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(D) Let the sides of the right-angled triangle be $x$ and $y$,and the hypotenuse be $h = 5$. By Pythagoras theorem,$x^2 + y^2 = 5^2 = 25$. The area of

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$5(\sqrt{2}+1)$

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(D) Let the sides of the right-angled triangle be $x$ and $y$,and the hypotenuse be $h = 5$. By Pythagoras theorem,$x^2 + y^2 = 5^2 = 25$. The area of the triangle is $A = \frac{1}{2}xy$. To maximize $A$,we maximize $A^2 = \frac{1}{4}x^2y^2$. Let $x^2 = 25 \cos^2 	heta$ and $y^2 = 25 \sin^2 	heta$,where $	heta \in (0, \pi/2)$. Then $A = \frac{1}{2} (5 \cos 	heta)(5 \sin 	heta) = \frac{25}{4} \sin(2	heta)$. The area $A$ is maximum when $\sin(2	heta) = 1$,which means $2	heta = \pi/2$,so $	heta = \pi/4$. Thus,$x = 5 \cos(\pi/4) = \frac{5}{\sqrt{2}}$ and $y = 5 \sin(\pi/4) = \frac{5}{\sqrt{2}}$. The triangle is an isosceles right-angled triangle. The perimeter $P = x + y + h = \frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}} + 5 = \frac{10}{\sqrt{2}} + 5 = 5\sqrt{2} + 5 = 5(\sqrt{2} + 1)$.

If the area of a right-angled triangle with hypotenuse $5$ is maximum,then its perimeter is

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