# 01 Arcs of quarter circles

**Example 01**

The figure shown below are circular arcs with center at each corner of the square and radius equal to the side of the square. It is desired to find the area enclosed by these arcs. Determine the area of the shaded region.

**Solution**

$\cos \alpha = \dfrac{10}{20}$

$\alpha = 60^\circ$

At corner A

$\beta = 90^\circ - \alpha = 90^\circ - 60^\circ$

$\beta = 30^\circ$

Area of rectangle ABDE

$A_{ABDE} = 20(10)$

$A_{ABDE} = 200 \, \text{ cm}^2$

Area of triangle ABC

$A_{ABC} = \frac{1}{2}(10)(20) \sin \alpha = 100 \sin 60^\circ$

$A_{ABC} = 50\sqrt{3} \, \text{ cm}^2$

Area of sector ACE

$A_{ACE} = \dfrac{\pi r^2 \beta_{deg}}{360^\circ} = \dfrac{\pi (20^2)(30^\circ)}{360^\circ}$

$A_{ACE} = \frac{100}{3}\pi \, \text{ cm}^2$

Area of CED

$A_{CED} = A_{ABDE} - A_{ABC} - A_{ACE}$

$A_{CED} = 200 - 50\sqrt{3} - \frac{100}{3}\pi$

$A_{CED} = 8.68 \, \text{ cm}^2$

A similar problem that involves the same area as A_{CED} was solved using integration. Look for the solution of area ABC in the following link: Solution by integration.

Required Area

$A_{required} = A_{square} - 8A_{CED}$

$A_{required} = 20^2 - 8(8.68)$

$A_{required} = 330.56 \, \text{ cm}^2$ *answer*