The square ABCD has side length 2*sqrt2.
A circle with centre A and radius 1 is drawn. A second circle with centre C is drawn so that it just
touches the 1st circle at point P on AC .
Determine the total area of the
regions inside the square but outside the two circles.
Source: one of Mayhem problems
The large circle around C has radius CP = CA-1 = sqrt 2 (2 sqrt 2) -1 =3
and, say, intersects AB at x and AD at y.
cos(<xCB) = (2 sqrt 2)/3, and thus
<xCB = 19.471 deg = <yCD (congruent triangles)
<xCy = 90 deg - ( <xCB + <yCD ) = 51.058 deg
Area of sector xCy = (<xCy/360) (Area of large circle)
Area of sector xCy = (51.058/360) (pi 3^2) = 4.01005 sq units
Note: xB = yD = sqrt [3^2 - (2 sqrt 2)^2] = 1, so the sum of area of the two triangles xCB and xCD
= 2 (1/2 base x height)
= 1 x 2 sqrt 2
= 2.828 sq units
Area inside Square but outside sectors and two triangles
= Area of Square - (1/4 small circle area + sector xCy area + area of the two triangles xCB and yCD)
(2 sqrt 2)^2 - (pi/4 +4.01005+ 2.828) = 8 - (0.785 + 4.01005 + 2.828)
= 0.3761 sq units
Edited on October 19, 2018, 3:04 am
soln
