Question

Draw a $△ABC$, right - angled at $B$, such that $AB=3cm,BC=4cm$. Now construct a $△PBQ$ similar to $\Delta ABC$, each of whose side is $\frac{7}{5}$ times the corresponding side of $\Delta ABC$.

Answer 1

$BQ=\frac{7}{5}BC=\frac{28}{5}=5.6$

$PB=\frac{7}{5}AB=\frac{21}{5}=4.2$

On $BA$ extended mark $P$ at $4.2$ cm from $B$ on same side of $A$

On $BC$ extended mark $Q$ at $5.6$ cm from $B$ on same side of $C$

Join $P,Q$

$\therefore △PBQ$ is formed.