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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

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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.

Jan 21, 2019

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Jan 30, 2019
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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

Here's a pic :    Jan 30, 2019
edited by CPhill  Jan 30, 2019
edited by CPhill  Jan 30, 2019