A triangle has vertices A, B, and C. Vertex A has an angle of $pi/12$ , vertex B has an angle of $(3pi)/8$ , and the triangle's area is $21$ . What is the area of the triangle's incircle?

Question

1294 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-01T05:34:36

Area of Incircle $A_i = color(green)(6.6)$

$hat A = pi/12, hatB = 3pi/8, hat C = pi - pi/12 - (3pi)/8 = (13pi)/24, A_t = 21$

$(1/2) bc sin A = (1/2) ca sin B = (1/2) a b sin C = A_t$

$bc = (2 * A_t) / sin A = (2 * 21) = sin (pi/12) = 162.28$

$ca = (2*21) / sin ((3pi)/8) = 45.46$

$ab = (2*21)/sin ((13pi)/24) = 42.36$

$abc = sqrt(ab* bc * ca) = sqrt (42.36 * 45.46 * 162.28) = 559$

$a = (abc) / (bc) = 559 / 162.28 = 3.44$

Similarly, $b = 559 / 45.46 = 12.3$

$c = 559 / 42.36 = 13.2$

Semi perimeter of triangle ABC

$p/2 = (a + b + c) / 2 = (3.44 + 12.3 + 13.2) / 2 = 14.47$

Radius of Incircle $r_i = A_t / (p/2) = 21 / 14.47 = 1.45$

Area of Incircle $A_i = pi r_i^2 = pi * 1.45^2 = color(green)(6.6)$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 pi/12 , vertex B has an angle of (3pi)/8 (3pi)/8 , and the triangle's area is 21 21 . What is the area of the triangle's incircle?