Question

Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and $\angle$ ABC = ${60}^{\circ }$. Then, construct a triangle whose sides are $\frac{5}{7}$ times the corresponding sides of $\Delta$ ABC. Justify your construction.

Answer 1

Steps of construction :

1. Construct a $\Delta$ ABC with given measurement.

2. Below BC, make an acute angle $\angle$ CBY.

3. Along BY, mark off seven points $B_1,B_2,B_3,B_4,B_5,B_6$ and $B_7$ such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$.

4. Join $B_{7}C$.

5. From $B_5$, draw $B_{5}D\parallel B_{7}C$, meeting BC at D.

6. From D, draw $DE\parallel CA$, meeting AB at E.

7. $\Delta$ EBD is the required triangle, each of whose sides is $\frac{5}{7}$ of the corresponding side of $\Delta$ ABC

Justification :

Since $DE\parallel CA\Delta ABC\sim \Delta EBD$

$\therefore \frac{EB}{AB}=\frac{DE}{CA}=\frac{BD}{BC}=\frac{5}{7}$