SOLUTION: I have a circle inside a regular pentagon. CE is from top of pentagon to the bottom. I know that CE is 1.4375. I need to find DE which is the radius of the circle. CE is a straigh

Algebra ->  Trigonometry-basics -> SOLUTION: I have a circle inside a regular pentagon. CE is from top of pentagon to the bottom. I know that CE is 1.4375. I need to find DE which is the radius of the circle. CE is a straigh      Log On


   

Question 391000: I have a circle inside a regular pentagon. CE is from top of pentagon to the bottom. I know that CE is 1.4375. I need to find DE which is the radius of the circle. CE is a straight line going all the way through middle of the circle.
Found 2 solutions by scott8148, Edwin McCravy:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
picture the pentagon divided into ten congruent right triangles
___ by lines drawn from the midpoint of a side to the opposite vertex

the "central" angle of the right triangle (in the center of the pentagon) is 36º (360º / 10)

the length of the side of the triangle from the right angle to the center of the pentagon is D/2
___ this same side , added to the hypotenuse of the right triangle , equals C

(D / 2) + h = C

from trigonometry ___ (D / 2) / h = cos(36º) ___ (D / 2) / cos(36º) = h

substituting ___ (D / 2) + [(D / 2) / cos(36º)] = C

D{1 + [1 / cos(36º)]} = 2C

D = 2C / {1 + [1 / cos(36º)]}

Answer by Edwin McCravy(20085) About Me  (Show Source):
You can put this solution on YOUR website!
I have a circle inside a regular pentagon. CE is from top of pentagon to the bottom. I know that CE is 1.4375. I need to find DE which is the radius of the circle. CE is a straight line going all the way through middle of the circle.


Draw DA and DB:



Angle ADC = %22360%B0%22%2F5 = %2272%B0%22
Angle BDA = %22360%B0%22%2F5 = %2272%B0%22

Therefore angle BDC = %2272%B0%22%2B%2272%B0%22=%22144%B0%22

Erase DA (to keep figure from being cluttered)

Draw the diagonal CB, and label angle BDC as having measure 144°



Since triangle CDB is isosceles and has vertex angle 144°,
its two congruent base angles are 36° each because 180°-144°= 36° and
each of the base angles is one-half of that.  So we label angle BCD
as 18°:




CEB is a right triangle.  Therefore 

BE%2FCE=tan(18°)

BE = CE*tan(18°) and since CE is given as 1.4375

BE = 1.4375*tan(18°) = 0.4670720633

Angle BDE is suplementary to angle CDB which 144°,
so angle BDE is 180°-144° = 36°.   Now I'll erase some more:



Now triangle DBE is a right triangle, so

BE%2FDB = sin(36°)

BE = DB*sin(36°)

DB = BE%2Fsin%28%2236%B0%22%29 = 0.4670720633%2Fsin%28%2236%B0%22%29 = 0.7946304565.

Since DB is a radius of the circle, that's what was required.

Answer radius = 0.7946 rounded to nearest ten thousandth.

Edwin