A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 , vertex B has an angle of (pi)/2 , and the triangle's area is 16 . What is the area of the triangle's incircle?

1 Answer
Jan 9, 2018

Area of Incircle color(purple)(A_i )= color(purple)(20.3114)

Explanation:

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A = pi/12, B = pi/2, C = (5pi)/12

A_t = 16 = (1/2) a.c sin B = (1/2) a.b sin C = (1/2) b.c sin A

a.c = 32/ sin (pi/2) = color (blue)(32)

a.b = 32 / sin ((5pi)/12) ~~ color(blue)(33.1288)

b.c = 32 / sin (pi/12) ~~ color(blue)(123.6385)

(a.c * a.b * b.c) = (abc)^2 ~~ 32 * 33.1288 * 123.6385

abc = sqrt(131071.8444) ~~ 362.0385

color(red)(a) = (abc) / (bc )= 362.0385 / 123.6385 = color(red)(2.9282)

color(red)(b) = (abc) / (ac )= 362.0385 / 32 = color(red)(11.3137)

color(red)(c) = (abc) / (AB )= 362.0385 / 33.1288 = color(red)(10.9282)

Radius of Incircle of Right Angled Triangle: According to this property, the radius of the incircle of the triangle can be found using the formula color(purple)(r = (a+c−b)/2), where ‘a’ and ‘c’ are the two sides of a triangle and ‘b’ is the hypotenuse.

color(green)(r )= (a + c - b) / 2 = (2.9282 + 10.9282 - 11.3137) / 2 = color(green)(1.2713)

Area of Incircle color(purple)(A_i )= pi r^2 = pi (1.2713)^2 = color(purple)(20.3114)