A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{π}{8}$ $pi/8$ , and the triangle's area is $14$ $14$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-27T02:20:45

$Area of inscribed circle = A_{i} = π r^{2} = 3.4165$ $color(maroon)("Area of inscribed circle " = A_i = pi r^2 = 3.4165$

$"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B$

$"Given " hat A = pi/12, hat B = pi/8, hat C = (19pi)/24, A_t = 14$

http://mathibayon.blogspot.com/2015/01/derivation-of-formula-for-radius-of-incircle.html#.WzLtzNIza70

$a b = (2 A_t) / sin C = 28 / sin ((19pi)/24) = 46$

$b c = (2 A_t) / sin A = 28 / sin (pi/12) = 108.18$

$c a = (2 A_t) / sin B = 28 / sin (pi/8) = 73.17$

$a = (a b c) / (b c) = sqrt(46 * 108.18 * 73.17) / 108.18 = 5.58$

$b = (a b c) / (c a) = sqrt(46 * 108.18 * 73.17) / 73.17 = 8.15$

$c = (a b c) / (a b) = sqrt(46 * 108.18 * 73.17) / 46 = 13.12$

$"Semi-perimeter " = s = (a + b + c) / 2 = 26.85 / 2 = 13.425$

$"Radius of inscribed circle " = r = A_t / s = 14 / 13.425 = 1.0428$

$"Area of inscribed circle " = A_i = pi r^2 = pi * 1.0428^2 = 3.4165$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12pi/12 , vertex B has an angle of pi/8 π8pi/8 , and the triangle's area is 14 1414 . What is the area of the triangle's incircle?