In the given figure, ∆ACB ∼ ∆APQ. If BC = 8 cm,

Solution:

It is given that $\triangle A C B \sim \triangle A P Q$.

$B C=8 \mathrm{~cm}, P Q=4 \mathrm{~cm}, B A=6.5 \mathrm{~cm}$ and $A P=2.8 \mathrm{~cm}$

We have to find $C A$ and $A Q$.

Since $\triangle A C B \sim \triangle A P Q$

$\Rightarrow \frac{B A}{A Q}=\frac{C A}{A P}=\frac{B C}{P Q}$

So 

$\frac{6.5 \mathrm{~cm}}{A Q}=\frac{8 \mathrm{~cm}}{4 \mathrm{~cm}}$

$A Q=\frac{6.5 \mathrm{~cm} \times 4 \mathrm{~cm}}{8 \mathrm{~cm}}$

$=3.25 \mathrm{~cm}$

Similarly

$\frac{C A}{A P}=\frac{B C}{P Q}$

$\frac{C A}{2.8 \mathrm{~cm}}=\frac{8 \mathrm{~cm}}{4 \mathrm{~cm}}$

$C A=2.8 \mathrm{~cm} \times 2 \mathrm{~cm}$

$=5.6 \mathrm{~cm}$

Hence, $C A=5.6 \mathrm{~cm}$ and $A Q=3.25 \mathrm{~cm}$