Area of intersection of four circles!

Geometry Level 5

The diagram above shows a square $ABCD$ with each side a. Four circles, each with radius a are centered at the vertices $A, B, C,D$ of the square. If $a=12 \text{ cm}$ then find out the shaded area (in $\text{ cm}^2$ ) of intersection of these circles. Give your answer to 3 decimal places.

The answer is 45.38113108.

2 solutions

Ahmed Moh AbuBakr
Apr 24, 2015

you can do that to get the four parts.. then add the square

> easy

Why u took angle of triangle as 30°

Aman Verma - 5 years, 5 months ago

Venture Hi
Apr 24, 2015

Please refer to photo.

Circle with center (0,12)= 144=x^2+(y-12)^2=12-sqrt(144-x^2) Circle with center(0,0) =144=x^2+y^2=sqrt(144-x^2)

The intersection points between the two circles= ( 6sqrt3, 6)

Area A= Integrate 12-sqrt(144-x^2) from 6sqrt3 to 0 = 18.133

Area B= Integrate sqrt(144-x^22) from 12 to 6sqrt3= 6.5222

Area C= 144- 1/4 pi 144=30.903

Area D = 30.903- ( A+B)=> 30.903-(18.133+6.522)= 6.248

Area E= Area C - 2 (Area D)= 30.903- 2 6.248=18.407

Area F , our answer, is 144 - ( 4 Area E + 4 Area D)= 144- (4 18.407 + 4 6.248)= $\boxed{45.38}$