MCQ
In a quadrilateral $ABCD, \angle\text{A} + \angle\text{C}$ is 2 times $\angle\text{B} + \angle\text{D}.$ If $\angle\text{A} = 140^\circ$ and $\angle\text{D} = 60^\circ,$ then $ZB = ?$
- ✓$60^\circ $
- B$120^\circ$
- C$80^\circ$
- DNone of these
✓
Answer
Correct option: A.
$60^\circ $
Given,
$ABCD$ is a parallelogram
$\angle\text{A} + \angle\text{C} = 2(\angle\text{B} + \angle\text{D})$
$\angle\text{A} = 40^\circ$
$∵ \angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ [angle sum property of quadrilateral]
$\Rightarrow \angle\text{A} + \angle\text{C} + \angle\text{B} + \angle\text{D} = 360^\circ$
$⇒ 2(\angle\text{B} + \angle\text{D})+ \angle\text{B} + \angle\text{D} = 360^\circ$
$⇒ 3(\angle\text{B} + \angle\text{D})= 360^\circ$
$\Rightarrow\angle\text{B}+\angle\text{D}=\frac{360^\circ}{3}=120^\circ$
$\because\ \angle\text{B}=60^\circ$ [given]
$\therefore\ \angle\text{B}=120^\circ-60^\circ=60^\circ$
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