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(B) Let the two equal sides of the isosceles right-angled triangle be $x \, cm$. Then,the hypotenuse is $\sqrt{x^2 + x^2} = \sqrt{2}x \, cm$. The peri

Vedclass · Accepted Answer

$(3-2 \sqrt{2}) p^2 \, cm^2$

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(B) Let the two equal sides of the isosceles right-angled triangle be $x \, cm$. Then,the hypotenuse is $\sqrt{x^2 + x^2} = \sqrt{2}x \, cm$. The perimeter is given as $x + x + \sqrt{2}x = 2p \, cm$. $x(2 + \sqrt{2}) = 2p \implies x = \frac{2p}{2 + \sqrt{2}}$. Rationalizing the denominator: $x = \frac{2p(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2p(2 - \sqrt{2})}{4 - 2} = p(2 - \sqrt{2})$. The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} x^2$. Area $= \frac{1}{2} [p(2 - \sqrt{2})]^2 = \frac{1}{2} p^2 (4 + 2 - 4\sqrt{2}) = \frac{1}{2} p^2 (6 - 4\sqrt{2}) = (3 - 2\sqrt{2}) p^2 \, cm^2$.

The perimeter of an isosceles right-angled triangle is $2p \, cm$. Its area is

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