Question

If the points $A\left( {0,0} \right),B\left( {12,0} \right),C\left( {12,2} \right),D\left( {6,7} \right)$ and $E\left( {0,5} \right)$ are the vertices of the pentagon $ABCDE$, then it’s area (in square units) is
1) $58$
2) $60$
3) $61$
4) $62$
5) $63$

Answer 1

Hint: In the question, we are given the vertices of a pentagon and are required to find the area of the pentagon using the same. So, to find the area of the pentagon we will use the formula to find the area of a pentagon if it’s vertices are given, which is,
Area of pentagon, $ = \dfrac{1}{2}\left[ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_5} + {x_5}{y_1}} \right) - \left( {{y_1}{x_2} + {y_2}{x_3} + {y_3}{x_4} + {y_4}{x_5} + {y_5}{x_1}} \right)} \right]$
Using this formula for the area of the pentagon, we will get the required answer.

Complete step-by-step solution:
Given, the vertices of the pentagon are,
$A\left( {0,0} \right),B\left( {12,0} \right),C\left( {12,2} \right),D\left( {6,7} \right),E\left( {0,5} \right)$
We are to find the area of the pentagon $ABCDE$.
The formula to find the area of pentagon if vertices are given is,
Area of pentagon $ = \dfrac{1}{2}\left[ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_5} + {x_5}{y_1}} \right) - \left( {{y_1}{x_2} + {y_2}{x_3} + {y_3}{x_4} + {y_4}{x_5} + {y_5}{x_1}} \right)} \right]$
So, we can write the given points as,
$A\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)$
$B\left( {{x_2},{y_2}} \right) = \left( {12,0} \right)$
$C\left( {{x_3},{y_3}} \right) = \left( {12,2} \right)$
$D\left( {{x_4},{y_4}} \right) = \left( {6,7} \right)$
$E\left( {{x_5},{y_5}} \right) = \left( {0,5} \right)$
Now, substituting the coordinates of all the points in the formula to find the area, we get,
Area of pentagon $ = \dfrac{1}{2}\left[ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_5} + {x_5}{y_1}} \right) - \left( {{y_1}{x_2} + {y_2}{x_3} + {y_3}{x_4} + {y_4}{x_5} + {y_5}{x_1}} \right)} \right]$
$ = \dfrac{1}{2}\left[ {\left\{ {\left( 0 \right)\left( 0 \right) + \left( {12} \right)\left( 2 \right) + \left( {12} \right)\left( 7 \right) + \left( 6 \right)\left( 5 \right) + \left( 0 \right)\left( 0 \right)} \right\} - \left\{ {\left( 0 \right)\left( {12} \right) + \left( 0 \right)\left( {12} \right) + \left( 2 \right)\left( 6 \right) + \left( 7 \right)\left( 0 \right) + \left( 5 \right)\left( 0 \right)} \right\}} \right]$Now, opening the brackets, we get,
$ = \dfrac{1}{2}\left[ {\left\{ {0 + 24 + 84 + 30 + 0} \right\} - \left\{ {0 + 0 + 12 + 0 + 0} \right\}} \right]$
Now, simplifying the terms, we get,
$ = \dfrac{1}{2}\left[ {\left\{ {138} \right\} - \left\{ {12} \right\}} \right]$
Again, opening the brackets, we get,
$ = \dfrac{1}{2}\left[ {138 - 12} \right]$
Carrying out the calculations, we get,
$ = \dfrac{1}{2}\left( {126} \right)$
$ = 63$
Therefore, the area of the pentagon is $63$ sq. units.
Hence, the correct option is 5.

Note: The above problem can also be solved by finding the measure of the sides of the pentagon. If the sides of pentagon are equal, then we can find it’s area by using the formula,
Area of pentagon$ = \dfrac{1}{4}\sqrt {5\left( {5 + 2\sqrt 5 } \right)} {a^2}$
But, if the length of the sides of the pentagon are different, then using geometry we can divide the pentagon into three triangles and then find the length of the sides of those three triangles and use Heron’s formula to find the area of each triangle and then add their areas to get the area of the pentagon.