In a circle touching the side $BC$ of $\Delta ABC$ at $P$ and sides $AB$ and $AC$ produced at $Q$ and $R$ respectively. If $AQ=10\text{cm}$, then the perimeter of $\Delta ABC$ is:

According to the properties of circles, lengths of tangents drawn from an external point to a circle are equal.
1. From point $A$:

$$AQ=AR=10\text{cm}$$

2. From point $B$:

$$BQ=BP$$

3. From point $C$:

$$CR=CP$$

The formula for the perimeter of $\Delta ABC$ is:

$$\text{Perimeter}=AB+BC+AC$$

We can rewrite the side $BC$ as $BP+PC$:

$$\text{Perimeter}=AB+(BP+PC)+AC$$

Substituting $BP=BQ$ and $PC=CR$ into the equation:

$$\text{Perimeter}=AB+BQ+CR+AC$$

$$\text{Perimeter}=(AB+BQ)+(AC+CR)$$

From the figure description, $AB+BQ=AQ$ and $AC+CR=AR$:

$$\text{Perimeter}=AQ+AR$$

Since $AQ=AR=10\text{cm}$:

$$\text{Perimeter}=10+10=20\text{cm}$$