Prove that the least perimeter of an isosceles triangle in which a circle of radious `r` can be inscribed is `6sqrt(3)r`

Explanation

AQ =r cot `alpha` =AP
AO =r cosec `alpha`

`(x)/(AO+ON) = " tan"alpha`
` x= ( r " cosec "alpha +r) " tan "alpha`
`x=r(" sec"alpha + " tan"alpha)
Perimeter = p =4x +2AQ
`p= 4r( sec`alpha` + tan`alpha`) + 2rcot`alpha`
p= r(4sec`alpha` + 4tan`alpha` + 2cot`alpha`)
`(dp)/(dalpha) =r[4sec alpha " tan"alpha + 4 sec^(2) alpha - 2"cosec"^(2)alpha]`
for max or min `(dp)/(dalpha) =0 rArr " " 2sin^(3) alpha + 3sin^(2) alpha -1 =0`
`rArr " " (sinalpha + 1) (2sin^(2) alpha+ sinalpha -1) =0`
`(sinalpha +1)^(2) (2sinalpha -1) =0 " "rArr " "sinalpha =1//2 " "rArr " " alpha =30^(@) =pi//6`
`P_("leat") =r [(4.2)/(sqrt(3)) +(4)/(sqrt(3)) +2sqrt(3)] =r [(8+4+6)/(sqrt(3)) ] =((6sqrt(3)sqrt(3)))/(sqrt(3)) r=6sqrt(3) r`