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 $\mathrm{cm})=$ ?
Answer & explanation
Correct answer: option 2
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.