Given a rectangle and an equilateral triangle. The rectangle and equilateral triangle have equal area and are inscribed in a circle. The circle has a radius 2.
Find the dimensions of the rectangle.
By drawing a diagonal in the rectangle, we form a diameter. In the case of the triangle, we use Ptolemy's Theorem. This tells us that the radius is 2/3 the height of the triangle. Therefore, the height of the triangle is 3.
Applying the Pythagorean Theorem to the rectangle, x2 + y2 = 16 [1]
Similarly with the triangle, 4a2 = a2 + 9 so a=3
Hence the area of the triangle is 33 so x2y2 = 27 [2]
From [1] we get y2 = 16- x2
Substituting this into [2] we get, x2 (16-x2) = 27
Leading to the quartic, x4 - 16x2 + 27 = 0
Similarly with the triangle, 4a2 = a2 + 9 so a=3
Hence the area of the triangle is 33 so x2y2 = 27 [2]
From [1] we get y2 = 16- x2
Substituting this into [2] we get, x2 (16-x2) = 27
Leading to the quartic, x4 - 16x2 + 27 = 0
Solving this as a quatratic in x2 we obtain the roots, x2 = 8 37
Using y2 = 16 – x2 we observe that y2 = 16 – ( 8 37).That is, when x2 = 8 37, y2 = 8 37, and vice versa.
Therefore the rectangle has dimensions, (8 37) by 8 37.
Using y2 = 16 – x2 we observe that y2 = 16 – ( 8 37).That is, when x2 = 8 37, y2 = 8 37, and vice versa.
Therefore the rectangle has dimensions, (8 37) by 8 37.