Find the area and perimeter of an isosceles right-angled triangle,

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a

Area of isosceles right $-$ angled triangle $=\frac{1}{2} \times$ Base $\times$ Height

$=\frac{1}{2} \times 10 \times 10$

$=50 \mathrm{~cm}^{2}$

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

If h denotes the hypotenuse, we have:

$h^{2}=a^{2}+a^{2}$

$\Rightarrow h=2 a^{2}$

$\Rightarrow h=\sqrt{2} a$

$\Rightarrow h=10 \sqrt{2} \mathrm{~cm}$

$\therefore$ Perimeter of the isosceles right-angled triangle $=2 a+\sqrt{2} a$

$=2 \times 10+1.41 \times 10$

$=20+14.1$

$=34.1 \mathrm{~cm}$