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

Question

1204 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-21T12:35:54

The area of triangl's incircle is #2.9# sq.unit.

#/_A = pi/8= 180/8=22.5^0 , /_B = pi/12=180/12= 15^0#

#:./_C= 180-(22.5+15)=142.5^0 ; A_t=12#

Area ,# A_t= 1/2*b*c*sinA or b*c=(2*12)/sin22.5 ~~ 62.72 #,

similarly ,#a*c=(2*12)/sin15 =92.73 #, and

#a*b=(2*12)/sin142.5 ~~ 39.42#

#(a*b)*(b*c)*(c.a)=(abc)^2= (39.42*62.72*92.73) # or

#abc=sqrt((39.42*62.72*92.73)) ~~ 478.82 #

#a= (abc)/(bc)=478.82/62.72~~7.63#

#b= (abc)/(ac)=478.82/92.73~~5.16#

#c= (abc)/(ab)=478.82/39.42~~12.15#

Semi perimeter : #S/2=(7.63+5.16+12.15)/2~~12.47#

Incircle radius is #r_i= A_t/(S/2) = 12/12.47~~0.96#

Incircle Area = #A_i= pi* r_i^2= pi*0.96^2 ~~ 2.9# sq.unit.

The area of triangl's incircle is #2.9# sq.unit. [Ans]