ABCD is a rectangle of length $$\ell$$ and breadth b. The breadth is two-seventh $$\left(\frac{2}{7}\right)$$ of its length $$\ell$$. P is the mid-point of side AB. As shown in the figure, APQ & BPR are the two quadrants while CXD is a semicircle with X as the mid-point of the arc CXD, then

This is a multiple-correct question, and the correct options are (a), (b) and (d). Take $$AB = b = 2$$ and $$AD = \ell = 7$$; the quadrants and semicircle all have radius 1, so with A at the origin, $$Q = (0, 1)$$, $$R = (2, 1)$$ and $$X = (1, 6)$$. Then $$CR = 7 - 1 = 6$$ and $$CR \colon AB = 3 \colon 1$$, while $$XQ = \sqrt{1 + 25} = \sqrt{26}$$ with $$AQ = 1$$. Since $$XQ = XR = \sqrt{26}$$, triangle XQR is isosceles, not scalene. The shaded area is $$14 - 2 \cdot \frac{\pi}{4} - \frac{\pi}{2} = 14 - \pi$$, so the ratio is $$14 \colon (14 - \pi)$$.