The maximum area of a right-angled triangle w | vedclass

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(A) Let the two legs of the right-angled triangle be $x$ and $y$. Given the hypotenuse is $h$,we have $x^2 + y^2 = h^2$. The area of the triangle is $

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$\frac{h^2}{4}$

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(A) Let the two legs of the right-angled triangle be $x$ and $y$. Given the hypotenuse is $h$,we have $x^2 + y^2 = h^2$. The area of the triangle is $A = \frac{1}{2}xy$. To maximize $A$,we maximize $A^2 = \frac{1}{4}x^2y^2$. Substituting $y^2 = h^2 - x^2$,we get $f(x) = \frac{1}{4}x^2(h^2 - x^2) = \frac{1}{4}(h^2x^2 - x^4)$. Taking the derivative with respect to $x$: $f'(x) = \frac{1}{4}(2h^2x - 4x^3)$. Setting $f'(x) = 0$,we get $2h^2x = 4x^3$,which implies $x^2 = \frac{h^2}{2}$,so $x = \frac{h}{\sqrt{2}}$. Then $y^2 = h^2 - \frac{h^2}{2} = \frac{h^2}{2}$,so $y = \frac{h}{\sqrt{2}}$. The maximum area is $A = \frac{1}{2} 	imes \frac{h}{\sqrt{2}} 	imes \frac{h}{\sqrt{2}} = \frac{h^2}{4}$.

The maximum area of a right-angled triangle with hypotenuse $h$ is

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