Question

Question 6
If angles A, B, C and D of the quadrilateral ABCD, taken in order are in the ratio 3: 7: 6: 4, then ABCD is a
(A) rhombus
(B) parallelogram
(C) trapezium
(D) kite

Answer 1

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 ${360}^{\circ }$.

$\therefore 3x+7x+6x+4x={360}^{\circ }\Rightarrow 20x={360}^{\circ }\Rightarrow x=\frac{{360}^{\circ }}{{20}^{\circ }}={18}^{\circ }$
$\therefore$ Angles of the quadrilateral are
$\angle A=3\times 18={54}^{\circ }\angle B=7\times 18={126}^{\circ }\angle C=6\times 18={108}^{\circ }$
And $\angle D=4\times 18={72}^{\circ }$
From figure, $\angle BCE={180}^{\circ }-\angle BCD$ [linear pair axiom]
$\Rightarrow \angle BCE={180}^{\circ }-{108}^{\circ }={72}^{\circ }$
Since, the corresponding angles are equal .
$\therefore BC||AD$
Now, sum of cointerior 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.