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?
+128460 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
