A triangle has vertices A, B, and C. Vertex A has an angle of $pi/2$ , vertex B has an angle of $( pi)/8$ , and the triangle's area is $16$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-26T15:12:29

Area of in circle $A_i = 6.7$ sq units

$A_t = (1/2) ab sin C =(1/2)b c sin A = (1/2) c a sin B$

$hat A = pi/2, hat B = pi/8, hat C = (3pi)/8$

$a b = (2 * 16) / sin ((3pi)/8) = 34.64$

$c a = (2*16)/ sin (pi/8) = 83.62$

$b c= (2*16)/ sin(pi/2) = 32$

$a = (abc) / (b c) = sqrt(34.64*83.62*32) / 32 = 9.51$

$b = (abc) / (c a) = sqrt(34.64* 83.62*32) / 83.62 = 3.64$

$c = sqrt(34.64*83.62*32)/34.64 = 8.79$

Semi perimeter $s = (a + b + c) / 2$

$s = (9.51 + 3.64 + 8.79) / 2 = 10.97$

Radius of in circle $r = A_t / s = 16 / 10.97 = 1.46$

Area of in circle $A_i = pi r^2 = pi * 1.46^2 = 6.7$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/2 pi/2 , vertex B has an angle of ( pi)/8 ( pi)/8 , and the triangle's area is 16 16 . What is the area of the triangle's incircle?