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 $ \Rightarrow \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ 

Area of the triangle formula while three sides given   $ \Rightarrow \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 answer:
The vertices of the given polygon is $A\left( {2,5} \right),B\left( {7,1} \right),C\left( {3, - 4} \right),D\left( { - 2,3} \right)$

Following is the diagram of polygon.

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: While calculating the area of the triangle, remember that s is half of the perimeter and not the perimeter. so do not forget to divide the perimeter by 2 .

The area cannot be negative so while taking the square root, do not consider negative values.