A trapezium ABCD, in which AB is parallel to CD, is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M, and OM = 2. If $\angle $AMB is 60°, the difference between the lengths of the parallel sides is X$\sqrt{3}$. Find the value of X.
(a) 2
(b) 3
(c) 5
(d) 1

Hint: To understand the question better, we will draw the figure of a trapezium ABCD inside the circle with centre O. We will draw the diagonals and label the point of intersection as M. We will prove triangle AMB and triangle DMC to be equilateral. Then, we will make a construction perpendicular from diagonal BD at point K to point O. After that, we will derive a relation of BD and K and find the difference between AB and CD.

Complete step-by-step answer:
Let us draw the required figure of trapezium ABCD inscribed in a circle with centre O. Then, we will draw the diagonals BD and CA of the trapezium.
We will also draw a perpendicular from O to BD at K.

It is given that OM = 2 and $\angle $AMB is 60°.
Since, ABCD is a trapezium, it is given that AB is parallel to CD.
And $\angle $AMB = $\angle $CMD = 60°, as they are vertically opposite angles.
Therefore, we can say that triangle DMC and triangle AMB are both similar and equilateral triangles.
Now, as we can see, OM is bisecting $\angle $AMB. Therefore, $\angle $OMK = 30°.
This means, in triangle OMK, sin30 = $\dfrac{\text{OK}}{\text{OM}}$, therefore OK = $\dfrac{1}{2}$OM = 1, since OM = 2.
Also, BD acts a chord to the circle and perpendicularly drawn from the centre of the circle to the chord, bisects the chord. Thus, K is the midpoint of BD.
$\Rightarrow $ BK = DK
In triangle, OMK, we apply Pythagora's theorem.
$\begin{align}
  & \Rightarrow \text{O}{{\text{M}}^{\text{2}}}=\text{O}{{\text{K}}^{\text{2}}}+\text{K}{{\text{M}}^{\text{2}}} \\
 & \Rightarrow {{2}^{2}}={{1}^{2}}+\text{K}{{\text{M}}^{2}} \\
 & \Rightarrow \text{K}{{\text{M}}^{2}}=3 \\
 & \Rightarrow \text{KM}=\sqrt{3} \\
\end{align}$
Now, AB – CD = BM – DM, since triangle AMB and triangle CMD are equilateral triangles.
$\Rightarrow $BM – CD = BK + KM – (DK – KM)
$\Rightarrow $BM – CD = 2KM
$\Rightarrow $BM – CD = 2$\sqrt{3}$
Therefore AB – CD = 2$\sqrt{3}$
$\Rightarrow $ X = 2
Hence, option (a) is the correct option.

Note: This question requires knowledge of triangles, similarity of triangles, circle and basic geometry. Students are advised to be thorough with these concepts. And students need to know the pythagora’s theorem as well.