prove that the least perimeter of an isoceles triangle in which a circle of radius r can be inscribed is r.6(3)^1/2

Let ABC is an isosceles triangle with AB=AC=x and a circle with centre O and radius r isinscribed in the triangle.O,A and O,E and O,D are joined.From ΔABF, AF2+BF2=AB2=>(3r)2+(y2)2=x2.......(i)Again,From ΔADO, (2r)2=r2+AD2=>3r2=AD2=>AD=3rNow, BD=BF and EC=FC (Since tangents drawn from an external point are equal)Now, AD+DB=x=>(3r)+(y2)=x=>y2=x3.......(ii)(3r)2+(x3r)2=x2=>9r2+x223rx+3r2=x2=>12r2=23rx=>6r=3x=>x=6r3.Now, From (ii),y2=63r3r=>y2=633r3r=>y2=(6333)r3=>y2=33r3=>y=23r.Perimeter=2x+y=2(63r)+23r=123r+23r=12r+6r3=183r=18×33×3r=63r.

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