In trapezium $ABCD$, $AB \parallel CD$, $\angle A = 75^{\circ}$. Then $\angle D$ is:

In trapezium $ABCD$, $AB \parallel CD$, $\angle A = 75^{\circ}$. Then $\angle D$ is:
  1. $75^{\circ}$
  2. $90^{\circ}$
  3. $105^{\circ}$
  4. $115^{\circ}$

Solution

Since $AB \parallel CD$, $\angle A$ and $\angle D$ are co-interior angles between the parallels, so they are supplementary: $\angle D = 180 - 75 = 105^{\circ}$.

Asked in: IMO

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