Question

Question 47
The angles P, Q, R and S of a quadrilateral are in the ratio 1 : 3 : 7 : 9. Then, PQRS is a
a) Parallelogram
b) Trapezium with $PQ\parallel RS$
c) Trapezium with $QR\parallel PS$
d) Kite

Answer 1

Let the angles be x, 3x, 7x and 9x, then,
$x+3x+7x+9x={360}^{\circ }$ [$\because$ sum of angles in any quadrilateral is ${360}^{\circ }$]
$\Rightarrow 20x={360}^{\circ }\Rightarrow x=\frac{{360}^{\circ }}{20}\Rightarrow x={18}^{\circ }$
Then, the angles P, Q, R and S are ${18}^{\circ },{54}^{\circ },{126}^{\circ }$ and ${162}^{\circ }$, respectively.
Since, $\angle P+\angle S={18}^{\circ }+{162}^{\circ }={180}^{\circ }$ and $\angle Q+\angle R={54}^{\circ }+{126}^{\circ }={180}^{\circ }$,
$So,PQ\parallel SR$.
$\therefore$ The quadrilateral PQRS is a trapezium with $PQ\parallel SR$.