Question

Construct a cyclic quadrilateral $ABCD$ such that $AB=6cm,\angle A={60}^{\circ },\angle ABD={30}^{\circ }$ and $\angle CBD={20}^{\circ }$. Then, perimeter of $△ABD$ is:

• A. $15.4cm$
• B. $13.1cm$
• C. $16.7cm$
• D. $14.2cm$

Answer 1

The correct option is D $14.2cm$

• Draw a line segment $AB$ of length $6cm$
• From vertex $A$ and base $AB$ draw a line at an angle of ${60}^o$.
  
• From vertex $B$ and base $AB$ draw a line at an angle of ${30}^o$ These two lines intersect each other at point $D$.
  
• Join $AD$.
  
• Draw perpendicular bisector of line $AD$ and $AB$. These perpendiculars bisector intersect each other at point $O$.
  
• From center as $O$, draw a circle of radius $OA$. This circle touches $D,A,B$.
  
• Taking $DB$ as base and $B$ as vertex draw a line at an angle ${20}^o$ from $DB$. This line cuts the circle at point $C$.
  
• Join $ABCD$.

$ABCD$ is a required quadrilateral.

$AB=6cm$ (given)

$BD=5.2cm$ (measurement)

$AD=3cm$ (measurement)

Perimeter of $△ABD$$=AB+BD+AD=6+5.2+3=14.2cm$