Question

Prove that the least perimeter of an isosceles triangle in which circle or radius r can be inscribed is 6root3 r

Answer 1

The solution has been attached 

Answer 2

Proved below.

Step-by-step explanation:

Given:

Let ABC is an isosceles triangle with AB  =AC = x and BC = y and a circle with centre O and radius r isinscribed in the triangle. Join OA and OE and OD.

From Δ ABF,

$AF^2+BF^2=AB^2$

⇒ $(3r)^2+(\frac{y}{2} )^2=x^2$         $[BF= \frac{BC}{2} ]$     [1]

Again, from ΔADO,

$AO^2=DO^2+AD^2$

$(2r)^2=r^2+AD^2$

$4r^2=r^2+AD^2$

$3r^2 = AD^2$

⇒ $AD =\sqrt{3}r$

Now, BE = BF and AD = AE   (Since tangents drawn from an external point are equal)           [2]

Now,

AB = BE + AE

x = BE + AE                [given]

x = BF + AD                [from 2]

$x = \sqrt{3}r+\frac{y}{2}$          

$\frac{y}{2} = x - \sqrt{3}r$           [3]

Putting value of Eq (3) in (1), we get

$(3r)^2+(x-\sqrt{3}r )^2=x^2$

$9r^2+x^2-2\sqrt{3}rx+ 3r^2=x^2$

$12 r^2=2\sqrt{3}rx$

$6r=\sqrt{3}x$

$x=\frac{6r}{\sqrt{3} }$

Now, From Eq (3)

$\frac{y}{2} = x - \sqrt{3}r$

$\frac{y}{2} = \frac{6r}{\sqrt{3} } - \sqrt{3}r$

$\frac{y}{2} = \frac{6\sqrt{3}}{3 } r - \sqrt{3}r$

$\frac{y}{2} =\frac{(6\sqrt{3}-3\sqrt{3})r }{3}$

$\frac{y}{2} =\frac{3\sqrt{3}r }{3}$

$y=2\sqrt{3} r$

Perimeter = 2x+y

                 = $2(\frac{6}{\sqrt{3} }r )+2\sqrt{3}r$

                 = $\frac{12}{\sqrt{3} }r +2\sqrt{3}r$

                 = $\frac{12r+ 6r}{\sqrt{3} }$

                 = $\frac{18r}{\sqrt{3} }$

                 = $\frac{18r \times \sqrt{3} }{\sqrt{3}\times\sqrt{3}}$

                 = $\frac{18\sqrt{3}r}{3}$

                 = $6\sqrt{3}r$

Hence proved.