Question

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?

Answer 1

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

Explanation:

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