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

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Question

(A) Let the right-angled triangle be $ABC$ with hypotenuse $AC = h$. Let $\angle A = 	heta$. Then the sides are $AB = h \cos 	heta$ and $BC = h \sin

Vedclass · Accepted Answer

$h^2 / 4$

Vedclass · Answer

(A) Let the right-angled triangle be $ABC$ with hypotenuse $AC = h$. Let $\angle A = 	heta$. Then the sides are $AB = h \cos 	heta$ and $BC = h \sin 	heta$. The area $A$ of the triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} (h \cos 	heta)(h \sin 	heta)$. $A = \frac{h^2}{2} \sin 	heta \cos 	heta = \frac{h^2}{4} (2 \sin 	heta \cos 	heta) = \frac{h^2}{4} \sin 2	heta$. For the area to be maximum,$\sin 2	heta$ must be maximum,which is $1$ when $2	heta = 90^{\circ}$ or $	heta = 45^{\circ}$. Thus,the maximum area is $\frac{h^2}{4} 	imes 1 = \frac{h^2}{4}$.

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

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