Question

The angles P, Q, R and S of a trapezium PQRS are in the ratio of 1:3:7:9. Then:

• A. $\angle P$ = ${18}^{\circ }$, $\angle Q$ = ${54}^{\circ }$, $\angle R$ = ${126}^{\circ }$ and $\angle S$ = ${162}^{\circ }$
• B. $\angle P$ = ${81}^{\circ }$, $\angle Q$ = ${45}^{\circ }$, $\angle R$ = ${126}^{\circ }$ and $\angle S$ = ${162}^{\circ }$
• C. $\angle P$ = $8^{\circ }$, $\angle Q$ = ${94}^{\circ }$, $\angle R$ = ${26}^{\circ }$ and $\angle S$ = ${62}^{\circ }$
• D. $\angle P$ = ${18}^{\circ }$, $\angle Q$ = ${54}^{\circ }$, $\angle R$ = ${216}^{\circ }$ and $\angle S$ = ${62}^{\circ }$

Answer 1

The correct option is A $\angle P$ = ${18}^{\circ }$, $\angle Q$ = ${54}^{\circ }$, $\angle R$ = ${126}^{\circ }$ and $\angle S$ = ${162}^{\circ }$

As the angles are in the ratio 1:3:7:9
We can take the angles as $x$, $3x$, $7x$ and $9x$.
$x+3x+7x+9x={360}^{\circ }$
$\Rightarrow 20x={360}^{\circ }$
$\Rightarrow x={18}^{\circ }$
$\angle P=x={18}^{\circ }$,
$\angle Q=3x={54}^{\circ }$
$\angle R=7x={126}^{\circ }$
$\angle S=9x={162}^{\circ }$