Question: from Rattanjeet Lamba, Canada
A rectangle of area 48 sq. ft. is inscribed in a circle of area 78.53981634 sq. ft. Find the length of the rectangle. Assume the length is longer than the width.
Answer: We have Pi(r2) = 78.53981634 from which we get r = 5. Since LW = 48, W = 48/L and 102 = L2 + (48/L)2 leading to L4 - 100L2 + 2304 = 0. Solving for the upper root of L2, we get L2 = 64 which gives L = 8.
Solution to bonus question
The dimensions of the rectangle is L = 12 and W = 5. Let x be the length of a hexagon side. Using the law of cosines, 2x2 - 2x2cos(120) = 25 from which we get x = 5/sqrt(3). Creating four corner triangles from the hexagon, we have a base of 5/2 and a height of 5/(2sqrt(3)). The area of one corner triangle is (5/2)(5/(2sqrt(3)))(1/2) = 25/(8sqrt(3)); thus the combined area of the four corner triangles is 25/(2sqrt(3)). The area of the rectangle is 5(5/(2sqrt(3)) + 5/sqrt(3) + 5/(2sqrt(3))) = 5(10/sqrt(3)) = 50/sqrt(3). The area of the hexagon is 50/sqrt(3) - 25/(2sqrt(3)) = 75/(2sqrt(3)) = 21.6506 after rounding to 4 decimals.