A circle is inscribed in $\triangle A B C$, touching $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$, respectively. If $\mathrm{AB}-\mathrm{BC}=4 \mathrm{~cm}, \mathrm{AB}$ $-\mathrm{AC}=2 \mathrm{~cm}$ and the perimeter of $\triangle A B C=32 \mathrm{~cm}$, then $\frac{B C}{2}($ in $\mathrm{cm})=$ ?

$\frac{13}{3}$

AB - BC = 4    ..(1)

AB - AC = 2    ..(2)

Perimeter of ABC = 32  

AB + BC + CA = 32    ..(3)

Adding equations (1), (2) and (3)

32 = 3AB - 6

3AB = 32 + 6

3AB = 38

AB = \(\frac{38}{3}\)

Putting it in equation (1), we get

\(\frac{38}{3}\) - BC = 4

\(\frac{BC}{2}\) = \(\frac{13}{3}\) cm.

Therefore, \(\frac{BC}{2}\) is \(\frac{13}{3}\) cm.