Question

A triangle has sides 6,7,and 8. The line through its incenter parallel to the shortest side is drawn to meet the other two sides at P and Q Then find the 7 times of length of the segment PQ ?

Answer 1

$\Delta =r\times s$ $\therefore \frac{21\times r}{2}$

$=\frac{1}{2}b\times h$=$\frac{6\times h}{2}=3h$ or $\frac{r}{h}=\frac{2}{7}$

Now $APQ$ and $ABC$ are similar.

Thus, $\frac{h-r}{h}=\frac{PQ}{6}$

$\Rightarrow 1-\frac{r}{h}=\frac{PQ}{6}$

$\Rightarrow 1-\frac{2}{7}=\frac{PQ}{6}$

$\Rightarrow \frac{5}{7}=\frac{PQ}{6}$

$\Rightarrow PQ=\frac{30}{7}$