Question

Construct a parallelogram ABCD, AB = 6 cm, $\angle$ ABC = ${80}^o$ and AC = 8 cm and find its area.

Answer 1

Given:
AB = 6 cm, $\angle$ ABC = ${80}^o$ and AC = 8 cm.
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 6 cm
Step 3 : At B on AB make $\angle$ ABX whose measure is ${80}^o$ .
Step 4 : With A as centre and radius 8 cm draw an arc. Let it cut $\overrightarrow{BX}$ at C.
Step 5 : Join $\overline{AC}$ .
Step 6 : With C as centre draw an arc of radius 6 cm.
Step 7 : With A as centre draw another arc with radius equal to the length of BC. Let the two arcs cut at D.
Step 8 : Join $\overline{AD}$ and $\overline{CD}$ .ABCD is the required parallelogram.
Step 9 : From C draw $\overline{CE}\perp \overline{AB}$ and measure the length of CE. CE = h = 6.4 cm. AB = b = 6 cm.
Calculation of area:
In the parallelogram ABCD, b = 6 cm and h = 6.4 cm.
Area of the parallelogram ABCD = b h = 6* 6.4 = 38.4 $c{m}^2$ .