Question
Construct a $\triangle\text{ABC},$ with BC = 7cm, $\angle\text{B}=60^\circ$ and AB = 6cm. 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 = 7cm.
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 BY, mark four points $B_1, B_2, B_3$ and $B_4$ such that $B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
Step 7. Join $\mathrm{CB}_4$.​​​​​​​
Step 8. From $B_3$, draw $B_3 C^{\prime} \| C B_4$ meeting $B C$ at $C^{\prime}$.​​​​​​​
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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