Question

Construct a parallelogram ABCD, when :

AB = 4.5 cm, $\angle B={120}^{\circ }$ and the distance between AB and DC = 3.0 cm.

Answer 1

Steps of construction:

$\Rightarrow$ Draw a line segment $AB=4.5$cm.

$\Rightarrow$ Construct an angle of $120$${}^{\circ }$ at B.

$\Rightarrow$ We have given that distance between AB and DC is 3cm, then $AD=BC=3$ cm.

$\Rightarrow$ Since it's parallelogram and AD $=$ BC then, AB $=$ CD.

$\Rightarrow$ With B as a center and radius $3$ cm then cut off at point C.

$\Rightarrow$ With C as center and radius CD $=4.5$ cm draw an arc.

$\Rightarrow$ Then with A as center and radius AD $=3$ cm draw an arc intersecting at point D.

$\Rightarrow$ Join AD and CD.