Question

Construct a cyclic quadrilateral $ABCD$ such that $AB=6cm,\angle A={60}^{\circ },\angle ABD={30}^{\circ }$ and $\angle CBD={20}^{\circ }$. Then, $l\left(BC\right)=?$
(Round-off your answer to the nearest integer)

Answer 1

• 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.

Length of side $BC=3.9cm\Rightarrow 4cm$