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Question
Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is `6sqrt3` r.
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Solution
Let ABC is an isosceles triangle with AB=AC=x and a circle with centre O and radius r is inscribed in the triangle. O,A and O,E and O,D are joined.From ΔABF,
`AF^2+BF^2=AB^2`
`⇒(3r)^2+(y2)^2=x^2 .....(1)`
Again,From ΔADO,`(2r)^2=r^2+AD^2`
`⇒3r^2=AD^2`
`⇒AD=sqrt3r`
Now, BD=BF and EC=FC (Since tangents drawn from an external point are equal) Now, AD+DB=x
`⇒(sqrt3r)+(y^2)=x`
`⇒y^2=x−sqrt3 .....(2)`
`∴(3r)^2+(x−sqrt3r)^2=x^2`
`⇒9r^2+x^2−2sqrt3rx+3r^2=x^2`
`⇒12r^2=2sqrt3rx`
`⇒6r=sqrt3x`
`⇒x=6r/sqrt3`
Now, From (2),
`y/2=6/sqrt3r−sqrt3r`
`⇒y/2=6/sqrt3r−sqrt3r`
`⇒y/2=((6sqrt3−3sqrt3)r)/3`
`⇒y/2=(3sqrt3r)/3`
`⇒y=2sqrt3r`
Perimeter=2x+y
`=2(6/sqrt3r)+2sqrt3r`
`=12/sqrt3r+2sqrt3r`
`=(12r+6r)/sqrt3`
`=18/sqrt3r`
`=(18xxsqrt3)/(sqrt3xxsqrt3)r`
`=6sqrt3r`
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