Problema Solution
The perimeter of square ABCD is 24√(8/9) cm. A circle with center A and radius 13/4 cm is drawn. A second circle with center C is drawn so that it just touches the first circle at a point on AC. Find the area of the region inside the square but outside the two circles.
C
Answer provided by our tutors
Let
d = the diagonal length of the square
a = the length of the side
Since the perimeter is 24√(8/9) we have:
4a = 24√(8/9)
a = 6√(8/9)
Using the Pythagorean Theorem we have:
d^2 = 2a^2
d^2 = 2*(6√(8/9))^2
d^2 = 64
d = 8 cm
The area of the region inside the square but outside the two circles is:
C = a^2 - [(13/4)^2*pi*(90/360) - (d - 13/4)^2*pi*(90/360))]
C = (6√(8/9))^2 - [(13/4)^2*pi*(90/360) - (8 - 13/4)^2*pi*(90/360))]
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C = 41.42 cm^2
The area of the region inside the square but outside the two circles is 41.42 cm^2.