Concept:Area of an equilateral triangle = \(\frac{\sqrt3}{4}\ \times\ a^2\)where a = side of the triangle.Calculation:Given:Let the side of the large triangle is 'a' then the side of the regular hexagon is \(\frac{a}{3}\)If each triangle's area = \(\frac{\sqrt3}{4}\ \times\ a^2\) then the area of regular hexagon = \(6\ \times\ \frac{\sqrt{3}}{4}\ \times\ \left ( \frac{a}{3} \right )^2\)Required ratio = \(\frac{Area \ of\ regular\ hexagon}{Area\ of\ equilateral \ triangle}\) = \(\frac{A_H}{A_T}\)? \(\frac{A_H}{A_T}\ =\ \frac{6\ \times\ \frac{\sqrt{3}}{4}\ \times\ \left ( \frac{a}{3} \right )^2}{\frac{\sqrt3}{4}\ \times\ a^2}\ =\ \frac{6}{9}\ =\ \frac{2}{3}\)? The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is \(\frac{2}{3}\)Download Solution PDFShare on Whatsapp