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

The maximum area of a right angled triangle with hypotenuse $h$ is
  1. $h^2 / 2 \sqrt{2}$
  2. $h^2 / 2$
  3. $h^2 / \sqrt{2}$
  4. $h^2 / 4$

Solution

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} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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