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Hexagon circumscribed (Posted on 2025-05-21) Difficulty: 3 of 5
Compute the area of the circle circumscribed about an equiangular hexagon with sides of lengths 20, 16, 20, 16, 20, 16.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution solution Comment 1 of 1
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

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