Question

In $△PQR,PQ=13,QR=14$ and $PR=15$. A altitude is drawn from $P$ and a parallel line is drawn as $LM$ which divides the triangle into two equal areas, then the length of $PL$ is

• A. $5$
• B. $5(3-\sqrt{6})$
• C. $5(3-\sqrt{7})$
• D. $5(3-\sqrt{8})$

Answer 1

The correct option is C $5(3-\sqrt{7})$

Using heron's formula to find the area of the triangle,
$s=\frac{13+14+15}{2}=21\Rightarrow Area=\sqrt{21\times (21-13)\times (21-14)\times (21-15)}=\sqrt{21\times 8\times 7\times 6}=84$
Now the area of the triangle can be written as,
$\frac{1}{2}\times PT\times 14=84\Rightarrow PT=12$
$QT=\sqrt{{13}^2-{12}^2}=5$
Assuming $RM=y$ and $LM=x$


Area of $△LMR$,
$\frac{1}{2}\times xy=42\Rightarrow xy=84$
Area of $△PQT\Rightarrow \frac{1}{2}\times 5\times 12=30$
Area of trapezium $PTML\Rightarrow \frac{1}{2}\times (12+x)\times (9-y)=42-30\Rightarrow 108-12y+9x-xy=12\times 2\Rightarrow 10-84-12y+9x=24\Rightarrow 9x=12y\Rightarrow 9\times \frac{84}{y}=12y\Rightarrow y=3\sqrt{7}\Rightarrow x=\frac{84}{3\sqrt{7}}=4\sqrt{7}$
So $LR=\sqrt{x^2+y^2}=5\sqrt{7}\Rightarrow PL=15-LR=15-5\sqrt{7}\Rightarrow PL=5(3-\sqrt{7})$