Question

[Key Concept: Place the right trapezium ($AB$ parallel to $CD$, $AB$ perpendicular to $AD$, $AB = 3CD$) on axes; area follows from the parallel sides and the height.]

Area of trapezium

$$ABCD = \frac{1}{2}(a + 3a)(2r) = 4$$

$$\Rightarrow \ ar = 1 $$

Equation of $BC$ is $y = - r^{2}\left( x - \frac{3}{r} \right)$

$$y + r^{2}x - 3r = 0$$

$\because\ BC$ is tangent to circle

$${\Rightarrow \frac{\left| r + r^{3} - 3r \right|}{\sqrt{1 + r^{4}}} = r }{\Rightarrow r^{4} + 4 - 4r^{2} = 1 + r^{4} }{\Rightarrow \ r = \frac{\sqrt{3}}{2} }$$