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.

noobieatmath Jan 21, 2019

#2**+1 **

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 :

CPhill Jan 30, 2019