# 10 Area common to three squares inside the regular hexagon

**Problem**

Three squares are drawn so that each will contain a side of regular hexagon as shown in the given figure. If the hexagon has a perimeter of 60 in., compute the area of the region common to the three squares. The required area is the shaded region in the figure.

**Solution**

$P = nx$

$60 = 6x$

$x = 10 \, \text{ in.}$

$y = x \sin 60^\circ = 10 \sin 60^\circ$

$y = 5\sqrt{3} \, \text{ in.} = 8.6602 \, \text{ in.}$

$b = x - y = 10 - 8.6602$

$b = 1.3398 \, \text{ in.}$

From triangle ABC

$\tan 30^\circ = \dfrac{b}{a/2}$

$a = \dfrac{2b}{\tan 30^\circ} = \dfrac{2(1.3398)}{\tan 30^\circ}$

$a = 4.6412 \, \text{ in.}$

$A = \frac{1}{2}a^2 \sin 60^\circ$

$A = \frac{1}{2}(4.6412^2) \sin 60^\circ$

$A = 9.327 \, \text{ in.}^2$ *answer*