10 Area common to three squares inside the regular hexagon

$60 = 6x$

$x = 10 \, \text{ in.}$
 

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

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

$b = x - y = 10 - 8.6602$

$b = 1.3398 \, \text{ in.}$
 

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