**Example 02**

Arcs of quarter circles are drawn inside the square. The center of each circle is at each corner of the square. If the radius of each arc is equal to 20 cm and the sides of the square are also 20 cm. Find the area common to the four circular quadrants. See figure below.

**Solution 02**

$A_{AECDB} = \frac{1}{4}\pi (20^2) = 100\pi \, \text{ cm}^2$

$A_{AECDB} = 314.16 \, \text{ cm}^2$

Area of sector AFCGB

$A_{AFCGB} = \frac{1}{6}\pi (20^2) = \frac{200}{3}\pi \, \text{ cm}^2$

$A_{AFCGB} = 209.44 \, \text{ cm}^2$

Area of segment AFCE

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

$A_{AFCE} = \frac{200}{3}\pi - 100\sqrt{3} \, \text{ cm}^2$

$A_{AFCE} = 36.23 \, \text{ cm}^2$

Area of BGCD

$A_{BGCD} = A_{AECDB} - A_{AFCGB} - A_{AFCE}$

$A_{BGCD} = 314.16 - 209.44 - 36.23$

$A_{BGCD} = 68.49 \, \text{ cm}^2$

Required Area

$A_{required} = A_{square} - 4A_{BGCD} = 20^2 - 4(68.49)$

$A_{required} = 126.04 \, \text{ cm}^2$ *answer*

**Another Solution**

$\cos (30^\circ + 15^\circ) = \dfrac{10}{x}$

$x = \dfrac{10}{\cos 45^\circ} = 10\sqrt{2} \, \text{ cm}$

Area of sector DAC

$A_{DAC} = \dfrac{\pi(20^2)(15^\circ + 15^\circ)}{360^\circ} = \dfrac{100\pi}{3} \text{ cm}^2$

$A_{DAC} = 104.72 \, \text{ cm}^2$

Area of triangle DBC

$A_{DBC} = \frac{1}{2}(20x) \sin 15^\circ = \frac{1}{2}(20)(10\sqrt{2}) \sin 15^\circ$

$A_{DBC} = 36.60 \, \text{ cm}^2$

Area of ABC

$A_{ABC} = A_{DAC} - 2A_{DBC} = 104.72 - 2(36.60)$

$A_{ABC} = 31.52 \, \text{ cm}^2$

Required Area

$A_{required} = 4A_{ABC} = 4(31.52)$

$A_{required} = 126.08 \, \text{ cm}^2$ *answer*

**Solution by Calculus**