Question

A regular hexagon of maximum possible area is cut off from an equilateral triangle. The ratio of the area of the triangle to the area of hexagon will be-

• A. $\sqrt{\frac{3}{2}}$
• B. $\frac{\sqrt{6}}{2}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

Let $3a$ be the side of equilateral triangle $PQR$.

Then, $arPQR=\frac{\sqrt{3}}{4}\times 3a\times 3a$

$=\frac{9\sqrt{3}}{4}{a}^2$

We know, that a regular hexagon is formed by joining 6 equilateral triangles. Hence side of regular hexagon inscribe from an equilateral $△PQR$ is $\frac{3a}{3}=a$.

ar. of hexagon $ABCDEA=6\times \frac{\sqrt{3}}{4}\times a^2=\frac{6\sqrt{3}}{4}{a}^2$

$\frac{ar△PQR}{arABCDEA}=\frac{\frac{a\sqrt{3}}{4}{a}^2}{\frac{6\sqrt{3}}{4}{a}^2}=\frac{9}{6}=\frac{3}{2}$