Question

The angles of a quadrilateral are in the 3 : 4 : 5 : 6. Show that the quadrilateral is a trapezium.

Answer 1

Let the angles of an quadrilateral be $3x,4x,5x$ and $6x$.

We know that sum of four angles of quadrilateral is ${360}^o$

$\therefore$ $3x+4x+5x+6x={360}^o$

$18x={360}^o$

$x={20}^o$

$3x=3\times {20}^o={60}^o$

$4x=4\times {20}^o={80}^o$

$5x=5\times {20}^o={100}^o$

$6x=6\times {20}^o={120}^o$

Let $A={60}^o,\angle B={80}^o,\angle C={100}^o$ and $\angle D={120}^o$

$\angle A+\angle D={60}^o+{120}^o={180}^o$

$\angle B+\angle C={80}^o+{100}^o={180}^o$

Now,$A$ and $D$ are supplementary

and $B$ and $C$ are supplementary

$\therefore$ $AB\parallel CD$

when $2$ opposite sides are parallel, the the quadrilateral is trapezium.