Question
Draw a $\triangle\text{ABC}$ with side $BC = 6\ cm, AB = 5\ cm$ and $\angle\text{ABC} = 60^\circ.$ Then construct a triangle whose sides are $\Big(\frac{3}{4}\Big)^\text{th}$ of the corresponding sides of the $\triangle\text{ABC}.$
✓
Answer
Steps of construction:
- Draw a line segment $BC = 6\ cm.$
- At $B$, draw a ray $BX$ making an angle of $60^\circ $ with $BC$ and cut off $BA = 5\ cm$.
- Join $AC$. Then $ABC$ is the triangle.
- Draw a ray BY making an acute angle with $BC$ and cut off $4$ equal parts making $BB_1 = B_1B_2B_2B_3 = B_3B_4.$
- Join $B_4$ and $C$.
- From $B_3$, draw $B_3C’$ parallel to $B_4C$ and $C’A’$ parallel to $CA$.
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