Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is `6sqrt3r`.

Options

A. `4sqrt(3r)`
B. `2sqrt(3r)`
C. `6sqrt(3r)`
D. `8sqrt(3r)`

Answer

3

Explanation

Let ABC be an isosceles triangle in which a circel of the is inscribed
Let `angleBAD =theta`(semi vertical angle)
In `triangleOAE,OA=OE cosec theta =r cosec theta ,AE =r cot theta` .Thus
`AD=OA+OD=r(cosec theta +1)`
In `triangleABD,BD=AD tan theta =r(cosec theta+1)tan theta`
`AB=AD sec theta =r(cosec theta +1)sec theta`
Now perimeter of the `triangle` ABC
S=AB+AC+BC=2AB+2BD
`=2r(cosec theta +1)(sec theta + tan theta)=(4r(1+sin theta)^(2))/(sin 2 theta)`
`therefore (dS)/(d theta) =4r[2(1 + sin theta)cos theta sin 2 theta -(1+sin theta)^(2)2 cos 2 theta ]//(sin 2 theta)^(2)`
`=8r(1+sin theta)[sin 2 theta cos theta-cos theta is theta - cos 2 theta]//(sin 2 theta)^(2)`
`=8r(1+sin theta)(sin theta -1 +2 sin^(2) theta)//(sin 2 theta)^(2)`
`16r(1+sin theta)^(2)(sin theta -1//2)//(sin 2 theta)^(2)`
for maximum or minimum of s. s `ds//d` that=0 or `sin theta =1//2`
Now if `theta` is little less and little greater than `pi//6` then sign of `ds//d theta` changes from -ve +ve Hence S is minimum when `theta=pi//6` which is the point of minima
Hence the least perimeter of the `triangle` is
`triangle =4r[1+sin(pi//6)]^(2)//sin (pi//3)=6sqrt(3r)`