A circle is inscribed in a rhombus the lengths of whose diagonals are 2 feet and 4 feet. What is the area, in square feet, of the region of the rhombus that is outside the circular region? Express your answer as a decimal to the nearest tenth.
Source: MathMovesU MATH CHALLENGE PROBLEM - 2012/01/06
The diagonals of the rhombus are perpendicular bisectors of each other, so they partition the rhombus into 4 congruent right
triangles with side lengths 1 foot and 2 feet. Applying the Pythagorean theorem to any of these triangles, we have each side
s of the rhombus satisfies s^2 = 1^2+2^2 = 5. Let r be the radius of the inscribed circle. Since the point of intersection of
the diagonals is the center of the circle, each triangle has a base of 1 foot and a height of 2 feet, or a base of s and
height of r. Thus twice the area of each triangle can be calculated in two ways, namely rs and 1*2 = 2. Equating these
expressions, we get rs = 2, and after squaring both sides of the equality, r^2*s^2 = 4. The area of the circle is then
pi*r^2 = 4*pi/s^2 = 4*pi/5. Now the area of the rhombus is 2*4/2 = 4. Hence the area of the region of the rhombus that is
outside the circular region is 4(1-pi/5) = 1.5 square feet.
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