Compute the area of the circle circumscribed about an equiangular hexagon with sides of lengths
20, 16, 20, 16, 20, 16.
Each angle of the hexagon is 120°. Each diagonal of the hexagon that skips over one vertex therefore forms a triangle with two sides that are 20 and 16 respectively with the angle connecting them being 120°.
The square of this smaller-length diagonal is given by the law of cosines as
20^2 +16^2 - 2*20*16*cos(120°)
The diagonal comes out to be 4*sqrt(61), so half the diagonal is 2*sqrt(61).
Since the alternate vertices of the hexagon form an equilateral triangle, this half-diagonal forms one leg of a right triangle whose hypotenuse is the desired radius, having one vertex at the center of the circumscribing circle. One angle of this triangle, at the vertex of the hexagon, is 30° (half the 60° of the equilateral triangle) and the other angle, at the circle's center, is 60°. The desired hypotenuse is 2*sqrt(61)/sin(60°) = 4*sqrt(183)/3 =~ 18.0369990112916.
That's the radius of the circumcircle. Its area is pi*16*183/9 = pi * 976/3 =~ 1022.06480996788.
Edited on May 21, 2025, 11:19 am
|
|
Posted by Charlie
on 2025-05-21 11:14:21 |
solution