A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #22 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-02-02T11:42:29

The area of incircle of triangle is #7.47# sq.unit.

#/_A = pi/8= 180/8=22.5^0 , /_B = pi/6=180/6= 30^0#

# /_C= 180-(22.5+30)=127.5^0: A_t=22 #

We know Area ,# A_t= 1/2*b*c*sinA# or

#b*c=(2*22)/sin22.5 ~~114.98 #, similarly ,#a*c=(2*22)/sin30#

#:.a*c = 88.0; a*b=(2*22)/sin127.5 ~~ 55.46#

#(a*b)*(b*c)*(c.a)=(abc)^2= (55.46*114.98*88.0)#

# =561165.55 :. abc =sqrt(561165.55)=749.11 #

#a= (abc)/(bc)=749.11/114.98~~6.52#

#b= (abc)/(ac)=749.11/88.00~~8.51#

#c= (abc)/(ab)=749.11/55.46~~13.51#

Semi perimeter : #S/2=(6.52+8.51+13.51)/2 ~~14.27#

Incircle radius is #r_i= A_t/(S/2) = 22.0/14.27 ~~1.54#

Incircle Area = #A_i= pi* r_i^2= pi*1.54^2 ~~ 7.47# sq.unit.

The area of incircle of triangle is #7.47# sq.unit. [Ans]