Question
Construct an isosceles triangle whose base is $9\ cm$ and altitude $5\ cm$. Construct another triangle whose side are $\frac34$ of the corresponding sides of the first isosceles triangle.
✓
Answer
Steps of construction:
1. Draw a line segment $B C=9 cm$
2. Draw peependicular bisector $P Q$ of $B C$ Meeting it at $M$.
3. Fron QP cut-off a distance $M A=5 cm$.
4. Join $A B$ and $A C$.
Thus, isosceles $\triangle ABC$ is obtained.
5. Below $B C$, make an acute $\angle C B X$.
6. Along BX , mark off 4 points $B_1, B_2, B_3, B_4$, such that $B B_1,=B_1 B_2=B_1 B_3=B_3 B_4$.
7. Join $B_4 C$
8. From $B_3$, draw $B_3 C^{\prime} \| B_4 C$, meeting BC at $C ^{\prime}$.
9. From $C^{\prime}$, draw $C^{\prime} A^{\prime} \| C A$, meeting $A B$ at $A^{\prime}$.
Thus, $\triangle A ^{\prime} BC ^{\prime}$ is the required triangle similar to $\triangle ABC$ Such that each side of $\triangle A ^{\prime} BC ^{\prime}$ is $\frac{3}{4}$ times the corresponding side of $\triangle ABC$.
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