Question 229897
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Construct a square by drawing 4 line segments connecting the centers of adjacent circles.  The line segments will intersect the points of tangency of the circles and you will note that the hypocycloid shape in the center for which you want the area is totally contained within this circle.  Since the shape is a square, the vertex angles are right angles, therefore the four circle sectors that are contained within the square are each one-fourth the area of one of the circles.  The area of the four circle sectors added together is therefore the area of one of the circles.  Since the diameter of each of the circles is 10, the radius of each of the circles is 5 and the area of a circle is 25&#960; mē.  The area of the square is 10 X 10 = 100, so the area of the hypocycloid is the area of the square minus the area of one of the circles, 100 - 25&#960; mē.  And the area of the hypocycloid plus a circle is just the area of the square, i.e. 100 mē


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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