Question

Construct a $\square ABCD$ such that $AB=5.2cm,BC=3.1cm,AD=4.7cm,AC=4.2cm$ and $BD=6.1cm$, then $m\angle DAB=$?

• A. ${76}^{\circ }$
• B. ${54}^{\circ }$
• C. ${32}^{\circ }$
• D. ${81}^{\circ }$

Answer 1

The correct option is A ${76}^{\circ }$

$\Rightarrow$ Draw a line segment $AB=5.2cm$

$\Rightarrow$ Take $B$ as a center with radius $3.1cm$ and draw an arc.

$\Rightarrow$ Take $A$ as a center with radius $4.2cm$ and draw an arc meeting previous arc at point $C$.

$\Rightarrow$ Again take $A$ as a center with radius $4.7cm$ and draw an arc.

$\Rightarrow$ Take $B$ as a center with radius $6.1cm$ and draw an arc meeting previous previous arc at point $D$.

$\Rightarrow$ Join $BC,CD,AD,AC$ and $BD$.

$\Rightarrow$ Thus, $ABCD$ is a required quadrilateral.

$\Rightarrow$ We have of measure angle of DAB.

$\therefore$ $m\angle DAB={76}^{\circ }$