(A) Let the angles of the quadrilateral be $2x, 4x, 5x,$ and $7x$ degrees.
We know that the sum of the interior angles of a quadrilateral is $360^{\circ}$.
Therefore,$2x + 4x + 5x + 7x = 360^{\circ}$.
$18x = 360^{\circ}$.
$x = 20^{\circ}$.
Now,calculating each angle:
$\angle A = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle B = 4 \times 20^{\circ} = 80^{\circ}$.
$\angle C = 5 \times 20^{\circ} = 100^{\circ}$.
$\angle D = 7 \times 20^{\circ} = 140^{\circ}$.
Since $\angle B + \angle C = 80^{\circ} + 100^{\circ} = 180^{\circ}$,the adjacent angles on side $BC$ are supplementary,which implies $AB \parallel CD$. $A$ quadrilateral with one pair of parallel sides is a trapezium.