MCQ
The maximum area of a right angled triangle with hypotenuse $h$ is
- A$\frac{{{h^2}}}{{2\sqrt 2 }}$
- B$\frac{{{h^2}}}{{2}}$
- C$\frac{{{h^2}}}{{\sqrt 2 }}$
- ✓$\frac{{{h^2}}}{{4}}$
✓
Answer
Correct option: D.
$\frac{{{h^2}}}{{4}}$
d
Let base $=b$
Let base $=b$
Altitude (or perpendicur) $ = \sqrt {{h^2} - {b^2}} $
Area, $\mathrm{A}=\frac{1}{2} \times$ base $\times$ altitude
$=\frac{1}{2} \times b \times \sqrt{h^{2}-b^{2}}$
$\Rightarrow \frac{d \mathrm{A}}{d b}=\frac{1}{2}\left[\sqrt{h^{2}-b^{2}}+b \cdot \frac{-2 b}{2 \sqrt{h^{2}-b^{2}}}\right]$
$=\frac{1}{2}\left[\frac{h^{2}-2 b^{2}}{\sqrt{h^{2}-b^{2}}}\right]$
Put $\frac{d \mathrm{A}}{d b}=0,$
$\Rightarrow \quad b=\frac{h}{\sqrt{2}}$
Maximum area $=\frac{1}{2} \times \frac{h}{\sqrt{2}} \times \sqrt{h^{2}-\frac{h^{2}}{2}}=\frac{h^{2}}{4}$
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