Question

Construct a quadrilateral $ABCD$ with $AB=5cm,BC=3.6cm,\angle ABC={80}^{\circ },\angle DAB={60}^{\circ }$ and $\angle ADC={120}^{\circ }$.

Answer 1

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

$\therefore$ $\angle A+\angle B+\angle C+\angle D={360}^o.$

$\therefore$ ${60}^o+{80}^o+\angle C+\angle {120}^o={360}^o$

$\therefore$ ${260}^o+\angle C={360}^o.$

$\therefore$ $\angle C={100}^o.$

$\Rightarrow$ Draw a line segment $AB=5cm$

$\Rightarrow$ Take $A$ as a center and draw an angle of ${60}^o$

$\Rightarrow$ Take $B$ as a center and draw an angle of ${80}^o$

$\Rightarrow$ Cut off $BC=3.6cm$

$\Rightarrow$ Take $C$ as a center and draw an angle of ${100}^o$, which cuts of $\angle A$ at $D.$

$\therefore$ $ABCD$ is a required quadrilateral.