Triangular Area

I decided to use an area approach here.

The height $h$ of isosceles $\triangle{ABC}$ is $h = \sqrt{(x - 1)^2 - (\dfrac{x}{2})^2} =$ $\dfrac{\sqrt{3x^2 - 8x + 4}}{2}$

$\implies$ The area $A_{\triangle{ABC}} = \dfrac{1}{2} x h = \dfrac{\sqrt{3x^2 - 8x + 4} * x}{4}$

and $A_{\triangle{ABC}}$ can also be expressed as $A_{\triangle{ABC}} = A_{\triangle{APC}} + A_{\triangle{APB}} + A_{\triangle{BPC}}$

$= \dfrac{3(3x - 2)}{4}$

$\implies 3(3x - 2) = \sqrt{3x^2 - 8x + 4} * x \implies 9(3x - 2)^2 = (x - 2)(3x - 2)(x^2)$ $\implies (2 - 3x)(x^3 - 2x^2 - 27x + 18) = 0 \implies (2 - 3x)(x - 6)(x^2 + 4x - 3) = 0$

$(x^2 + 4x - 3) = 0 \implies x = \sqrt{7} - 2$ (dropping the negative root) and

$x = \sqrt{7} - 2 \implies |\overline {\rm AB}| = \sqrt{7} - 3 < 0 \therefore x = \sqrt{7} - 2$ is not a valid solution to the problem.

$x = \dfrac{2}{3} \implies |\overline {\rm AB}| < 0 \therefore x = \dfrac{2}{3}$ is not a valid solution to the problem.

Using $x =6 \implies |\overline {\rm AC}| = 6, |\overline {\rm AB}| = |\overline {\rm BC}| = 5$ and height $h = 4 \implies$

$A_{\triangle{ABC}} = \boxed{12}$ .

Let the base length of the isosceles triangle be $a$ , then the leg length is $a-1$ . Then its height of triangle $h = \sqrt{ (a-1)^2 - \left(\dfrac a2\right)^2} = \dfrac {\sqrt{(3a-2)(a-2)}}2$ and its area $A= \dfrac {a\sqrt{(3a-2)(a-2)}}4$ . Since the incircle of the triangle has a radius of $\dfrac 32$ , the area is also given by $A=\dfrac {3(3a-2)}4$ . Therefore,

$\begin{aligned} \frac {a\sqrt{(3a-2)(a-2)}}4 & = \frac {3(3a-2)}4 \\ a\sqrt{a-2} & = 3\sqrt{3a-2} & \small \blue{\text{Squaring both sides}} \\ a^3 - 2a^2 & = 27a - 18 \\ a^3 - 2a^2 - 27a + 18 & = 0 \\ (a-6)(a^2 + 4a-3) & = 0 \end{aligned}$

$\implies a = \begin{cases} 6 & \implies A = \dfrac {3(3(6)-2)}4 = \boxed{12} \\ \sqrt 7 - 2 & \implies A < 0 \small \red{\text{ Unacceptable}} \\ - \sqrt 7 - 2 & \implies A < 0 \small \red{\text{ Unacceptable}} \end{cases}$