Question
The area of the pentagon whose vertices are $$\left( {4,\,1} \right),\,\left( {3,\,6} \right),\,\left( { - 5,\,1} \right),\,\left( { - 3,\, - 3} \right)$$ and $$\left( { - 3,\,0} \right)$$ is :
A.
$$30\,{\text{uni}}{{\text{t}}^2}$$
B.
$$60\,{\text{uni}}{{\text{t}}^2}$$
C.
$$120\,{\text{uni}}{{\text{t}}^2}$$
D.
none of these
Answer :
$$30\,{\text{uni}}{{\text{t}}^2}$$
Solution :
As the points are in order, the area
\[\begin{array}{l}
= \left| {\frac{1}{2}\left\{ {\left| \begin{array}{l}
4\,\,\,1\\
3\,\,\,6
\end{array} \right| + \left| \begin{array}{l}
\,\,\,\,3\,\,\,\,6\\
- 5\,\,\,1
\end{array} \right| + \left| \begin{array}{l}
- 5\,\,\,\,\,\,\,\,1\\
- 3\,\, - 3
\end{array} \right| + \left| \begin{array}{l}
- 3\,\,\, - 3\\
- 3\,\,\,\,\,\,\,\,\,0
\end{array} \right| + \left| \begin{array}{l}
- 3\,\,\,\,0\\
\,\,\,\,4\,\,\,\,\,1
\end{array} \right|} \right\}} \right|\\
= 30\,{\rm{uni}}{{\rm{t}}^2}
\end{array}\]