A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #(pi)/12 #, and the triangle's area is #1 #. What is the area of the triangle's incircle?

Question

A triangle has vertices A, B, and C.  Vertex A has an angle of #pi/8 #, vertex B has an angle of #(pi)/12  #, and the triangle's area is #1 #.  What is the area of the triangle's incircle?

Autumn Armstrong · Accepted Answer

Area of triangle’s Incircle #A_i = color(green)0.2424#

Three angles are #A = pi / 8, B = pi / 12, C = pi - pi/8 - pi/12 = (19pi)/24#

Given Area of triangle A_t = 1#

#bc = A_t / ((1/2) sin A) = (2 * 1) / sin (pi / 8) = 5.2263#

#ca = A_t / ((1/2) sin B) = 2 / sin (pi / 12) = 7.7274#

#ab = A_t / ((1/2) sin C) = 2 / sin ((19pi)/24) = 3.2854#

#sqrt(ab * bc * ca)= (abc) = sqrt(5.2263 * 7.7274 * 3.2854) = 11.5188#

#a = (abc) / (bc) = 11.5188 / 5.2263 = 2.204#

#b = (abc) / (ca) = 11.5188 / 7.7274 = 1.4906#

#c = (abc) / (AB) = 11.5188 / 3.2854 = 3.5061#

Semi perimeter #s = (a + b + c) / 2 = (2.204 + 1.4906 + 3.5061) / 2 ~~ 3.6#

Inradius #I_r = A_t / (s/2) = 1 / 3.6#

Area of Incircle #A_i = pi (I_r)^2 = pi * (1/3.6)^2 = color(green)(0.2424)#

Maverick Smith · Answer

undefined To find the area of the incircle of a triangle, you can use the formula:

$$ \text{Area of incircle} = \text{Area of triangle} \times \frac{s}{s_0} $$

where $ s $ is the semiperimeter of the triangle and $ s_0 $ is the semiperimeter of the triangle's incircle.

First, find the semiperimeter $ s $ of the triangle:

$$ s = \frac{a + b + c}{2} $$

where $ a $, $ b $, and $ c $ are the side lengths of the triangle. Since the angles are given, you can use the law of sines to find the side lengths:

$$ \frac{a}{\sin(\frac{\pi}{8})} = \frac{b}{\sin(\frac{\pi}{12})} = \frac{c}{\sin(\pi - \frac{\pi}{8} - \frac{\pi}{12})} $$

Now, solve for $ a $, $ b $, and $ c $. Once you have the side lengths, you can find $ s $ and then the area of the incircle using the formula above.