Question

Question 2
Draw a parallelogram ABCD in which BC=5 cm, AB = 3 and $\angle ABC={60}^{\circ }$ divide it into triangle BCD and ABD by the diagonal BD. Construct the triangle BD’C' similar to $\Delta BDC$ with scale factor $\frac{4}{3}$. Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA, Is A’ BC’ D’ a parallelogram?

Answer 1

1. Firstly we draw a line segment, then either of one end of the line segment with length 5 cm and making an angle ${60}^{\circ }$with this end. We know that in parallelogram both opposite sides are equal and parallel, then again draw a line with 5 cm making an angle with ${60}^{\circ }$ from other end of the line segment. Now, join both parallel lines by a line segment whose measurement is 3cm, we get a parallelogram. After that, we draw a diagonal and get a triangle BDC.

2. Now , we construct the triangle $B{D}^{{}^{\prime }}{C}^{{}^{\prime }}$ similar to $\Delta BDC$ with scale factor $\frac{4}{3}$.

3. With B as center and radius equal to 5 cm draw an arc cut the point C on BY.

4. Again draw a ray AZ making an acute $\angle ZA{X}^{{}^{\prime }}={60}^{\circ }$ $[\because BY\parallel AZ,\therefore YB{X}^{{}^{\prime }}=ZA{X}^{{}^{\prime }}={60}^{\circ }]$.

5. With A as center and radius equal to 5 cm draw an arc cut the point D on AZ.



6. Now join CD and finally a parallelogram ABCD.

7. Join BD which is a diagonal of parallelogram ABCD.

8. From B draw any ray BX downwards making an acute $\angle CBX$.

9. Locate A points B₁, B₂, B₃, B₄ on BX, such that $B{B}_1=B_2{B}_3=B_3{B}_4$.

10. Join $B_{3}C$ and from $B_4$ draw a line $B_4{C}^{{}^{\prime }}\parallel B_{3}C$ intersecting the extended line segment BC at $C^{{}^{\prime }}$.

11. From point $C^{{}^{\prime }}$ draw $C^{{}^{\prime }}{D}^{{}^{\prime }}\parallel CD$ intersecting the extend line segment BD at $D^{{}^{\prime }}$ then, $\Delta D^{{}^{\prime }}B{C}^{{}^{\prime }}$ is the required triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$.

12. Now draw a line segment $D^{{}^{\prime }}{A}^{{}^{\prime }}$ parallel to DA , where $A^{{}^{\prime }}$ lies on extended side BA i.e a ray $B{X}^{{}^{\prime }}$.

13. Finally we observe that $A^{{}^{\prime }}B{C}^{{}^{\prime }}{D}^{{}^{\prime }}$ is a parallelogram in which $A^{{}^{\prime }}{D}^{{}^{\prime }}=6.5cm$ $A^{{}^{\prime }}B=4cm$ and $\angle A^{{}^{\prime }}B{D}^{{}^{\prime }}={60}^{\circ }$ divide it into triangles $B{C}^{{}^{\prime }}{D}^{{}^{\prime }}$ and $A^{{}^{\prime }}B{D}^{{}^{\prime }}$ by the diagonal $B{D}^{{}^{\prime }}$.