Question

We draw a triangle with length of sides $5cm,8cm$ and $11cm$ and draw a parallelogram equal in area to that triangle and having an angle ${60}^{\circ }$. [Let us write instruction process and proof]

Answer 1

Steps of Construction:
1. We draw a triangle $ABC$ of given dimensions.
2. We draw a perpendicular bisector of side $AB$ which intersects $AB$ at $D$.
3. Join C to D.
4. Draw a line parallel to $AB$ through $C$ by completing the parallelogram $ACQB$.
5. From $D$, construct $\angle GDB={60}^{\circ }$.
6. The line cuts parallel line through $C$ at $E$.
7. From $E$, cut an arc of length DB on the parallel line.
8. Mark the point as F.
9. Join $F$ to $B$.

10. DBFE is the required parallelogram.