# 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 ->  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

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 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)   (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(8909)   (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 = = Angle BDA = = Therefore angle BDC = 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 =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 = sin(36°) BE = DB*sin(36°) DB = = = 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```