Given an equilateral triangle inscribed in a circle with a radius R.

Evaluate the dimensions (height, width) of an inscribed rectangle with the same area as the triangle.

WLOG let R=1

Then the triangle has an altitude of 3/2, edges of sqrt(3) and area of sqrt(27)/4.

Let L and W be the sides of the rectangle. The rectangle has area L*W and an diagonal of 2, which forms a right triangle with a pair of sides. Then:

L*W = sqrt(27)/4

L^2 + W^2 = 4

Substituting and simplifying yields:

L^4 - 4L^2 + 27/16 = 0

The two positive roots of the quartic equation are the dimensions of the rectangle: sqrt[8+sqrt(37)]/2 and sqrt[8-sqrt(37)]/2.

Then for general radius R: **L = R*sqrt[8+sqrt(37)]/2 and W = R*sqrt[8-sqrt(37)]/2**