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

In a right-angled isosceles triangle, two sides are equal in length and meet at a 90° angle.

Let the two equal sides each have length $a$ cm.

The hypotenuse is given as $A$ cm.

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By the Pythagorean theorem:

$a^2 + a^2 = A^2$

$2a^2 = A^2$

$a^2 = \dfrac{A^2}{2}$

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The two equal sides are perpendicular to each other, so they serve as the base and height of the triangle.

Area $= \dfrac{1}{2} \times \text{base} \times \text{height}$

Area $= \dfrac{1}{2} \times a \times a$

Area $= \dfrac{1}{2} \times a^2$

Substituting $a^2 = \dfrac{A^2}{2}$:

Area $= \dfrac{1}{2} \times \dfrac{A^2}{2}$

Area $= \dfrac{A^2}{4}$

Therefore, the area of the triangle is $\dfrac{A^2}{4}$ square cm.