# 06 Circular arcs inside and tangent to an equilateral triangle

**Example 06**

The figure shown below is an equilateral triangle of sides 20 cm. Three arcs are drawn inside the triangle. Each arc has center at one vertex and tangent to the opposite side. Find the area of region enclosed by these arcs. The required area is shaded as shown in the figure below.

**Solution 06**

$\angle DAB = 30^\circ$, $\angle EBA = 30^\circ$, $\angle ABD = 60^\circ$

$r^2 + 10^2 = 20^2$

$r^2 = 300$

$r = 17.32 \, \text{ cm}$

$BF = AD = r = 17.32 \, \text{ cm}$

$EF^2 + 10^2 = r^2$

$EF^2 + 100 = 300$

$EF = 14.14 \, \text{ cm}$

$BE = BF - EF = 17.32 - 14.14$

$BE = 3.18 \, \text{ cm}$

Sine law for triangle ABE

$\dfrac{r}{\sin \angle EBA} = \dfrac{BE}{\sin \angle EAB}$

$\dfrac{17.32}{\sin 30^\circ} = \dfrac{3.18}{\sin \angle EAB}$

$\sin \angle EAB = \dfrac{3.18 \sin 30^\circ}{17.32}$

$\angle EAB = 5.27^\circ$

$\angle DAE = \angle DAB - \angle EAB = 30^\circ - 5.27^\circ$

$\angle DAE = 24.73^\circ$

$A_{BED} = A_{ABD} - A_{ABE} - A_{AED}$

$A_{BED} = \frac{1}{2}(10)(20)\sin 60^\circ - \frac{1}{2}(20)(3.18) \sin 30^\circ - \dfrac{\pi(17.32^2)(24.73^\circ)}{360^\circ}$

$A_{BED} = 86.60 - 15.9 - 64.74$

$A_{BED} = 5.96 \, \text{ cm}^2$

Required Area

$A_{required} = A_{ABC} - 6A_{BED}$

$A_{required} = \frac{1}{2}(20^2) \sin 60^\circ - 6(5.96)$

$A_{required} = 137.44 \, \text{ cm}^2$ *answer*