A, B, C, D, and E are points on a circle of radius 2 in counterclockwise order. We know AB= BC= DE = 2 and CD = EA. Find [ABCDE].
Enter your answer in the form x+ysqrt(z) in simplest radical form.
Let O be the center of the circle.....we will have 5 triangles that comprise [ABCDE ]
Three of these AOB, BOC and DOE will be equilateral triangles with areas of [2^2 * sqrt (3) / 4] = √ (3) units^2
So...their combined areas = 3√ (3) units^2
The arc of the circle covered by the chords AB, BC and CD will be 180°
So....the chords CD and EA occupy 90° each
So...triangles COD and EOA will be be right isosceles triangles each with an area of (1/2)(2)^2 = 2
So....their combined areas = 4 units^2
So [ ABCDE ] = [ 4 + 3√3 ] units^2
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