Construct a $\vartriangle ABC$ in which $CA= 6\ cm$, $AB= 5\ cm$ and $\angle BAC= 45^{o}$, Then construct a triangle whose sides are $\frac {3}{5}$ of the corresponding sides of ABC.

Given: $CA= 6\ cm$, $AB= 5\ cm$ and $\angle BAC= 45^{o}$,

To do: To construct a $\vartriangle ABC$ in which $CA= 6\ cm$, $AB= 5\ cm$ and $\angle BAC= 45^{o}$, Then To construct a triangle whose sides are $\frac {3}{5}$ of the corresponding sides of ABC.

1. Draw AB=5 cm. With A as center, draw $\angle BAC= 45^{o}$. Join BC. ABC is thus formed.

2. Draw AX such that $\angle BAX$ is an acute angle.

3. Cut 5 equal arcs such that $AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}$

4. Join $A_{5}$ to $B$ and draw a line through $A_{3}$ parallel to $A_{5}B$ which meets AB at B'

Here, $AB'=\frac{3}{5}AB$.

Now draw a line through $B'$ parallel to $BC$ which joins $AC$ at $C'$.

Here,

 $B'C'=\frac{3}{5}BC$ and $AC'=\frac{3}{5}AC$

Thus, $\vartriangle AB'C'$ is the required triangle.