• # question_answer A circle is inscribed in an equilateral triangle and a square is inscribed in that circle. The ratio of the areas of the triangle and the square is A)  $3\sqrt{3}:1$   B)  $\sqrt{3}:4$ C)  $\sqrt{3}:8$     D)  $3\sqrt{3}:2$

Correct Answer: D

Solution :

[d] In the given figure ABC is an equilateral $\Delta$ of a side with a circle inscribed in it and a square inscribed in the circle. AD, BO and CO are the angle bisectors of$\angle A$, $\angle B$ and $\angle c$ and 0 is the centre of the circle. We know that the angle bisector from the vertex of an equilateral triangle is the perpendicular bisector of the opposite side. AD is the perpendicular bisector of BC. $\Rightarrow \,\,\,BD=\frac{a}{2}\,\,and\,\,\angle DOB=\frac{1}{2}\angle B=\frac{1}{2}\times 60{}^\circ =30{}^\circ$ Now in $\Delta BOD$ $\tan 30{}^\circ =\frac{OD}{BD}=\frac{Radius\,of\,circle}{\frac{a}{2}}$ $\Rightarrow$Radius of circle$=\frac{1}{\sqrt{3}}\times \frac{a}{2}=\frac{2}{2\sqrt{3}}$ Now in right $\Delta EDG$ $E{{G}^{2}}+G{{D}^{2}}=E{{D}^{2}}$(Pythagoras theorem) $2\,\,{{(EG)}^{2}}=2\,\,{{(OD)}^{2}}={{\left( \frac{a}{\sqrt{3}} \right)}^{2}}=\frac{{{a}^{2}}}{3}$ Side of the square $=\sqrt{\frac{{{a}^{2}}}{6}}=\frac{a}{\sqrt{6}}$ Now $ar$ $(\Delta ABC):\,\,ar\,\,(\Delta EFG)$ $\frac{\frac{\sqrt{3}}{4}{{a}^{2}}}{\frac{a}{\sqrt{6}}\times \frac{a}{\sqrt{6}}}=\frac{\frac{\sqrt{3}}{4}}{\frac{1}{6}}=3\sqrt{3}:2$

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