Question

Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and angle ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD‘ C‘ similar to triangle BDC with scale factor 4/3. Draw the line segment D‘A‘ parallel to DA where A’ lies on extended side BA. Is A’BC’D’ a parallelogram?

Answer 1

Steps of constructions:

1. Draw a line AB=3 cm.

2. Draw a ray BY making an acute ∠ABY=60°.

3. With centre B and radius 5 cm, draw an arc cutting the point C on BY.

4. Draw a ray AZ making an acute ∠ZAX’=60°.(BY||AZ, ∴ ∠YBX’=ZAX’=60°)

5. With centre A and radius 5 cm, draw an arc cutting the point D on AZ.

6. Join CD

7. Thus we obtain a parallelogram ABCD.

8. Join BD, the diagonal of parallelogram ABCD.

9. Draw a ray BX downwards making an acute ∠CBX.

10. Locate 4 points B₁, B₂, B₃, B₄ on BX, such that BB₁=B₁B₂=B₂B₃=B₃B₄.

11. Join B₄C and from B₃C draw a line B4C’||B3C intersecting the extended line segment BC at C’.

12. Draw C’D’|| CD intersecting the extended line segment BD at D’. Then, ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC.

13. Now draw a line segment D’A’|| DA, where A’ lies on the extended side BA.

14. Finally, we observe that A’BC’D’ is a parallelogram in which A’D’=6.5 cm A’B = 4 cm and ∠A’BD’= 60° divide it into triangles BC’D’ and A’BD’ by the diagonal BD’.