In trapezium $PQRS$, $PQ \parallel RS$. If $\angle P = 110^{\circ}$ and $\angle Q = 80^{\circ}$, then…

In trapezium $PQRS$, $PQ \parallel RS$. If $\angle P = 110^{\circ}$ and $\angle Q = 80^{\circ}$, then $\angle R + \angle S$ equals:
  1. $170^{\circ}$
  2. $180^{\circ}$
  3. $190^{\circ}$
  4. $210^{\circ}$

Solution

Total $= 360^{\circ}$. So $\angle R + \angle S = 360 - (110+80) = 360 - 190 = 170^{\circ}$.

Asked in: IMO

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