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

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(D) Let the base be $b$ and the hypotenuse be $h$.Then the altitude (perpendicular) is $\sqrt{h^2 - b^2}$.The area $A$ of the triangle is given by $A

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the base be $b$ and the hypotenuse be $h$.Then the altitude (perpendicular) is $\sqrt{h^2 - b^2}$.The area $A$ of the triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude} = \frac{1}{2} b \sqrt{h^2 - b^2}$.To maximize the area,we 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 $b^2 = \frac{h^2}{2}$,or $b = \frac{h}{\sqrt{2}}$.Substituting this value into the area formula:$A_{max} = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \sqrt{h^2 - \frac{h^2}{2}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \frac{h}{\sqrt{2}} = \frac{h^2}{4}$.

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

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