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 \[=\frac{1}{2}\times \frac{h}{\sqrt{2}}\times \sqrt{{{h}^{2}}-\frac{{{h}^{2}}}{2}}=\frac{{{h}^{2}}}{4}\]You need to login to perform this action.
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