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

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Answer 1 · 2018-02-25T11:20:59

The area of the triangle's incircle is #11.66# sq.unit.

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

# /_C= 180-(90+30)=60^0 ; Delta_a=24#

This is a right triangle of #(30,60,90) #

In right triangle #(30,60,90) #,if base is #b=x# , hypotenuse is

#a=2x# and perpendicular is #c=sqrt3*x#

Area of the triangle ,# A_t= 1/2*b*c=24 or 24= 1/2*x*sqrt3*x#

or #x^2=48/sqrt3 or x =sqrt(16*sqrt3) ~~ 27.71 :. b ~~ 5.264#

unit #c= sqrt3*x ~~ 9.12# unit ,# a=2x~~ 10.53# unit

Semi perimeter is #S/2=(5.264+9.12+10.53)/2 ~~ 12.46#

Incircle radius is #r_i= A_t/(S/2) = 24/12.46~~1.93#

Incircle Area = #A_i= pi* r_i^2= pi*1.93^2 ~~11 .66# sq.unit

The area of the triangle's incircle is #11.66# sq.unit [Ans]