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

From the area of the square, we can deduce that length of the side of the square is $$6$$ cm.

$$OB$$ is the radius of the circle and can be found as it's also the diagonal of the square which is $$6\sqrt{2}$$ cm.

$$LQ = \sqrt{OQ^2 + OL^2} = \sqrt{6^2 + (6\sqrt{2})^2} =\sqrt{108} cm$$

Consider $$\triangle PQD, \triangle LQO$$

$$\angle QOL = \angle QDP = 90\degree$$ (Angle made by the diameter on any point of the circle is 90 \degree)

$$\angle DQP = \angle OQL$$ (Common)

Thus, by AA similarity criterion, $$\triangle PQD \sim \triangle LQO$$

$$\dfrac{DQ}{OQ} = \dfrac{PQ}{QL} = \dfrac{2\times 6\sqrt{2}}{\sqrt{108}} $$

$$ DQ =  6\sqrt{2}\dfrac{2\times 6\sqrt{2}}{\sqrt{108}} = 8\sqrt{3}$$ cm

$$ x = 24$$