+0  
 
0
56
1
avatar

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

------------------------------------------------------------------------------------------------------------------

Very simple question, but the answer is incorrect. We can do better.smiley

 Dec 24, 2020
 #1
avatar+114320 
+2

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)

 

 

cool cool cool

 Dec 24, 2020

26 Online Users

avatar