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Prove that the Least Perimeter of an Isosceles Triangle in Which a Circle of Radius R Can Be Inscribed is - Mathematics and Statistics

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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 equalNow, 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`

Concept: Tangents and Normals
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