Given a triangle whose sides are #24# #cm#, #30# #cm#, and #36# #cm#. Find the radius of a circle which is tangent to the shortest and longest side of the triangle and whose center lies on the third side?

1 Answer
Jan 29, 2018

#r=(9sqrt7)/2~~11.91# cm

Explanation:

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Given : #AB=24, BC=30 and AC=36#,
Let #D and E# be the tangancy point on #AB and AC#, respectively, and let #O and r# be the center and the radius of the circle, respectively, as shown in the figure.
Use Heron's formula to find the area of #DeltaABC#,
#A_(DeltaABC)=sqrt(s(s-a)(s-b)(s-c)#
where #a,b and c# are the side lengths of the triangle, and
#s=(a+b+c)/2#
#=> s=(24+30+36)/2=45#
#=> A_(DeltaABC)=sqrt(45(45-24)(45-30)(45-36))=sqrt(45*21*15*9)=135sqrt7 " cm^2#
Now, #A_(DeltaAOB)=1/2*AB*r=1/2*24*r=12r#
#A_(DeltaAOC)=1/2*AC*r=1/2*36*r=18r#
#=> A_(DeltaAOB)+A_(DeltaAOC)=A_(DeltaABC)#
#=> 12r+18r=135sqrt7#
#=> 30r=135sqrt7#
#=> r=(135sqrt7)/30=(9sqrt7)/2~~11.91# cm