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The maximum area of a right angled triangle with hypotenuse $h$ is
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2188 Upvotes
Verified Answer
The correct answer is:
$h^2 / 4$
Let a right angled triangle $\mathrm{OAB}$, with $\mathrm{h}$ hypoteneous
Let $\angle \mathrm{OAh}=\theta$
now $\mathrm{OA}=\mathrm{h} \cos \theta$ and $\mathrm{OB}=\mathrm{h} \sin \theta$
$$
\begin{aligned}
& \Rightarrow \text { Area }=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}=\frac{1}{2} \mathrm{~h}^2\left(\frac{1}{2} \sin \theta \cos \theta\right) \\
& =\frac{1}{4} \mathrm{~h}^2 \sin 2 \theta, \sin 2 \theta, \text { for maximum } \theta=45^{\circ} \\
& \Rightarrow \frac{1}{4} \mathrm{~h}^2
\end{aligned}
$$
Let $\angle \mathrm{OAh}=\theta$
now $\mathrm{OA}=\mathrm{h} \cos \theta$ and $\mathrm{OB}=\mathrm{h} \sin \theta$
$$
\begin{aligned}
& \Rightarrow \text { Area }=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}=\frac{1}{2} \mathrm{~h}^2\left(\frac{1}{2} \sin \theta \cos \theta\right) \\
& =\frac{1}{4} \mathrm{~h}^2 \sin 2 \theta, \sin 2 \theta, \text { for maximum } \theta=45^{\circ} \\
& \Rightarrow \frac{1}{4} \mathrm{~h}^2
\end{aligned}
$$
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