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

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(D) Let the base be $b$.The altitude (or perpendicular) is $\sqrt{h^2-b^2}$.The area $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude} = \fra

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

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(D) Let the base be $b$.The altitude (or perpendicular) is $\sqrt{h^2-b^2}$.The area $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude} = \frac{1}{2} b \sqrt{h^2-b^2}$.To find the maximum area,differentiate $A$ with respect to $b$:$\frac{dA}{db} = \frac{1}{2} \left[ \sqrt{h^2-b^2} + b \cdot \frac{-2b}{2\sqrt{h^2-b^2}} ight] = \frac{1}{2} \left[ \frac{h^2-b^2-b^2}{\sqrt{h^2-b^2}} ight] = \frac{h^2-2b^2}{2\sqrt{h^2-b^2}}$.Setting $\frac{dA}{db} = 0$,we get $h^2 - 2b^2 = 0$,which implies $b = \frac{h}{\sqrt{2}}$.Substituting $b = \frac{h}{\sqrt{2}}$ into the area formula:$A = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \sqrt{h^2 - \frac{h^2}{2}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \frac{h}{\sqrt{2}} = \frac{h^2}{4}$.

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

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