Question

Construct a $△ABC$ in which $AB=4cm,\angle B={60}^{\circ }$ and altitude $CL=3cm$. Construct a $△ADE$ similar to $△ABC$ such that each side of $△ADE$ is $\frac{3}{2}$ times that of the corresponding side of $△ABC$.

Answer 1

Steps of construction
1. Draw a line segment $AB=4cm$.
2. Construct $\angle ABP=60$.
3. Draw a line $GH\parallel AB$ at a distance of $3cm$, intersecting $BP$ at $C$.
4. Join $CA$. Thus, $△ABC$ is obtained.
5. Extend $AB$ to $D$ such that $AD=\frac{3}{2}AB=\left(\frac{3}{2}\times 4\right)cm=6cm$.
6. Draw $DE\parallel BC$, cutting $AC$ produced at $E$. Then, $△ADE$ is the required triangle similar to $△ABC$ such that each side of $△ADE$ is $\frac{3}{2}$ times the corresponding side of $△ABC$.
Proof: Since $DE\parallel BC$, we have : $△ADE\sim △ABC$
$\therefore \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}=\frac{3}{2}$.