Question

Find the area and perimeter of an isosceles right-angled triangle, each of whose equal sides measures 10 cm. Take $\sqrt{2}$ = 1.414.

Answer 1

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{isosceles}\mathrm{right}-\mathrm{angled}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times 10\times 10 \\ =50{\mathrm{cm}}^2 \end{gathered}$$

The hypotenuse of an isosceles right-angled triangle can be obtained using Pythagoras' theorem.

If h denotes the hypotenuse, we have:

$$\begin{gathered} h^2=a^2+a^2 \\ \Rightarrow h=2a^2 \\ \Rightarrow h=\sqrt{2}a \\ \Rightarrow h=10\sqrt{2}\mathrm{cm} \end{gathered}$$

∴ Perimeter of ​the isosceles right-angled triangle = $2a+\sqrt{2}a$
$$\begin{gathered} =2\times 10+1.414\times 10 \\ =20+14.14 \\ =34.14\mathrm{cm} \end{gathered}$$