A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(5pi)/12 #, and the triangle's area is #21 #. What is the area of the triangle's incircle?

1 Answer
Jul 11, 2016

Area of triangle's incircle is #6.7(1dp)# sq.unit

Explanation:

#/_A = 180/12=15^0 ; /_B= 58180/12=75^0 ; /_C=180-(75+15)=90^0#
Area of triangle #A_r=21#(given). Let a,b,c be the sides of triangle.We know #(a*b*sinC)/2=A_r or a*b=42/sin90=42#similarly
# b*c=42/sin15=162.28# and # c*a=42/sin75=43.48:. ab*bc*ca=(abc)^2=42*162.28*43.48 or abc=544.38 :.# #a=544.38/162.28=3.355 ;b =544.38/43.48=12.52 ; c= 544.38/42=12.96#So semi perimeter#S=(3.355+12.52+12.96)/2 =14.42# Incircle radius is#A_r/S= 21/14.42=1.46# Area of triangle's incircle is #pi*1.46^2=6.7(1dp)# sq.unit [Ans]