Question

Draw parallelogram ABCD with AC = 9 cm, BD = 7 cm and $\angle$ AOB = ${120}^o$ where $\overline{AC}$ and $\overline{BD}$ intersect at O and find its area.

Answer 1

Given:
AC = 9 cm, BD = 7 cm and $\angle$ AOB = ${120}^o$
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AC = 9 cm.
Step 3 : Mark O the midpoint of $\overline{AC}$ .
Step 4 : Draw a line $\overleftrightarrow{XY}$ through O which makes $\angle$ AOY = ${120}^o$ .
Step 5 : With O as centre and 3.5 cm as radius draw two arcs on $\overleftrightarrow{XY}$ on either sides of $\overline{AC}$ cutting $\overrightarrow{OX}$ at D and $\overrightarrow{OY}$ at B.
Step 6 : Join $\overline{AB},\overline{BC},\overline{CD}$ and $\overline{DA}$ . ABCD is the required parallelogram.
Step 7 : From D draw $\overline{DE}\perp \overline{AB}$ and measure the length of DE. DE = h = 4 cm. AB = b = 7 cm.
Calculation of area:
In the parallelogram ABCD, b = 7 cm and h = 4 cm.
Area of the parallelogram ABCD = b *h = 7 *4 = 28 $c{m}^2$ .