Question

The ratio of angles A, B, C and D of the quadrilateral ABCD are 3:7:6:4 when it is taken in order. Find the shape of the quadrilateral.

• A. Rhombus
• B. Parallelogram
• C. Trapezium
• D. None of these

Answer 1

The correct option is C

Trapezium

Given, ratio of angles of quadrilateral ABCD is 3: 7: 6: 4

Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively.

We know that, sum of all angles of a quadrilateral is ${360}^{\circ }$ .

∴ 3x + 7x + 6x + 4x = ${360}^{\circ }$

$\Rightarrow 20x={360}^{\circ }$

$\Rightarrow x=\frac{{360}^{\circ }}{{20}^{\circ }}={18}^{\circ }$
$\therefore \text{Angles of the quadrilateral are}$
$\angle A=3\times {18}^{\circ }={54}^{\circ }$
$\angle B=7\times {18}^{\circ }={126}^{\circ }$
$\angle C=6\times {18}^{\circ }={108}^{\circ }$
$And\angle D=4\times {18}^{\circ }={72}^{\circ }$
$Fromfigure,\angle BCE={180}^{\circ }-\angle BCD\left[linearpairaxiom\right]$
$\Rightarrow \angle BCE={180}^{\circ }-{108}^{\circ }={72}^{\circ }$
$\text{Since, the corresponding angles are equal.}$
$\therefore BC\left|\right|AD$
$\text{Now, sum of co-interior angles,}$
$\angle A+\angle B={126}^{\circ }+{54}^{\circ }={180}^{\circ }$
$and\angle C+\angle D={108}^{\circ }+{72}^{\circ }={180}^{\circ }$

Hence, ABCD is a trapezium.