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 square itself has area 8.
The quadrant of the circle centered on A has area pi/4.
The diagonal of the square has length 4, so the radius of the circular quadrant centered on C is 3, and that quadrant has area 9*pi/4. However the radius exceeds the length of a side of the square by 4-2*sqrt(2).
The arc of this quadrant cuts AB sqrt(9-8) units from B, that is 1 unit from B; likewise it cuts AD 1 unit from D.
The circular sector has angle pi/2 - 2*arctan(1/(2*sqrt(2))) -- note one set of parentheses could have been left out if we gave precedence to multiplication over division. This makes the area of the sector 9*(pi/2 - 2*arctan(1/(2*sqrt(2))))/2. The combined areas of the two right triangles within the quadrant and the square is 2*sqrt(2).
So the sought area remaining in the square is
8 - pi/4 - 9*(pi/2 - 2*arctan(1/(2*sqrt(2))))/2 - 2*sqrt(2)
~= .37612342636639
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Posted by Charlie
on 2018-10-18 12:43:00 |
solution
