Question
Draw a $\triangle\text{ABC}$ in which base $BC = 6cm, AB = 5cm$ and $\angle\text{ABC}=60^\circ.$ Then construct another triangle whose sides are of the corresponding sides of $\triangle\text{ABC}.$
✓
Answer
Steps of construction:
Justification of the construction:$\because\text{B}_4\text{C }||\text{ B}_3\text{C}'$ [By construction]
$\therefore\ \frac{\text{BB}_3}{\text{BB}_4}=\frac{\text{BC}'}{\text{BC}}$
[Basic Proportionality Theorem] But $\frac{\text{BB}_3}{\text{BB}_4}=\frac{3}{4}$ [By condtruction]$\therefore\ \frac{\text{BC}'}{\text{BC}}=\frac{3}{4}\ \dots(\text{i})$
$\because\text{CA }||\text{ C}'\text{A}'$ [By construction]
$\therefore\ \triangle\text{BC}'\text{A}'\sim\triangle\text{BCA}$
From equation (i),$\frac{\text{A}'\text{B}}{\text{AB}}=\frac{\text{A}'\text{C}'}{\text{AC}}=\frac{\text{BC}'}{\text{BC}}=\frac{3}{4}$
[Basic Propotionality Theorem]
- Draw a triangle $ABC$ with side $BC = 6cm, AB = 5cm$ and $\angle\text{ABC}=60^\circ.$
- Draw a ray $BX,$ which makes an acute angle $\angle\text{CBX}$ below the line BC.
- Locate four points $B_1, B_2, B_3$ and $B_4$ on BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4.$
- Join $B_4C$ and draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ to $C’.$
- Draw a line through C’ parallel to the line CA to intersect BA at A’.
Justification of the construction:$\because\text{B}_4\text{C }||\text{ B}_3\text{C}'$ [By construction]
$\therefore\ \frac{\text{BB}_3}{\text{BB}_4}=\frac{\text{BC}'}{\text{BC}}$
[Basic Proportionality Theorem] But $\frac{\text{BB}_3}{\text{BB}_4}=\frac{3}{4}$ [By condtruction]$\therefore\ \frac{\text{BC}'}{\text{BC}}=\frac{3}{4}\ \dots(\text{i})$
$\because\text{CA }||\text{ C}'\text{A}'$ [By construction]
$\therefore\ \triangle\text{BC}'\text{A}'\sim\triangle\text{BCA}$
From equation (i),$\frac{\text{A}'\text{B}}{\text{AB}}=\frac{\text{A}'\text{C}'}{\text{AC}}=\frac{\text{BC}'}{\text{BC}}=\frac{3}{4}$
[Basic Propotionality Theorem]
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