**Question:**

Construct a parallelogram *ABCD,* in which diagonal* AC* = 3.8 cm, diagonal* BD *= 4.6 cm and the angle between *AC* and *BD* is 60°.

**Solution:**

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:

Step 1: Draw $A C=3.8 \mathrm{~cm}$

Step 2: Bisect $A C$ at $O$.

Step 3: Make $\angle C O X=60^{\circ}$

Produce $X O$ to $Y$.

Step 4:

$O B=\frac{1}{2}(4.6) \mathrm{cm}$ $O B=2.3 \mathrm{~cm}$ and $O D=\frac{1}{2}(4.6) \mathrm{cm}$ $O D=2.3 \mathrm{~cm}$

Sten 5. Join $A B, B C, C D$ and $A D$

Thus, *ABCD* is the required parallelogram.