A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(pi)/2 #, and the triangle's area is #3 #. What is the area of the triangle's incircle?

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Answer 1 · 2017-12-02T11:11:41

Area of triangle's incircle is #0.95# sq.unit.

#/_A = pi/12= 180/12=15^0 , /_B = pi/2=180/2= 90^0 #

#:. /_C= 180-(90+15)=75^0 ; A_t=3#

We know Area ,# A_t= 1/2*b*c*sinA or b*c=(2*3)/sin15 ~~23.18#.

Similarly #a*c=(2*3)/sin90 = 6.0 #, and #a*b=(2*3)/sin75~~ 6.21 #

#(a*b)*(b*c)*(c.a)=(abc)^2= (6.21*23.18*6.00) ~~863.69 #

#:.abc=sqrt(863.69) ~~29.39 ; a=(abc)/(bc)=29.39/23.18~~1.27#

#b= (abc)/(ac)=29.39/6.0~~4.90 # and

#c= (abc)/(ab)=29.39/6.21~~4.73#

Semi perimeter : #S/2=(1.27+4.90+4.73)/2~~5.45#

Incircle radius is #r_i= A_t/(S/2) = 3/5.45~~0.55#

Incircle Area = #A_i= pi* r_i^2= pi*0.55^2 ~~ 0.95# sq.unit

Area of triangle's incircle is #0.95# sq.unit [Ans]