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.
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 $\triangle \mathrm{ABC}$.
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