A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 , vertex B has an angle of ( pi)/6 , and the triangle's area is 27 . What is the area of the triangle's incircle?

1 Answer
sankarankalyanam
Jun 22, 2018

color(maroon)("Area of incircle " = pi r^2 = pi * (1.71)^2 = 9.186

Explanation:

![http://mathibayon.blogspot.com/2015/01/http://derivation-of-formula-for-radius-of-incircle.html](https://useruploads.socratic.org/qByQYJn5SEeAPjtKhB4j_incircle%20radius.png)

hat A = pi/8, hat B pi/6, hat C = (17pi) / 24), A_t = 27

A_t = (1/2) ab sin C = (1/2(bc sin A = (1/2) ac sin B

ab =( 2A_t) /sin C = (2 * 27) / sin ((17pi)/24) = 68.07

bc = (2 * 27) / sin (pi/8) = 141.11

ca = (2 * 27) / sin (pi/6) = 108

a = sqrt(abc)^2 / (bc) =sqrt (68.07 * 141.11 * 108) / 141.11 = 7.22

b = sqrt(abc)^2 / (ac) =sqrt (68.07 * 141.11 * 108) / 108 = 9.43

c = sqrt(abc)^2 / (ab) =sqrt (68.07 * 141.11 * 108) / 68.07 = 14.96

"Semiperimeter " = s = (a = b + c) / 2 = (7.22 + 9.43 + 14.96) / 2 = 15.81

color(chocolate)("Incircle radius " = r = A_t / s = 27 / 15.81 = 1.71

color(maroon)("Area of incircle " = pi r^2 = pi * (1.71)^2 = 9.186

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