# Given the radius of the circle inscribing the hexagon is r express the shaded area in terms of r?

Oct 30, 2016

$0.5036 \cdot {r}^{2}$

#### Explanation:

$O A = O C = O D = O B = O E = O F = r$
$\angle O C D = {360}^{\circ} / 6 = {60}^{\circ} \implies \Delta O C D$ is equilateral
Area $\Delta O C D = \frac{\sqrt{3}}{4} \cdot {r}^{2}$

The 3 green areas in sector $O C D$ are the same.

One green area ${A}_{G} = \pi {r}^{2} \cdot \frac{60}{360} - \frac{\sqrt{3}}{4} \cdot {r}^{2}$
$\implies {A}_{G} = \frac{2 \pi - 3 \sqrt{3}}{12} \cdot {r}^{2}$

The black area ${A}_{B}$ in sector $O C D = \pi {r}^{2} \cdot \frac{60}{360} - 3 {A}_{G}$
$\implies {A}_{B} = \frac{\pi {r}^{2}}{6} - 3 \left(\frac{2 \pi - 3 \sqrt{3}}{12}\right) \cdot {r}^{2}$
$= \left(\frac{2 \pi - 6 \pi + 9 \sqrt{3}}{12}\right) \cdot {r}^{2}$
$= \left(\frac{9 \sqrt{3} - 4 \pi}{12}\right) \cdot {r}^{2}$

Now, let the shaded area in your diagram be ${A}_{S}$
$\implies {A}_{S} = 2 \cdot {A}_{B}$
$\implies {A}_{S} = 2 \cdot \left(\frac{9 \sqrt{3} - 4 \pi}{12}\right) \cdot {r}^{2}$

$\implies {A}_{S} = \left(\frac{9 \sqrt{3} - 4 \pi}{6}\right) \cdot {r}^{2} = 0.5036 \cdot {r}^{2}$

Oct 30, 2016

Given the radius of the circle inscribing the hexagon is r

The area of the circle ${\Delta}_{c} = \pi {r}^{2}$

The area of the each of 6 equilateral triangles having length of each side r is ${\Delta}_{t} = \frac{\sqrt{3}}{4} {r}^{2}$

Area of each of the segment of circle marked X
$= {\Delta}_{x} = \frac{1}{6} {\Delta}_{c} - {\Delta}_{t}$

From the figure it is obvious that the area of Yellow shaded region in the given figure is

${\Delta}_{y} = 2 {\Delta}_{t} - 4 \times {\Delta}_{x}$

$\implies {\Delta}_{y} = 2 {\Delta}_{t} - 4 \times \left(\frac{1}{6} {\Delta}_{c} - {\Delta}_{t}\right)$

$\implies {\Delta}_{y} = 6 {\Delta}_{t} - \frac{2}{3} {\Delta}_{c}$

$\implies {\Delta}_{y} = 6 \times \frac{\sqrt{3}}{4} {r}^{2} - \frac{2}{3} \times \pi {r}^{2}$

$\implies {\Delta}_{y} = \frac{3 \sqrt{3}}{2} {r}^{2} - \frac{2}{3} \times \pi {r}^{2}$

$\implies {\Delta}_{y} = \frac{\left(9 \sqrt{3} - 4 \pi\right)}{6} {r}^{2}$