Question:
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 + y√z in the simplest radical form.
Answer:
This is a regular pentagon inscribed in a circle....I assume you want the area of ABCDE??
If so...... the area is 5 (1/2)(2)^2sin (72) = 10sin (72) = 10 √ [ 5/8 + √5/8 ] units^2
If you want the perimeter...we have that the half side lengh =
2sin(36)
And we have 10 half side lengths comprising the perimeter....so....the perimeter =
10 * 2 * sin(36) = 20√ [ 5/8 - √5/8 ] units
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Very simple question, but the answer is incorrect. We can do better.
+128448 Let O be the center of the circle
We can split this irregular pentagon into 5 triangles
Triangles AOB , BOC and DOE are all equilateral with sides of 2
Their combined areas are
3 (1/2) ( 2)^2 sin 60° = 6sqrt (3) /2 = 3sqrt (3)
And triangles COD and EOA each have sides of 2 and an included angle of [ 360 - 3(60)] / 2 =
90°
Their combined areas = 2* (1/2) (2)^2 sin (90°) = 4
So
[ABCDE ] = 4 + 3sqrt (3) (units^2)
