# Construct an isosceles triangle whose base is 8 cm and altitude 4 cm

Question:

Construct an isosceles triangle whose base is $8 \mathrm{~cm}$ and altitude $4 \mathrm{~cm}$ and then another triangle whose sides are $\mathbf{1} \frac{\mathbf{1}}{\mathbf{2}}$ times the corresponding sides of the isosceles triangle.

Solution:

Steps of construction :

1. Construct isosceles $\triangle \mathrm{ABC}$ such that $\mathrm{BC}=8 \mathrm{~cm}$ and altitude $\mathrm{AD}=4 \mathrm{~cm}$.

2. Draw any ray $\mathrm{BX}$ making an acute angle with $\mathrm{BC}$

3. Take $\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}$ on $\mathrm{BX}$ such that $\mathrm{BB}_{1}=\mathrm{B}_{1} \mathrm{~B}_{2}=\mathrm{B}_{2} \mathrm{~B}_{3}$.

4. Join $\mathrm{B}_{2} \mathrm{C}$.

5. Through $\mathrm{B}_{3}$ draw $\mathrm{B}_{3} \mathrm{C}^{\prime} \| \mathrm{B}_{2} \mathrm{C}$ and let it intersect $\mathrm{BC}$ (produced) at $\mathrm{C}^{\prime}$.

6. Through $\mathrm{C}^{\prime}$ draw, $\mathrm{C}^{\prime} \mathrm{A}^{\prime} \| \mathrm{CA}$ and let it intersect $\mathrm{BA}$ (produced) at $\mathrm{A}^{\prime}$.

Now, $\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}$ is the required triangle whose side are $\mathbf{1} \frac{\mathbf{1}}{\mathbf{2}}$ times the corresponding sides of the $\triangle \mathrm{ABC}$.