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(D) Let the two equal sides of the isosceles right-angled triangle be $a$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} \

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$20 \sqrt{2}$

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(D) Let the two equal sides of the isosceles right-angled triangle be $a$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Since it is an isosceles right-angled triangle,the base and height are equal,so $	ext{Area} = \frac{1}{2} a^2$.Given $	ext{Area} = 200 \, cm^2$,we have:$200 = \frac{1}{2} a^2$$a^2 = 400$$a = 20 \, cm$.The hypotenuse $h$ of a right-angled triangle is given by $h = \sqrt{a^2 + a^2} = a\sqrt{2}$.Substituting the value of $a$:$h = 20 \sqrt{2} \, cm$.

An isosceles right-angled triangle has an area of $200 \, cm^2$. The length of its hypotenuse is ..... $cm$.

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