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Quadrilateral \(ABCD\) is inscribed in a circle with segment \(AC\) a diameter of the circle. If \(m\angle DAC = 30^\circ\) and \(m\angle BAC = 45^\circ\), the ratio of the area of \(ABCD\) to the area of the circle can be expressed as a common fraction in simplest radical form in terms of \(\pi\) as \(\frac{a+\sqrt{b}}{c\pi}\), where  a, b, and c are positive integers. What is the value of a+b+c ?

 Jan 27, 2021
 #1
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The ratio is (3 + sqrt(6))/(2*pi), so a + b + c = 3 + 6 + 2 = 11.

 Jan 27, 2021
 #2
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We  can  choose any  convenient  value  for  the  diameter.....let  this  =  2...so  the radius = 1

 

Angles  ADC  and  ABC  will be right angles

 

So   in right  triangle ADC  angle  DAC = 30°   and angle  DCA   = 60°

 

So  DC  =  1       AD  =  sqrt (3)

 

So area of  triangle ADC  = (1/2) (1) (sqrt (3)   = sqrt (3)/2

 

And  triangle  ABD  will  be isosceles  right  with  AB  = BC  =  2/sqrt (2)   = sqrt (2)

 

So  the area of  triangle   ABD  =   (1/2) [sqrt (2)]^2   =  1

 

So   [ ABCD ]   =  1 + sqrt (3) / 2     =  [ 2 + sqrt (3) ]  /2

 

And the area of the  circle  =  pi 1^2  =  pi

 

So     [ABCD ]   / area of  circle  =       (2 + sqrt (3) )  /  ( 2 *  pi )  =   (2 + sqrt (3)  / (2pi)

 

So   a +  b +  c  =  2 + 3 + 2   =   7

 

Here's a pic :

 

 

cool cool cool

 Jan 27, 2021
edited by CPhill  Jan 27, 2021

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