrm{cm} \\ \Rightarrow \mathrm{Hypotenuse}=12\sqrt{2}\mathrm{cm} \\ \mathrm{Thus}, \\ \mathrm{Perimeter}=12+12+12\sqrt{2}=12\left(2+\sqrt{2}\right)\mathrm{cm} \end{gathered}$$

Hence, its perimeter is $\underline{12\left(2+\sqrt{2}\right)\mathrm{cm}}.$