The area of the right-angled isosceles triangle having hypotenuse A cm is:

$\frac{A^2}{4}$

The correct answer is Option (3) → $\frac{A^2}{4}$

Given: A right-angled isosceles triangle with hypotenuse = $A$ cm.

In a right-angled isosceles triangle, the two perpendicular sides are equal. Let each side be $x$.

By Pythagoras theorem:

$x^2 + x^2 = A^2$

$2x^2 = A^2$

$x^2 = \frac{A^2}{2}$

Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times x = \frac{1}{2}x^2$

Substitute $x^2$:

Area = $\frac{1}{2} \cdot \frac{A^2}{2} = \frac{A^2}{4}$