Question

We draw a triangles which $AB=6cm,BC=9cm,\angle ABC={55}^{\circ }$, let us draw a parallelogram equal in area to that triangle having an angle ${60}^{\circ }$ and length of one side is $\frac{1}{2}$ of $AC$.

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 $AC$ at $D$.
3. Join $B$ to $D$.
4. Draw a line parallel to $AC$ through $B$ by completing the parallelogram $AQBC$.
5. From D, construct $\angle GDA={60}^{\circ }$.
6. The line cuts parallel line through B at $E$.
7. From E, cut an arc of length DA on the parallel line.
8. Mark the point as $F$.
9. Join $F$ to $A$.

10. ADEF is the required parallelogram.