Question

Construct a quadrilateral ABCD with $AB=6.5\text{cm}$, $AD=5\text{cm}$, $CD=5\text{cm}$, $\angle$ BAC = ${40}^{\circ }$ and $\angle$ ABC = ${50}^{\circ }$ , and also find its area.

Answer 1

Given:
AB = 6.5 cm, AD = 5 cm, CD = 5 cm, $\angle BAC$ = ${40}^{\circ }$ and $\angle ABC$ = ${50}^{\circ }$.
Steps for construction:
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment $AB=6.5\text{cm}$.
Step 3 : At A on $\overline{AB}$ make $\angle BAX$whose measure is ${40}^{\circ }$ and at B on $\overline{AB}$ make $\angle ABY$whose measure is ${50}^{\circ }$ . They meet at $C$.
Step 4 : With A and C as centres draw two arcs of radius $5\text{cm}$ and let them cut at $D$.
Step 5 : Join $\overline{AD}$ and $\overline{CD}$. $ABCD$ is the required quadrilateral.
Step 6 : From $D$ draw $\overline{DE}\perp \overline{AC}$ and from $B$ draw $\overline{BC}\perp \overline{AC}$. Then measure the lengths of $BC$ and $DE$. $BC=h_1=4.2\text{cm},DE=h_2=4.3\text{cm}$ and $AC=d=5\text{cm}.$

Calculation of area:
In the quadrilateral ABCD, $d=5\text{cm},h_1=4.2\text{cm}$ and $h_2=4.3\text{cm}$.
Area of the quadrilateral $ABCD$ = $\frac{1}{2}d(h_1+h_2)$
=$\frac{1}{2}\left(57.8\right)(4.2+4.3)$
=$\frac{1}{2}\times 5\times 8.5$
=$21.25{\text{cm}}^2$