Question

Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, $\angle$ CAD = ${80}^o$ and $\angle$ ACD = ${40}^o$ .

Answer 1

Draw line $AB=5.5$ cm

Draw an arc of radius $6.5$ cm from point $A$ and an other arc of radius $4.5$ cm from point $B$ both meet at point $C$

Join $C$ to $A$ and $B$

Draw an angle of ${40}^0$ at point $A$ and an other angle of ${40}^0$ at point $C$. Both cut at point $D$

Join $D$ to $C$ and $A$

Hence $ABCD$ is the required quadrilateral.

In triangle $ABC$,

$s=\frac{6.5+5.5+4.5}{2}=8.25$

Then area of ABC$=\sqrt{8.25(8.25-6.5)(8.25-5.5)(8.25-4.5)}=\sqrt{8.25\times 1.75\times 2.75\times 3.75}=\sqrt{148.89}=12.20$

In triangle ACD

$\frac{AD}{\sin {40}^0}=\frac{6.5}{\sin {100}^0}\Rightarrow AD=\frac{6.5\times 0.64}{0.98}=4.25cm$

Then area triangle ACD =$\frac{1}{2}absin\theta =6.5\times 4.25sin{40}^0=27.63\times 0.64=17.68c{m}^2$

So area of ABCD quadrilateral $=12.20+17.68=29.88$ sq cm