Question

Area of the regular hexagon whose diagonal joins $(2,4)$ and $(6,7)$ is

• A. $\frac{75\sqrt{3}}{8}$
• B. $\frac{7\sqrt{3}}{16}$
• C. $\frac{25\sqrt{3}}{16}$
• D. $\frac{6\sqrt{3}}{5}$

Answer 1

Answer: (A)

We have $A(2,4)$ and $B(6,7)$ as end points of diagonal of the hexagon.
$\therefore AB=5$
So, length of the line segment joining centre of the hexagon and one of the vertices is $\frac{5}{2}$.
$\therefore$ Area of hexagon $=6\times$ Area of equilateral triangle having side length $\frac{5}{2}$
$=6\times \frac{\sqrt{3}}{4}\times \frac{25}{4}$
$=\frac{75\sqrt{3}}{8}$