#### Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 23 maths.

$6\sqrt{3}r$

Hint: Using Pythagoras theorem solve the problem easily

Given: To prove the least perimeter of an isosceles triangle in which a circle of radius r

Can be inscribed in $6\sqrt{3}r$

Solution:

Let ABC an isoceles triangle with AB = AC =x and BC =y and a circle with centre O and radius r inscribed in triangle ABC.

AO= 2r, OF =r

Using phythagoras theorem in triabgle ABF

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

$(3r)^2+\left ( \frac{y}{2} \right )^2=x^2$                            ..........(1)

Again for $\Delta ADO$,

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

$3r^2=AD^2$

$AD=\sqrt{3}r$

BD=BF, EC=FC

Now, AD + DB = x

$\sqrt{3}r+ \left (\frac{y}{2} \right )=x$

$\frac{y}{2} =x =\sqrt{3}r$                        ........(2)

\begin{aligned} &\therefore(3 r)^{2}+(x-\sqrt{3} r)^{2}=x^{2} \\ &9 r^{2}+x^{2}-2 \sqrt{3} r x+3 r^{2}=x^{2} \\ &12 r^{2}=2 \sqrt{3} r x \\ &6 r=\sqrt{3} x \\ &x=\frac{6 r}{\sqrt{3}} \end{aligned}

From (2)

\begin{aligned} &\frac{y}{2}=\frac{6 r}{\sqrt{3}}-\sqrt{3} r \\ &=\frac{(6 \sqrt{3}-3 \sqrt{3}) r}{3}=\frac{3 \sqrt{3}}{3} r \\ &y=2 \sqrt{3} r \end{aligned}

Perimeter = 2x +y

\begin{aligned} &=2\left(\frac{6 r}{\sqrt{3}}\right)+2 \sqrt{3} r \\ &=\frac{12 r}{\sqrt{3}}+2 \sqrt{3} r \\ &=\frac{12 r+6 r}{\sqrt{3}}=\frac{18 r}{\sqrt{3}} \\ &=\frac{18 \times \sqrt{3} r}{\sqrt{3} \times \sqrt{3}} \end{aligned}

Perimeter $=6\sqrt{3}$