Question

Suppose a square ABCD has a side length of 1 unit. The arcs BD and AC are circular arcs with centers at A and D, respectively. Find the area of the shaded region.

Answer 1

Answer:

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$

Step-by-step explanation:

The missing figure is shown in the attachment

The area of the shaded region = Area of Square - (Area of sector AOB + Area of equilateral  triangle BOC + Area of sector COD)

Area of Sector AOB=Area of Sector COD=$\frac{30}{360}*\pi*1^2=\frac{\pi}{12}$

Area of equilateral triangle =$\frac{1}{2}*r*\frac{\sqrt{3} }{2}*r= \frac{1}{2}*1*\frac{\sqrt{3} }{2}*=\frac{\sqrt{3} }{4}$

Area of shade region =$1^2-\frac{\sqrt{3} }{4}-\frac{\pi}{12}*2$

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$