A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/6 #, and the triangle's area is #18 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-30T11:27:29

Area of the triangle's incircle is #4.81# sq.unit.

#/_A = pi/12= 180/12=15^0 , /_B = pi/6=180/6= 30^0 #

#:. /_C= 180-(30+15)=135^0 ;A_t=18#. Area of triangle,

# A_t= 1/2*b*c*sin A or b*c=(2*18)/sin 15 ~~ 139.09 #,

similarly ,#a*c=(2*18)/sin 30=72.0 #, and

#a*b=(2*18)/sin 135 ~~ 50.91 #

#(a*b)*(b*c)*(c.a)=(abc)^2= (139.09*72.0*50.91) # or

#abc=sqrt((139.09*72*50.91)) = 714.03 #

#a= (abc)/(bc)=714.03/139.09~~5.13#

#b= (abc)/(ac)=714.03/72.0~~9.92#

#c= (abc)/(ab)=714.03/50.91~~14.03#

Semi perimeter : #S/2=(5.13+9.92+14.03)/2~~14.54#

Incircle radius is #r_i= A_t/(S/2) = 18/14.54~~1.24#

Incircle Area = #A_i= pi* r_i^2= pi*1.24^2 ~~4.81# sq.unit.

Area of the triangle's incircle is #4.81# sq.unit. [Ans]