Question

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${}^{\circ }$, 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

Answer 1

The correct option is A 2

$\Rightarrow$ In given figure $ABCD$ is a trapezium.

$\Rightarrow$ $AB\parallel CD$ [Given]

$\Rightarrow$ Given $\angle AMB={60}^{\circ }$

$\Rightarrow$ $\angle CMD={60}^{\circ }$ [Vertically opposite angles]

$\Rightarrow$ Then $△AMB$ and $△CMD$ are equalateral triangles.

$\Rightarrow$ In figure draw $OK$ perpendicular to $BD$.

$\Rightarrow$ $OM$ bisects $\angle AMB$ then,

$\Rightarrow$ $\angle OMK={30}^{\circ }$

$\Rightarrow$ Hence, $OK=\frac{OM}{2}=1$ [$\because$ OM=2]

$\Rightarrow$ In right angled triangle KMO,

$\Rightarrow$ $O{M}^2=O{K}^2+K{M}^2$ [Pythagoras theorem]

$\Rightarrow$ $2^2=1^2+K{M}^2$

$\Rightarrow$ $K{M}^2=4-1$

$\therefore$ $KM=\sqrt{3}$ -- ( 1 )

$\Rightarrow$ In figure we cal also observe that, K is mid-point of BD.

$\Rightarrow$ $AB-CD=BM-MD=BK+KM-(DK-KM)=2KM$,

$\Rightarrow$ $AB-CD=2\sqrt{3}$ [From ( 1 )]

$\therefore$ Value of $x$ is $2$.