Question:
The maximum area of a right angled triangle with hypotenuse $h$ is :
The maximum area of a right angled triangle with hypotenuse $h$ is :
Updated On: Apr 1, 2026
- $\frac{h^{2}}{2\sqrt{2}}$
- $\frac{h^{2}}{2}$
- $\frac{h^{2}}{\sqrt{2}}$
- $\frac{h^{2}}{4}$
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The Correct Option is D
Solution and Explanation
Let base $= b$
Altitude (or perpendicular) $p \rightarrow \left(q\,\rightarrow \,p\right) $$=\sqrt{h^{2}-b^{2}}$
Area, $A=\frac{1}{2}\times base \times$ altitude
$=\frac{1}{2}\times b\times\sqrt{h^{2}-b^{2}}$
$\Rightarrow \frac{dA}{db}=\frac{1}{2}\left[\sqrt{h^{2}-b^{2}}+b. \frac{-2b}{2\sqrt{h^{2}-b^{2}}}\right]$
$=\frac{1}{2}\left[\frac{h^{2}-2b^{2}}{\sqrt{h^{2}-b^{2}}}\right]$
Put $\frac{dA}{db}=0, \Rightarrow b=\frac{h}{\sqrt{2}}$
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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Concepts Used:
Maxima and Minima
What are Maxima and Minima of a Function?
The extrema of a function are very well known as Maxima and minima. Maxima is the maximum and minima is the minimum value of a function within the given set of ranges.
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