Question

Find the area of a polygon with the given vertices?$\left( {2,5} \right),\left( {7,1} \right),\left( {3, - 4} \right),\left( { - 2,3} \right)$

Answer 1

Hint: Here we find the area of the polygon with given vertices. First we find the distance of the vertices using distance formula and using the formula for finding the area of a triangle with given three vertices. Here we have a polygon so two areas of the triangle occur so add them to find the area of the polygon.

Formula used: Distance formula $ \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Area of the triangle formula while three sides given $\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $
Where $a,b$ and $c$ are lengths of the sides and $s = \dfrac{{a + b + c}}{2}$ (half the perimeter)

Complete step-by-step solution:
The vertices of the given polygon is $A\left( {2,5} \right),B\left( {7,1} \right),C\left( {3, - 4} \right),D\left( { - 2,3} \right)$
Now we find the distance of the sides by using distance formula$ \Rightarrow \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Here we have putting the value one by one and we get,
$AB = \sqrt {{{\left( {7 - 2} \right)}^2} + {{\left( {1 - 5} \right)}^2}} = \sqrt {{5^2} + {{( - 4)}^2}} = \sqrt {25 + 16} = \sqrt {41} \approx 6.4$
Then,
$BC = \sqrt {{{\left( {3 - 7} \right)}^2} + {{\left( { - 4 - 1} \right)}^2}} = \sqrt {{{( - 4)}^2} + {{( - 5)}^2}} = \sqrt {16 + 25} = \sqrt {41} \approx 6.4$
Then,
$CD = \sqrt {{{\left( { - 2 - 3} \right)}^2} + {{\left( {3 + 4} \right)}^2}} = \sqrt {{{( - 5)}^2} + {{(7)}^2}} = \sqrt {25 + 49} = \sqrt {74} \approx 8.6$
Then,
$AD = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {3 - 5} \right)}^2}} = \sqrt {{{( - 4)}^2} + {{( - 2)}^2}} = \sqrt {16 + 4} = \sqrt {20} \approx 4.47$
Then,
$AC = \sqrt {{{\left( {3 - 2} \right)}^2} + {{\left( { - 4 - 5} \right)}^2}} = \sqrt {{1^2} + {{( - 9)}^2}} = \sqrt {1 + 81} = \sqrt {82} \approx 9.06$
Now find the area of two triangle we find the first triangle $\Delta ABC = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $where $s = \dfrac{{a + b + c}}{2}$here $a = AB = 6.4,\,\,b = BC = 6.4,\,\,c = AC = 9.06$
Substitute into the formula and add them and divided by $2$ we get,
$ \Rightarrow s = \dfrac{{6.4 + 6.4 + 9.06}}{2} = 10.93$
$\Delta ABC = \sqrt {10.93\left( {10.93 - 6.4} \right)\left( {10.93 - 6.4} \right)\left( {10.93 - 9.06} \right)} $
Simplifying and we get approximate value,
$ \approx 20.48$
Now for find the area of the second triangle$\Delta ADC = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ we get,
Now $s = \dfrac{{a + d + c}}{2}$ here $a = AD = 4.47,\,\,b = CD = 8.6,\,\,c = AC = 9.06$ substitute into the formula and
Add them and divided by $2$ we get,
\[s = \dfrac{{4.47 + 8.6 + 9.06}}{2} = 11.065\]
$\Delta ADC = \sqrt {11.065\left( {11.065 - 4.47} \right)\left( {11.065 - 8.6} \right)\left( {11.065 - 9.06} \right)} $
Simplifying and we get approximate value,
$ \approx 19$
Now the area of the polygon $ABCD = $ Area of $\Delta ABC + $Area of $\Delta ADC$

Therefore we get, Area of $ABCD = 20.48 + 19 = 39.48$square units.

Note: We can also use the formula for finding the area of the polygon here our polygon is quadrilateral so we use the formula $\dfrac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_1} + {x_4}{y_1}} \right) - \left( {{x_2}{y_1} + {x_3}{y_2} + {x_1}{y_3} + {x_1}{y_4}} \right)} \right\}$
Here the vertices are$\left( {2,5} \right),\left( {7,1} \right),\left( {3, - 4} \right),\left( { - 2,3} \right)$
$ \Rightarrow \dfrac{1}{2}\left\{ {\left( {2(1) + 7( - 4) + 3(5) + ( - 2)(5)} \right) - \left( {7(5) + 3(1) + 2( - 4) + 2(3)} \right)} \right\}$
On simplify the term and we get,
$ \Rightarrow \dfrac{1}{2}\left\{ {\left( {2 - 28 + 15 - 10} \right) - \left( {35 + 3 - 8 + 6} \right)} \right\}$
On adding the term and we get,
$ \Rightarrow \dfrac{1}{2}\left\{ { - 21 - 26} \right\}$
On adding the term and we get
$ \Rightarrow \dfrac{1}{2}\left\{ { - 47} \right\}$
Let us divide the term and we get,
$ \Rightarrow - 23.5$
Here area is not in negative sign so we take$\left| { - 23.5} \right| = 23.5$. This solution varies from our main solution because our main solution is finding approximately. This formula is used to find the exact solution.