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

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Answer 1 · 2018-08-05T10:26:54

The area of the triangle's incircle is #7.63# sq.unit

#/_A = pi/12= 180/12=15^0 , /_B = (5 pi)/12= 75^0 #

#:. /_C= 180-(75+15)=90^0 ; A_t=24#

Area ,# A_t= 1/2*b*c*sin A or b*c=(2*24)/sin 15~~ 185.46 #,

similarly ,#a*c=(2*24)/sin 75 =49.69#, and

#a*b=(2*24)/sin 90 = 48.0 #

#(a*b)*(b*c)*(c.a)=(abc)^2= (185.46*49.69*48) # or

#abc=sqrt( 185.46*49.69*48) ~~ 665.11 #

#a= (abc)/(bc)=665.11/185.46~~ 3.59# unit

#b= (abc)/(ac)=665.11/49.69~~ 13.39# unit

#c= (abc)/(ab)=665.11/48~~ 13.86 # unit

Semi perimeter : #S/2=(3.59+13.39+13.86)/2~~15.42#

Incircle radius is #r_i= A_t/(S/2) = 24/15.42~~1.56# unit

Incircle Area = #A_i= pi* r_i^2= pi*1.56^2 ~~7.63# sq.unit

The area of the triangle's incircle is #7.63# sq.unit [Ans]