Question

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

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

Answer 1

The correct option is A $\frac{75\sqrt{3}}{8}$

$AD-\sqrt{(6-2{)}^2(7-4{)}^2}.\sqrt{16+9}=5$....(1)

Let $AB=a$

in $△AFB$ $\angle A={120}^0$

$\angle BAP=\frac{120}{2}={60}^0$

$\cos \angle BAP=\frac{AP}{AB}$

$\frac{1}{2}=\frac{AP}{AB}\Rightarrow AP=\frac{AB}{2}=\frac{a}{2}$

$sin\angle BAP=\frac{BP}{AB}=\frac{\sqrt{3}}{2}$

$BP=\frac{\sqrt{3}a}{2}\Rightarrow FB=2\times BP=\sqrt{3}a$

$AD=AP+PQ+QD$

$AD=\frac{a}{2}+a+\frac{a}{2}=2a=5$

$\Rightarrow a=\frac{5}{2}$

area of $ABCDEF=area\left(△AFB\right)+$

area $\left(FBCE\right)+$

area $\left(△EDC\right)$

$=(\frac{1}{2}\times \sqrt{3}a\times \frac{a}{2})+(\sqrt{3}a\times a)+(\frac{1}{2}\times \sqrt{3}a\times \frac{a}{2})$

$=\frac{\sqrt{3}{a}^2}{2}+\sqrt{3}{a}^2=\frac{3\sqrt{3}}{2}{a}^2$

$areaofABCDEF\frac{3\sqrt{3}{a}^2}{2}=\frac{3\sqrt{3}}{2}\left(\frac{25}{4}\right)=\frac{75\sqrt{3}}{8})$