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A circle is inscribed in an equilateral triangle of side length a. Then, the area of any square inscribed in the circle is 

Option: 1

\frac{a^2}{6}


Option: 2

\frac{a^2}{4}


Option: 3

\frac{a^2}{3}


Option: 4

\frac{a^{^{2}}}{2}


Answers (1)

best_answer

\mathrm{\operatorname{Sin} 60=\frac{A D}{a} \Rightarrow A D=\frac{a \sqrt{3}}{2}}

Incentre of the equilateral triangle coincides with centroid G 

Let the side length of square PQRS be x.

In the right-angled \mathrm{\Delta QGR}

\mathrm{Q R^2=G Q^2+G R^2}

Or \mathrm{x^2=2\left(\frac{a}{2 \sqrt{3}}\right)^2=\frac{a^2}{2 \times 3}=\frac{a^2}{6}}

Area of square \mathrm{=\frac{a^2}{6}}

 

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Rishi

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