MCQ
The diagonals $AC$ and $BD$ of a rectangle $ABCD$ intersect each other at $P.$ If $\angle\text{ABD} = 50^\circ,$ then $\angle\text{DPC} =\ ?$
- A$70^\circ $
- B$100^\circ$
- C$90^\circ$
- ✓$80^\circ$
✓
Answer
Correct option: D.
$80^\circ$
Given,$ABCD$ is a rectangle
Diagonals $AC\ \&\ BD$ intersect each other at $P$
$\angle\text{ABD} = 50^\circ$
$∵$ diagonals of rectangle bisect each other and are equal in length
$⇒ \angle\text{ABD} = \angle\text{PDC}$ [alternate angles]
$⇒ \angle\text{PDC}= \angle\text{PCD} = 50^\circ$
In $\triangle\text{DPC}$
$⇒ \angle\text{DPC} + \angle\text{PCD} + \angle\text{PDC} = 180^\circ$
$⇒ \angle\text{DPC} + 50^\circ + 50^\circ= 180^\circ$
$⇒ \angle\text{DPC} = 180^\circ - 100^\circ = 80^\circ$
Diagonals $AC\ \&\ BD$ intersect each other at $P$
$\angle\text{ABD} = 50^\circ$
$∵$ diagonals of rectangle bisect each other and are equal in length
$⇒ \angle\text{ABD} = \angle\text{PDC}$ [alternate angles]
$⇒ \angle\text{PDC}= \angle\text{PCD} = 50^\circ$
In $\triangle\text{DPC}$
$⇒ \angle\text{DPC} + \angle\text{PCD} + \angle\text{PDC} = 180^\circ$
$⇒ \angle\text{DPC} + 50^\circ + 50^\circ= 180^\circ$
$⇒ \angle\text{DPC} = 180^\circ - 100^\circ = 80^\circ$
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