Question

A triangle has side lengths $6,7,8.$ The line through its incentre and parallel to the shortest side is drawn to meet the other two sides at $P$ and $Q.$ The length of the segment $PQ$ is

• A. $\frac{12}{5}$
• B. $\frac{15}{4}$
• C. $\frac{30}{7}$
• D. $\frac{33}{9}$

Answer 1

The correct option is C $\frac{30}{7}$


Area of $△ABC\left(\Delta \right)=\frac{6h}{2}=3h$ (where $h$ is the altitude from $A$)
Also, $\Delta =\frac{21r}{2}$ (using $\Delta =r\cdot s)$
$\therefore \frac{21r}{2}=3h\Rightarrow \frac{r}{h}=\frac{2}{7}$

Now, triangle $APQ$ and triangle $ABC$ are similar.
$\Rightarrow \frac{h-r}{h}=\frac{PQ}{BC}\Rightarrow 1-\frac{r}{h}=\frac{PQ}{6}\Rightarrow 1-\frac{2}{7}=\frac{PQ}{6}\Rightarrow PQ=\frac{30}{7}$