The sum of the lengths of the hypotenuse and | vedclass

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(C) Let the hypotenuse be $h$ and one side be $x$. Given $h + x = c$ (constant),so $h = c - x$.In a right-angled triangle,the third side is $\sqrt{h^2

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$\frac{\pi}{3}$

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(C) Let the hypotenuse be $h$ and one side be $x$. Given $h + x = c$ (constant),so $h = c - x$.In a right-angled triangle,the third side is $\sqrt{h^2 - x^2} = \sqrt{(c-x)^2 - x^2} = \sqrt{c^2 - 2cx}$.The area $A$ of the triangle is $A = \frac{1}{2} 	imes x 	imes \sqrt{c^2 - 2cx}$.To maximize $A$,we maximize $A^2 = \frac{1}{4} x^2 (c^2 - 2cx) = \frac{1}{4} (c^2 x^2 - 2cx^3)$.Let $f(x) = c^2 x^2 - 2cx^3$. Differentiating with respect to $x$:$f'(x) = 2c^2 x - 6cx^2 = 2cx(c - 3x)$.Setting $f'(x) = 0$,we get $x = \frac{c}{3}$ (since $x 
eq 0$).Now,$h = c - \frac{c}{3} = \frac{2c}{3}$.Let $	heta$ be the angle between the hypotenuse and the side $x$. Then $\cos 	heta = \frac{x}{h} = \frac{c/3}{2c/3} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

The sum of the lengths of the hypotenuse and one side of a right-angled triangle is constant. The area of the triangle will be maximum if the angle between them is:

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