Applications Of Derivatives Question 412
Question: The maximum area of a right angled triangle with hypotenuse h is:
Options:
A) $ \frac{h^{2}}{2\sqrt{2}} $
B) $ \frac{h^{2}}{2} $
C) $ \frac{h^{2}}{\sqrt{2}} $
D) $ \frac{h^{2}}{4} $
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Answer:
Correct Answer: D
Solution:
[d] Let base = b Altitude (or perpendicular) $ =\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}[ \sqrt{h^{2}-b^{2}}+b.\frac{2b}{2\sqrt{h^{2}-b^{2}}} ] $
$ =\frac{1}{2}[ \frac{h^{2}-2b^{2}}{\sqrt{h^{2}-b^{2}}} ] $ 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} $
BETA
