You have 8 sticks. Three of them can form the sides of a right triangle with area 50.

But you can use all eight to form a cyclic (inscribable within a circle) octagon all of whose angles are equal. What would be the area of such an octagon?

In order form a cyclic octagon where all angles are equal, opposite sides should be of equal length with four sides of length s and four sides of length h (s may or may not equal h).

A right triagle where two legs are of equal length (s) will have a hypotenuse (h) of length s x SQRT(2).

As the area of the isoceles right triangle equals 50, the legs of equal length (s) would equal 10 with a hypotenuse (h) of length 10 x SQRT(2).  s can be determined using the standard formula for the area of a triangle:
1/2 x s [length] x s [height] = 50 [area of a triangle]
s² = 50 x 2 = 100
s = 10

In calculating the area of the irregular octagon with four sides of length 10 and four sides of length 10 x SQRT(2), I find the area to be 700.

Edited on April 13, 2009, 6:24 pm

Posted by Dej Mar on 2009-04-13 17:55:41