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Equilateral triangle ABC has side length 6 cm and is inscribed in circle P. Congruent smaller circles centered at D, E and F are inscribed in the three regions between an arc of circle P and a side of triangle ABC, as shown. If segments AF, BE and CD all intersect P, what is the total area of all three small circles?

 

 Jul 28, 2022
 #1
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Let M be the mid-point of AB

 

Then triangle PMB  is a 30-60-90 right triangle with MB  = 3   PM = sqrt 3   and   PB = sqrt (12)  = the radius of the large circle  = R

 

So  the radius of one of the smaller circles =  

 

[ R  - PM ]  / 2  =   [  sqrt (12)  - sqrt (3) ]  / 2  =   

 

So......the total area of the three small circles = 

 

3 * pi *    ( [  sqrt (12) - sqrt (3) ] / 2 )^2  =

 

(3/4) pi * [ 12 - 2sqrt (36) + 3]  =

 

(3/4)pi  * [ 12 - 12 + 3]  =

 

(3/4)pi (3)  =

 

(9/4) pi cm^2

 

 

cool cool cool

 Jul 28, 2022

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