Construct a parallelogram ABCD in which BC = 5 cm,

Question:

Construct a parallelogram ABCD in which BC = 5 cm, ∠BCD = 120° and CD = 4.8 cm.

Solution:

Steps of construction:

Step 1: Draw BC= $5 \mathrm{~cm}$

Step 2: Make an $\angle B C D=120^{\circ}$

Step 2: With $C$ as centre draw an arc of $4.8 \mathrm{~cm}$, name that point as $D$

Step 3: With $D$ as centre draw an $\operatorname{arc} 5 \mathrm{~cm}$, name that point as $A$

Step 4: With $B$ as centre draw another $\operatorname{arc} 4.8 \mathrm{~cm}$ cutting the previous arc at $A$.

Step 5: Join $A D$ and $A B$

​then, ABCD is a required parallelogram.

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