Question
Draw a $\triangle\text{ABC}$ with side BC = 6cm, AB = 5cm 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:
i. Draw a line segment $B C=6 cm$.
ii. At $B$, draw a ray $B X$ making an angle of $60^{\circ}$ with $B C$ and cut off $B A=5 cm$.
iii. Join $A C$. Then $A B C$ is the triangle.
iv. Draw a ray $B Y$ making an acute angle with $B C$ and cut off 4 equal parts making $B B_1=B_1 B_2 B_2 B_3=B_3 B_4$.
v. Join $B_4$ and $C$.
vi. From $B_3$, draw $B_3 C^{\prime}$ parallel to $B_4 C$ and $C^{\prime} A^{\prime}$ parallel to $C A$.
Then $\triangle\text{A}'\text{BC}'$ is the required triangle.
i. Draw a line segment $B C=6 cm$.
ii. At $B$, draw a ray $B X$ making an angle of $60^{\circ}$ with $B C$ and cut off $B A=5 cm$.
iii. Join $A C$. Then $A B C$ is the triangle.
iv. Draw a ray $B Y$ making an acute angle with $B C$ and cut off 4 equal parts making $B B_1=B_1 B_2 B_2 B_3=B_3 B_4$.
v. Join $B_4$ and $C$.
vi. From $B_3$, draw $B_3 C^{\prime}$ parallel to $B_4 C$ and $C^{\prime} A^{\prime}$ parallel to $C A$.
Then $\triangle\text{A}'\text{BC}'$ is the required triangle.
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