8. As shown in Figure 4, the area of $\triangle A B C$ is $S$, and the perimeter is $p$. The three sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$ are outside $\triangle A B C$ and are at a distance of $h$ from the corresponding sides. Then the perimeter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\qquad$ , and the area is $\qquad$
8. $\left(1+\frac{p h}{2 S}\right) p,\left(1+\frac{p h}{2 S}\right)^{2} S$.

It is easy to see that the line $A A^{\prime}$ bisects both $\angle A^{\prime}$ and $\angle B A C$, so the incenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle A B C$ is the same. Let the inradii of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle A B C$ be $r^{\prime}$ and $r$, respectively, then
$$
r^{\prime}=r+h \text {, and } r=\frac{2 S}{p} \text {. }
$$

Since $\triangle A^{\prime} B^{\prime} C^{\prime} \backsim \triangle A B C$, and the similarity ratio is
$$
\frac{r^{\prime}}{r}=\frac{r+h}{r}=1+\frac{p h}{2 S},
$$

therefore, the perimeter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\left(1+\frac{p h}{2 S}\right) p$, and the area is $\left(1+\frac{p h}{2 S}\right)^{2} S$.