In the adjoining figure, POQ is the diameter of the semicircle with centre O. OABC is a square whose area is $$36 \text{ cm}^2$$. If $$QD = x \text{ cm}$$, the value of $$x\sqrt{3}$$ is

The square has side 6, so with O as origin the vertices are $$A(-6, 0)$$, $$B(-6, 6)$$ and $$C(0, 6)$$, and since $$B$$ lies on the semicircle the radius is $$OB = 6\sqrt{2}$$. The line joining $$Q(6\sqrt{2}, 0)$$ to $$C$$ meets the arc again at $$D(-2\sqrt{2}, 8)$$, so $$QD = \sqrt{(8\sqrt{2})^2 + 8^2} = \sqrt{192} = 8\sqrt{3}$$. Hence $$x\sqrt{3} = 8\sqrt{3} \times \sqrt{3} = 24$$.