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 = ?
- A60º
- B120º
- C80º
- DNone of these
✓
Answer
- 60º
Solution:
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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