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 $98$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-02-28T16:04:28

Area of Incircle $A_i = 40.72$ sq units

$hatA = pi/2, hatB = pi/8, hat C = pi - pi/2 - pi/8 = (3pi)/8, A_t = 98$

Next we have to find the hypotenuse and the other side of the right triangle.

$(1/2) bc sin A = A_t$

$bc = (2A_t)/sin A = 196 / sin (pi/2) = 196$

$ca = 196 / sin B = 296 / sin (pi/8) = 512.1727$

$ab = 196 / sin C = 196 / sin ((3pi)/8) = 212.1489$

$:. ab * bc * ca = (196 * 512.1727 * 212.1489)$

$abc = sqrt (196 * 512.1727 * 212.1489) = 4614.84$

$c = (abc) / (AB) = 4614.84 / 212.1489 = 21.75$

$b = (abc) / (ca) = 4614.84 / 512.1727 = 9$

$a = (abc) / (bc) = 4614.84 / 196 = 23.55$

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Radius of Incircle

#r_i = (-a + b + c ) / 2 = (9 + 21.75 - 23.55) / 2 ~~ 3.6

Area of Incircle $A_i = pi r_i^2 = pi * 3.6^2 = 40.72$ sq units

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 98 98 . What is the area of the triangle's incircle?