Question

In figure, P and Q are two lamp posts. If the area of the $\Delta PDC$ is same as that of the rectangle $ABCD$, find the distance between the two lamp posts.

• A. $2\sqrt{14}$ $cm$
• B. $\sqrt{14}$ $cm$
• C. $8\sqrt{14}$ $cm$
• D. $6\sqrt{14}$ $cm$

Answer 1

The correct option is A $2\sqrt{14}$ $cm$

In $△PDC$, we have $a=10cm,b=12cm$ and $c=6cm$
Semi perimeter, $s=\frac{a+b+c}{2}=\frac{10cm+12cm+6cm}{2}=14cm$
Area of $△PDC=\sqrt{14\times (14-10)\times (14-12)\times (14-6)}c{m}^2$
$=\sqrt{14\times 4\times 2\times 8}$

$=8\sqrt{14}{cm}^2$
Also, Area of $△PDC=\frac{1}{2}\times base\times height=\frac{1}{2}\times CD\times PM$
$\Rightarrow \frac{1}{2}\times 12\times PM=8\sqrt{14}\Rightarrow PM=\frac{4\sqrt{14}}{3}$
Again, it is given that, area of rectangle = area of triangle
$⟹AB\times QM=8\sqrt{14}⟹12\times QM=8\sqrt{14}⟹QM=\frac{2\sqrt{14}}{3}$
Therefore, distance between two lamp posts, $PQ=PM+QM=\frac{4\sqrt{14}}{3}+\frac{2\sqrt{14}}{3}$
$=\frac{6\sqrt{14}}{3}=2\sqrt{14}cm$