document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #277339 by Edwin McCravy(20085) $\"\"$ $\"About$

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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.
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document.write( "Draw DA and DB:\r\n" );
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document.write( "Angle ADC =  = \r\n" );
document.write( "Angle BDA =  = \r\n" );
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document.write( "Therefore angle BDC = \r\n" );
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document.write( "Erase DA (to keep figure from being cluttered)\r\n" );
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document.write( "Draw the diagonal CB, and label angle BDC as having measure 144°\r\n" );
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document.write( "Since triangle CDB is isosceles and has vertex angle 144°,\r\n" );
document.write( "its two congruent base angles are 36° each because 180°-144°= 36° and\r\n" );
document.write( "each of the base angles is one-half of that.  So we label angle BCD\r\n" );
document.write( "as 18°:\r\n" );
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document.write( "CEB is a right triangle.  Therefore \r\n" );
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document.write( "=tan(18°)\r\n" );
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document.write( "BE = CE*tan(18°) and since CE is given as 1.4375\r\n" );
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document.write( "BE = 1.4375*tan(18°) = 0.4670720633\r\n" );
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document.write( "Angle BDE is suplementary to angle CDB which 144°,\r\n" );
document.write( "so angle BDE is 180°-144° = 36°.   Now I'll erase some more:\r\n" );
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document.write( "Now triangle DBE is a right triangle, so\r\n" );
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document.write( " = sin(36°)\r\n" );
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document.write( "BE = DB*sin(36°)\r\n" );
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document.write( "DB =  =  = 0.7946304565.\r\n" );
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document.write( "Since DB is a radius of the circle, that's what was required.\r\n" );
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document.write( "Answer radius = 0.7946 rounded to nearest ten thousandth.\r\n" );
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document.write( "Edwin

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