A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12, vertex B has an angle of pi/8 π8, and the triangle's area is 8 8. What is the area of the triangle's incircle?

1 Answer
Jun 20, 2018

color(purple)("Area of incircle " = A_i = pi * r^2 = 2.44 " sq units"Area of incircle =Ai=πr2=2.44 sq units

Explanation:

http://jwilson.coe.uga.edu/NCTM%20Boston%20Project/Heron/Introduction.htmlhttp://jwilson.coe.uga.edu/NCTM%20Boston%20Project/Heron/Introduction.html

hat A = pi/12, hat B = pi/8, hat C = (19pi) / 24, A_t = 8ˆA=π12,ˆB=π8,ˆC=19π24,At=8

A_t = (1/2) a b sin C = (1/2) bc sin A = (1/2) ca sin BAt=(12)absinC=(12)bcsinA=(12)casinB

ab = (2 A_t) / sin C = 16 / sin ((19pi)/24) = 26.28ab=2AtsinC=16sin(19π24)=26.28

bc = A_t / sin A = 16/ sin (pi/12) = 61.82bc=AtsinA=16sin(π12)=61.82

ca = A_t / sin C = 16 / sin(pi/8) = 41.81ca=AtsinC=16sin(π8)=41.81

a = (abc) / (bc) = sqrt(26.28 * 61.82 * 41.81) / 61.82 = 4.22a=abcbc=26.2861.8241.8161.82=4.22

260.63
b = (abc) / (ac) = sqrt(26.28 * 61.82 * 41.81) / 41.81 =6.24b=abcac=26.2861.8241.8141.81=6.24

c = (abc) / (ab) = sqrt(26.28 * 61.82 * 41.81) / 33.94 = 7.68c=abcab=26.2861.8241.8133.94=7.68

"Semiperimeter " = s = (a + b + c) / 2 = (4.22 + 6.24 + 7.68) / 2 = 9.07Semiperimeter =s=a+b+c2=4.22+6.24+7.682=9.07

"Incircle radius " = r = A_t / s = 8 / 9.07Incircle radius =r=Ats=89.07

color(purple)("Area of incircle " = A_i = pi * (8/9.07)^2 = 2.44 " sq units"Area of incircle =Ai=π(89.07)2=2.44 sq units