A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(5pi)/8 #, and the triangle's area is #25 #. What is the area of the triangle's incircle?

1 Answer
sankarankalyanam
Jun 21, 2018

#color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.065^2 = 0.0133#

Explanation:

The distances from the incenter to each side are equal to the inscribed circle's radius. The area of the triangle is equal to 12×r×(the triangle's perimeter), 1 2 × r × ( the triangle's perimeter ) , where r is the inscribed circle's radius.

First to find the perimeter of the triangle.

#hat A = pi/12, hat B = (5pi)/8, hat C = (7pi)/24, A_t = 25#

Area of triangle #A_t = (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sin B#

#ab = (25 * 2) / sin ((7pi)/24) = 63.02#

Similarly, #bc = (25 * 2) / sin (pi/12) = 193.19#

#ca = (25 * 2) / sin ((5pi/8) = 54.12#

#ab * bc * ca = (abc)^2 = 63.02 * 193.19 * 54.12#

#abc = sqrt(63.02 * 193.19 * 54.12) ~~ 810.98#

#a = (abc) / (bc) = 810.98 / 193.19 = 4.2#

Likewise, #b = (abc) / (ca) = 810.98 / 54.12 = 14.98#

#c = (abc) / (ab) = 810.98 / 63.02 = 12.87#

#"Perimeter of the triangle " p = a + b + c = 4.2 + 14.98 + 12.87 = 32.05#

#"Radius of incircle " r = A_t / (12 * p) = 25 / (12 * 32.05) = 0.065#

#color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.065^2 = 0.0133#

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