If angles $P,Q,R$ and $S$ of the quadrilateral $PQRS$ taken in order are in the ratio $3:7:6:4$ then $PQRS$ is a

The correct option is B: trapezium

$\Rightarrow$ $\angle P:\angle Q:\angle R:\angle S=3:7:6:4$ [ Given ]

$\Rightarrow$ Let $\angle P=3x,\angle Q=7x,\angle R=6x$ and $\angle S=4x$.

In quadrilateral PQRS, we have

$\Rightarrow$ $\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$ [Sum of all angles of quadrilateral is ${360}^{\circ }$]

$\Rightarrow$ $3x+7x+6x+4x={360}^{\circ }$

$\Rightarrow$ $20x={360}^{\circ }$

$\therefore$ $x={18}^{\circ }$

$\Rightarrow$ $\angle P=3\times {18}^{\circ }={54}^{\circ }$

$\Rightarrow$ $\angle Q=7\times {18}^{\circ }={126}^{\circ }$

$\Rightarrow$ $\angle R=6\times {18}^{\circ }={108}^{\circ }$

$\Rightarrow$ $\angle S=4\times {18}^{\circ }={72}^{\circ }$

Now, $\angle P+\angle Q={54}^{\circ }+{126}^{\circ }={180}^{\circ }$

$\Rightarrow$ and $\angle R+\angle S={108}^{\circ }+{72}^{\circ }={180}^{\circ }$

$\Rightarrow$ If transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.

$\therefore$ $PS\parallel QR$

$\therefore$ Quadrilateral $PQRS$ is a trapezium.