Question
Construct a $\triangle\text{ABC},$ with $BC = 7\ cm$, $\angle\text{B}=60^\circ$ and $AB = 6\ cm$. Construct another triangle whose sides are $\frac34\text{times}$ the corresponding sides of $\triangle\text{ABC}.$
✓
Answer
Steps of Construction:
Step 1. Draw a line segment $BC = 7\ cm.$
Step 2. At B, draw $\angle\text{XBC}=60^\circ.$
Step 3. With B as centre and radius 6cm, draw an arc cutting the ray $BX$ at $A.$
Step 4. Join AC. Thus, $\triangle\text{ABC}$ is the required triangle.
Step 5. Below BC, draw an acute angle $\angle\text{YBC}.$
Step 6. Along $B Y$, mark four points $B_1, B_2, B_3$ and $B_4$ such that $B B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
Step 7. Join $\mathrm{CB}_4$.
Step 8. From $\mathrm{B}_3$, draw $\mathrm{B}_3 \mathrm{C}^{\prime} \| \mathrm{CB}_4$ meeting BC at $\mathrm{C}^{\prime}$.
Step 9. From C', draw $A'C' || AC $ meeting $AB$ in $A'.$
Here, $\triangle\text{ABC}$ is the required triangle whose sides are $\frac34\text{times}$ the corresponding sides of $\triangle\text{ABC}.$
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