Question

In $△PQR,\angle PQR={30}^{\circ },\angle PRQ={75}^{\circ }$ and $QR=8cm$. Let us draw a rectangle equal in area to that triangle.

Answer 1

Steps of Construction:
1. We draw a triangle $PQR$ of given dimensions.
2. We draw a perpendicular bisector of side $QR$ which intersects $QR$ at $D$.
3. Join P to $D$.
4. Draw a line parallel to $AC$ through $B$ by completing the parallelogram $AQBC$.
5. As we have to draw a rectangle. The perpendicular bisector through D can be used.
6. The perpendicular bisector cuts parallel line through $P$ at $E$.
7. From E, cut an arc of length DR on the parallel line.
8. Mark the point as $F$.
9. Join $F$ to $R$.

10. EFRD is the required rectangle.