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  • If a rectangle is inscribed in an equilateral triangle of side length as shown in the figure, then the square of the largest area of such a rectangle is_______

If a rectangle is inscribed in an equilateral triangle of side length 2 \sqrt{2} as shown in the figure, then the square of the largest area of such a rectangle is__________.
 

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Let height of rectangle = x

\frac{x}{y}= \tan 60^{\circ}\Rightarrow y= \frac{x}{\sqrt{3}}
\therefore Length\, of\, rectangle= 2\sqrt{2}-2y= 2\sqrt{2} -\frac{2x}{\sqrt{3}}
Area\, o\! f\, rectangle= x\left ( 2\sqrt{2} -\frac{2x}{\sqrt{3}} \right )
                                        = 2\sqrt{2}x-\frac{2}{\sqrt{3}}x^{2}
F\! or\, max\, area\Rightarrow 2\sqrt{2}-\frac{4}{\sqrt{3}}x= 0\left ( {f}'\left ( x \right )= 0 \right )   
                              \Rightarrow x= \sqrt{\frac{3}{2}}
Area= 2\sqrt{2}\sqrt{\frac{3}{2}}-\frac{2}{\sqrt{3}}\, \frac{3}{2}= 2\sqrt{3}-\sqrt{3}= \sqrt{3}
Square\, o\! f\, Area= 3

Posted by

Kuldeep Maurya

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