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 #15 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-01-03T11:31:20

The area of the triangle's incircle is #4.01# sq.unit.

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

# :. /_C= 180-(30+15)=135^0 : A_t=15#

We know Area ,# A_t= 1/2*b*c*sinA or b*c=(2*15)/sin15# or

# b*c ~~ 115.91#, similarly ,# A_t= 1/2*a*c*sinB# or

#a*c=(2*15)/sin30 = 60 #, and #a*b=(2*15)/sin135 ~~ 42.43 #

#(a*b)*(b*c)*(c.a)=(abc)^2= (42.43*115.91*60) # or

#abc=sqrt(42.43*115.91*60) ~~ 543.20(2dp) #

#a= (abc)/(bc)=543.20/115.91~~4.69#

#b= (abc)/(ac)=543.20/60~~9.05#

#c= (abc)/(ab)=543.20/42.43~~12.80#

Semi perimeter : #S/2=(4.69+9.05+12.80)/2 ~~13.27#

Incircle radius is #r_i= A_t/(S/2) = 15/13.27~~ 1.13#

Incircla Area = #A_i= pi* r_i^2= pi*1.13^2 ~~4.01# sq.unit

The area of the triangle's incircle is #4.01# sq.unit [Ans]