Question

Construct a triangle of sides $4cm$, $5cm$ and $6cm$, then construct a triangle similar to it, whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer 1

Draw a line segment $AB=4cm$

Draw an arc at $5cm$ from point $A$.

Draw an arc at $6cm$ from point $B$ so that it intersects the previous arc.

Joint the point of intersection from $A$ and $B$.

This gives the required $△ABC$.

Dividing the base in $2:3$ ratio:

Draw a ray $AX$ at an acute angle from $AB$.

Plot three points on $AX$ so that $A{A}_1=A_1{A}_2=A_2{A}_3$
Join $A_3$ to $B$.

Draw a line from point $A_2$ so that this line is parallel to $A_{3}B$ and intersects $AB$ at point $B^{\prime }$.

Draw a line from point $B^{\prime }$ parallel to $BC$ so that this line intersects $AC$ at point $C^{\prime }$.

$\Delta A{C}^{\prime }{B}^{\prime }$ is the required triangle.