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(C) In an isosceles right-angled triangle,let the two equal sides be $a$.By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^2 + a^

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$16\, cm^{2}$

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(C) In an isosceles right-angled triangle,let the two equal sides be $a$.By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^2 + a^2} = a\sqrt{2}$.Given that the hypotenuse $h = 8\, cm$,we have $a\sqrt{2} = 8$.Thus,$a = \frac{8}{\sqrt{2}} = 4\sqrt{2}\, cm$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Since the base and height are both $a$,$	ext{Area} = \frac{1}{2} 	imes a^2$.Substituting the value of $a$,$	ext{Area} = \frac{1}{2} 	imes (4\sqrt{2})^2 = \frac{1}{2} 	imes (16 	imes 2) = \frac{1}{2} 	imes 32 = 16\, cm^{2}$.

Find the area of an isosceles right-angled triangle whose hypotenuse is $8\, cm$.

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