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 sq. units
B) 60 sq. units
C) 120 sq. units
D) None of these

Answer 1

Hint:
Here, we have to find the area of the pentagon whose vertices are given. We will substitute the vertices of the pentagon in the formula of the area of the pentagon. We will simplify the expression using the formula of determinants. Then we will simplify it further to get the required answer.

Formula used:
We will use the given formulas:
1) Area of the pentagon whose vertices are \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right),\left( {{x_4},{y_4}} \right)\] and \[\left( {{x_5},{y_5}} \right)\] is given by the formula \[A = \dfrac{1}{2}\left[ {\left| {\begin{array}{*{20}{l}}{{x_1}}&{{y_1}}\\{{x_2}}&{{y_2}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_2}}&{{y_2}}\\{{x_3}}&{{y_3}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_3}}&{{y_3}}\\{{x_4}}&{{y_4}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_4}}&{{y_4}}\\{{x_5}}&{{y_5}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_5}}&{{y_5}}\\{{x_1}}&{{y_1}}\end{array}} \right|} \right]\]
2) Determinant is given by \[\left| {\begin{array}{*{20}{l}}{{x_1}}&{{y_1}}\\{{x_2}}&{{y_2}}\end{array}} \right| = {x_1}{y_2} - {y_1}{x_2}\]

Complete step by step solution:
We are given a pentagon with vertices \[\left( {4,1} \right),\left( {3,6} \right),\left( { - 5,1} \right),\left( { - 3, - 3} \right)\]and \[\left( { - 3,0} \right)\].
Now, we will find the area of the pentagon using the area of the pentagon formula.
Substituting \[\left( {4,1} \right),\left( {3,6} \right),\left( { - 5,1} \right),\left( { - 3, - 3} \right)\]and \[\left( { - 3,0} \right)\] in the formula \[A = \dfrac{1}{2}\left[ {\left| {\begin{array}{*{20}{l}}{{x_1}}&{{y_1}}\\{{x_2}}&{{y_2}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_2}}&{{y_2}}\\{{x_3}}&{{y_3}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_3}}&{{y_3}}\\{{x_4}}&{{y_4}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_4}}&{{y_4}}\\{{x_5}}&{{y_5}}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{{x_5}}&{{y_5}}\\{{x_1}}&{{y_1}}\end{array}} \right|} \right]\], we get
\[A = \dfrac{1}{2}\left[ {\left| {\begin{array}{*{20}{l}}4&1\\3&6\end{array}} \right| + \left| {\begin{array}{*{20}{l}}3&6\\{ - 5}&1\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{ - 5}&1\\{ - 3}&{ - 3}\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{ - 3}&{ - 3}\\{ - 3}&0\end{array}} \right| + \left| {\begin{array}{*{20}{l}}{ - 3}&0\\4&1\end{array}} \right|} \right]\]
Now using the formula of determinant \[\left| {\begin{array}{*{20}{l}}{{x_1}}&{{y_1}}\\{{x_2}}&{{y_2}}\end{array}} \right| = {x_1}{y_2} - {y_1}{x_2}\] , we get
\[ \Rightarrow A = \dfrac{1}{2}\left[ {\left( {4 \cdot 6 - 1 \cdot 3} \right) + \left( {3 \cdot 1 - 6 \cdot - 5} \right) + \left( { - 5 \cdot - 3 - - 3 \cdot 1} \right) + \left( { - 3 \cdot 0 - - 3 \cdot - 3} \right) + \left( { - 3 \cdot 1 - 0 \cdot 4} \right)} \right]\]
By simplifying the expression, we get
\[ \Rightarrow A = \dfrac{1}{2}\left[ {\left( {24 - 3} \right) + \left( {3 + 30} \right) + \left( {15 + 3} \right) + \left( {0 - 9} \right) + \left( { - 3 - 0} \right)} \right]\]
\[ \Rightarrow A = \dfrac{1}{2}\left[ {21 + 33 + 18 - 9 - 3} \right]\]
By adding the terms, we will get
\[ \Rightarrow A = \dfrac{1}{2}\left[ {60} \right]\]
Dividing 60 by 2, we get
\[ \Rightarrow A = 30\] sq. units
Therefore, the area of the pentagon is 30 sq. units.

Thus the option (A) is correct.

Note:
We know that the pentagon is a five sided polygon. A polygon is a shape where it has \[n\] sides. A polygon is a two dimensional figure. The area of the polygon is the space occupied by a two dimensional figure. Thus, the area of the polygon can be used to find the area of the polygon which has \[n\]number of sides. We need to also keep in mind that determinants can be found for square matrices only.