• # question_answer The maximum area of a right angled triangle with hypotenuse h is :     JEE Main  Online Paper (Held On 22 April 2013) A)  $\frac{{{\operatorname{h}}^{2}}}{2}\sqrt{2}$                              B)  $\frac{{{\operatorname{h}}^{2}}}{2}$ C)  $\frac{{{\operatorname{h}}^{2}}}{\sqrt{2}}$                                                D)  $\frac{{{\operatorname{h}}^{2}}}{4}$

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