A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #98 #. What is the area of the triangle's incircle?

1 Answer
Jun 22, 2018

#color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59#

Explanation:

http://mathibayon.blogspot.com/2015/01/derivation-of-formula-for-radius-of-incircle.html

#hat A = pi/2, hat B pi/6, hat C = pi/3, A_t = 98#

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

#ab =( 2A_t) /sin C = (2 * 98) / sin (pi/3) = 226.32#

#bc = (2 * 98) / sin (pi/2) = 196#

#ca = (2 * 98) / sin (pi/6) = 392#
4169.97
#a = sqrt(abc)^2 / (bc) =sqrt (226.32 * 196 * 392) / 196 =21.28#

#b = sqrt(abc)^2 / (ac) =sqrt (226.32 * 196 * 392) / 392 =10.64#

#c = sqrt(abc)^2 / (ab) =sqrt (226.32 * 196 * 392) / 226.32 =18.43#

#"Semiperimeter " = s = (a = b + c) / 2 = (21.28 + 10.64 + 18.43) / 2 = 25.18#

#color(chocolate)("Incircle radius " = r = A_t / s = 98 / 25.18 = 3.89#

#color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59#