please solve it urgent

\begin{array}{l} Let ABC is an isosceles triangle with AB=AC= x and a circle with centre O and radius r is \\ inscribed in the triangle . \\ O,A and O,E and O,D are joined . \\ From Δ ABF, \\ {AF}^{2} + {BF}^{2} = {AB}^{2} \\ = > {(3 r)}^{2} + {(\frac{y}{2})}^{2} = x^{2} ....... (i) \\ Again,From Δ ADO, {(2 r)}^{2} = r^{2} + {AD}^{2} \\ = > 3 r^{2} = {AD}^{2} \\ = > AD= \sqrt{3} r \\ Now, BD=BF and EC=FC (Since tangents drawn from an external point are equal) \\ Now, AD+DB= x \\ = > (\sqrt{3} r) + (\frac{y}{2}) = x \\ = > \frac{y}{2} = x - \sqrt{3} ....... (ii) \end{array} \begin{array}{l} ∴ {(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{array} \begin{array}{l} Now, From (ii), \\ \frac{y}{2} = \frac{6}{\sqrt{3}} r - \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 = 2 x + y \\ = 2 (\frac{6}{\sqrt{3}} r) + 2 \sqrt{3} r \\ = \frac{12}{\sqrt{3}} r + 2 \sqrt{3} r \\ = \frac{12 r + 6 r}{\sqrt{3}} \\ = \frac{18}{\sqrt{3}} r \\ = \frac{18 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} r \\ = 6 \sqrt{3} r . \end{array}