A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, 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 · 2018-01-26T14:54:33

The area of the triangle's incircle is #8.75# sq.unit.

#/_A = pi/2= 180/2=90^0 , /_B = pi/6=180/6= 30^0 #

#:. /_C= 180-(90+30)=60^0 ; A_t=18#

We now Area ,# A_t= 1/2*b*c*sinA or b*c=(2*18)/sin90 = 36 #,

similarly ,#a*c=(2*18)/sin30 = 72 #, and

#a*b=(2*18)/sin60 ~~41.57;(a*b)*(b*c)*(c.a)=(abc)^2#

#= (36*72*41.57) :.abc=sqrt((36*72*41.57))=328.25#

#a= (abc)/(bc)=328.25/36~~9.12# unit

#b= (abc)/(ac)=328.25/72~~4.56# unit

#c= (abc)/(ab)=328.25/41.57~~7.90# unit

Semi perimeter : #S/2=(9.12+4.56+7.90)/2=10.79# unit

Incircle radius is #r_i= A_t/(S/2) = 18/10.79=1.67# unit

Incircla Area = #A_i= pi* r_i^2= pi*1.67^2 =8.75# sq.unit

The area of the triangle's incircle is #8.75# sq.unit [Ans]