Question
Draw a triangle $\triangle\text{ABC}$ in which AB = 5cm, BC = 6cm and ABC= 60º. Construct a triangle similar to $\triangle\text{ABC}$ with scale factor $\frac{5}{7}.$ Justify the construction.
✓
Answer
Scale factor $\frac{5}{7}<1,$ so the resulting $\triangle$ will be smaller than $\triangle\text{ABC}.$
Steps of construction:
- Draw AB=5cm.
- Draw $\triangle\text{ABC}=60^\circ,$ cut BC = 6cm and join AC.
- Draw acute $\angle\text{BAX}$ and mark it equispaced marks $A_1, A_2, ..., A_7$as shown in figure.
- Join $A_7B$ and draw $A_5B' || A_7$ B. B' is on segment AB.
$\triangle\text{AB}'\text{C}'\sim\triangle\text{ABC}$ with scale factor $\frac{5}{7}.$
Justification:
In $\triangle\text{AA}_5\text{B}'\text{and AA}_7\text{B},$ $\text{A}_7\text{B}\ ||\ \text{A}_5\text{B}'$
$\therefore\ \angle\text{A}_5=\angle\text{A}_7\ \ [\text{Correponding}\ \angle\text{s}]$
$\angle\text{BAA}_5=\angle\text{BAA}_7\ \ [\text{Common}]$
$\therefore\ \triangle\text{AA}_5\text{B}'\sim\triangle\text{AA}_7\text{B}$ [By AA criterion of similarity]
$\Rightarrow\ \frac{\text{AB}'}{\text{AB}}=\frac{\text{AA}_5}{\text{AA}_7}=\frac{5\text{x}}{7\text{x}}=\frac{5}{7}\ \ ...(\text{i})$
where $\text{x}=\text{AA}_1=\text{A}_1\text{A}_2=...\text{A}_6\text{A}_7$
Similarly, $\triangle\text{AB}'\text{C}'\sim\triangle\text{ABC}$ [By AA criterion of similarity]
$\Rightarrow\ \frac{\text{AB}'}{\text{AB}}=\frac{\text{AC}'}{\text{AC}}=\frac{\text{B}'\text{C}'}{\text{BC}}$
$\Rightarrow\ \frac{5}{7}=\frac{\text{AC}'}{\text{AC}}=\frac{\text{B}'\text{C}'}{\text{BC}}$
Hence, $\triangle\text{AB}'\text{C}'\sim\triangle\text{ABC}$ with scale factor $\frac{5}{7}.$
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