Find the area

I have divided the quadrilateral into two triangles.

Pythagorean theorem in the green one gives us the length of the diagonal of the quadrilateral as $\sqrt{36^2+27^2}=45$

The yellow triangle is isosceles and its height, marked with a dashed line, is from Pythagorean theorem $\sqrt{45^2-18^2}=9\sqrt{21}$

Area of green triangle $A_g=\frac{1}{2}\times 36\times 27=486$

Area of yellow triangle $A_y=\frac{1}{2}\times36\times 9\times \sqrt{21}=162\sqrt{21}$

Total area $A=A_g+A_y=\boxed{486+162\sqrt{21}}$

Apply pythagorean theorem

$x^2=36^2+27^2$ $\implies$ $x^2=1296+729$ $\implies$ $x^2=2025$ $\implies$ $x=\sqrt{2025}$ $\implies$ $x=45$

Compute $A_1$ :

$A_1=\dfrac{1}{2}bh=\dfrac{1}{2}(36)(27)=486$

Compute $A_2$ using the Heron's Formula :

$s=\dfrac{36+45+45}{2}=63$ $~~$ $\implies$ $A_2=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{63(63-36)(63-45)(63-45)}=\sqrt{63(27)(18)(18)}=162\sqrt{21}$

Finally, $A=A_1+A_2=$ $\color{#3D99F6}\large\boxed{486+162\sqrt{21}}$ $\boxed{\text{answer}}$