Question

Draw a $\Delta ABC$ in which AB = 5 cm , BC = 6 cm and $\angle ABC={60}^{\circ }$. Construct a triangle similar to ABC with scale factor $\frac{5}{7}$. Justify the construction.

Answer 1

Steps of construction

1. Draw a line segment AB= 5 cm
2. From Point B, draw $\angle ABY={60}^{\circ }$ on which take BC = 6cm
3. Join AC, $\Delta ABC$ is the required triangle.
4. From A, draw any ray AX downwards marking an acute angle.
5. Mark 7 points $B_1,B_2,B_3,B_4,B_5,B_6$ and $B_7$ on AX. Such that $A{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$
6. Join $B_{7}B$ and from $B_5$ draw $B_{5}M\parallel B_{7}B$ intersecting AB at M
7. From point M draw $MN\parallel BC$ intersecting AC at N. then $\Delta AMN$ is the required triangle whose sides are equal to $\frac{5}{7}$ of the corresponding sides of the $\Delta ABC$
Justification
Here $B_{5}M\parallel B_{7}B$ (by construction)

$\therefore \frac{AM}{BM}=\frac{5}{2}$

Now, $\frac{AB}{AM}=\frac{AM+MB}{AM}$

$=1+\frac{MB}{AM}=1+\frac{2}{5}=\frac{7}{5}$

Also, $MN\parallel BC$

$\therefore \Delta AMN\sim \Delta ABC$

Therefore $\frac{AM}{AB}=\frac{AN}{AC}=\frac{NM}{BC}=\frac{5}{7}$