Question

construct a parallelogram ABCD in which AB=6 cm,<A=60° and BC=5cm​

Answer 1

Answer:

Draw AB = 6 cm.

At A draw ∠QAB = 60°.

From AQ cut AD = 3 cm.

From D, draw an arc of radius 6 cm.

From B, draw an arc of radius 3 cm which meets first arc at C.

Join CD and BC. Thus ABCD is the required ||gm.

Bisect ∠DAB, so that bisector meets CD at P.

Join PB and measure ZAPB. ∴ ∠APB = 90°.

Hope it's help u

please mark as Brainliest answer