In trapezium $ABCD$ with $AB \parallel CD$, $\angle A + \angle D$ equals
In trapezium $ABCD$ with $AB \parallel CD$, $\angle A + \angle D$ equals
- $180^{\circ}$
- $90^{\circ}$
- $120^{\circ}$
- $360^{\circ}$
Solution
Co-interior angles between parallel sides.
Asked in: MH-SSC-9
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