Question

Construct a cyclic quadrilateral $ABCD$ such that $AB=3.4cm,BC=4.2cm,AD=2.7cm$ and $\angle ABC={60}^{\circ }$. Then, diameter of the circle circumscribing the cyclic quadrilateral is:

• A. $2.2cm$
• B. $4.4cm$
• C. $3.3cm$
• D. $5.5cm$

Answer 1

The correct option is B $4.4cm$

• Draw a line segment $BC$ of length $4.2cm$.
• Draw a line at an angle of ${60}^o$ by taking $BC$ as base and $B$ as vertex.
  
• Cut an arc of $3.4cm$ from $B$. This arc cuts the line at point $A$.
  
• Join $AB$.
  
• Draw perpendicular bisectors of $AB$ and $BC$. These perpendicular bisectors cuts each other at point $O$.
  
• Draw a circle of radius $OB$ and center $O$, this circle touches $A,B,C$.
  
• Mark an arc of length $2.7cm$ on circle from point $A$. This arc cuts the circle at point $D$.
  
• Join $ABCD$.
  
• $ABCD$ is the required cyclic quadrilateral.
  

We know, $R=\frac{a}{2\sin A}$

In $△ADC$

$\angle ADC={120}^{o}AC=3.8cm\Rightarrow R=\frac{AC}{2\sin D}=\frac{3.8\times 2}{2\times \sqrt{3}}=2.194$

Diameter $=2R=2\times 2.194=4.4cm$