question_answer

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) \[\frac{{{h}^{2}}}{\sqrt{2}}\]

D) \[\frac{{{h}^{2}}}{4}\]

Correct Answer: D

Solution :

[d] Let base = b Altitude (or perpendicular) \[=\sqrt{{{h}^{2}}-{{b}^{2}}}\] Area, \[A=\frac{1}{2}\times base\times altitude\] \[=\frac{1}{2}\times b\times \sqrt{{{h}^{2}}-{{b}^{2}}}\] \[\Rightarrow \frac{dA}{db}=\frac{1}{2}\left[ \sqrt{{{h}^{2}}-{{b}^{2}}}+b.\frac{2b}{2\sqrt{{{h}^{2}}-{{b}^{2}}}} \right]\]                         \[=\frac{1}{2}\left[ \frac{{{h}^{2}}-2{{b}^{2}}}{\sqrt{{{h}^{2}}-{{b}^{2}}}} \right]\] Put \[\frac{dA}{db}=0,\Rightarrow b=\frac{h}{\sqrt{2}}\] Maximum area \[=\frac{1}{2}\times \frac{h}{\sqrt{2}}\times \sqrt{{{h}^{2}}-\frac{{{h}^{2}}}{2}}=\frac{{{h}^{2}}}{4}\]