MCQ
If $\angle\text{A},\angle\text{B},\angle\text{C}$ and $\angle\text{D}$ of a quadrilateral $ABCD$ taken in order, are in the ratio $3 : 7 : 6 : 4$ then $ABCD$ is a:
- ARhombus.
- BKite.
- ✓Trapezium.
- DParallelogram.
✓
Answer
Correct option: C.
Trapezium.
Let the common multiple be $x.$
$\therefore$ The angle measure $3x, 7x, 6x$ and $4x.$
Since the sum of the angles of a quadrilateral is $360^\circ ,$ we have
$3x + 7x + 6x + 4x = 360$
$\Rightarrow 20x = 360$
$\Rightarrow x = 18^\circ $
$\therefore$ The angles of the quadrilateral are
$3x = 3(18) = 54^\circ $
$7x = 7(18) = 126^\circ $
$6x = 6(18) = 108^\circ $ and
$4x = 4(18) = 72^\circ $
Now, $54 + 126 = 180^\circ $ and $108 + 72 = 180^\circ $
So, the angles are interior angles and hence we get one pair of parallel sides of $ABCD.$
Hence, $ABCD$ is a trapezium.
$\therefore$ The angle measure $3x, 7x, 6x$ and $4x.$
Since the sum of the angles of a quadrilateral is $360^\circ ,$ we have
$3x + 7x + 6x + 4x = 360$
$\Rightarrow 20x = 360$
$\Rightarrow x = 18^\circ $
$\therefore$ The angles of the quadrilateral are
$3x = 3(18) = 54^\circ $
$7x = 7(18) = 126^\circ $
$6x = 6(18) = 108^\circ $ and
$4x = 4(18) = 72^\circ $
Now, $54 + 126 = 180^\circ $ and $108 + 72 = 180^\circ $
So, the angles are interior angles and hence we get one pair of parallel sides of $ABCD.$
Hence, $ABCD$ is a trapezium.
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