In an isosceles right-angled triangle, the length of the hypotenuse is $20$. Find the perimeter and the area of the triangle.

(N/A) Let the two equal sides of the isosceles right-angled triangle be $x$. According to the Pythagorean theorem, $x^2 + x^2 = (20)^2$.
$2x^2 = 400$, which implies $x^2 = 200$.
Therefore, $x = \sqrt{200} = 10\sqrt{2}$.
The perimeter of the triangle is the sum of all sides: $P = x + x + 20 = 2x + 20 = 2(10\sqrt{2}) + 20 = 20\sqrt{2} + 20$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times x = \frac{1}{2} \times x^2$.
Substituting $x^2 = 200$, we get Area $= \frac{1}{2} \times 200 = 100$.