Answer:
Area of Figure = 54 [tex]unit^2[/tex]
Step-by-step explanation:
Given Graph of a polygon with 7 side.
To find: Area of figure
First we write coordinates of all vertices from left bottom going in clock wise direction.
Say, Vertex A ( -3, -3 ) , Vertex B ( 4, -3 ) , Vertex C ( 4, 3 ) , Vertex D ( 3, 4 ) , Vertex E ( -2, 3 ) , Vertex F ( -4, 5 ) , Vertex G ( -5, 3 )
Now we divide the given figure in a trapezium ABCG and 2 triangle ΔCDE , ΔEFG
Area of Trapezium = [tex]\frac{1}{2}\times height\times (sum of parallel sides)[/tex]
From Figure, Height , BC = 6 unit and Parallel side, AB = 7 unit , CG = 9 units
⇒ Area of Trapezium = [tex]\frac{1}{2}\times BC\times(AB+CG)[/tex]
= [tex]\frac{1}{2}\times 6\times(7+9)[/tex]
= [tex]3\times(16)[/tex]
= 48 [tex]unit^2[/tex]
Now for area of triangle we use Area of triangle using coordinates. Given by,
[tex]Area\: of\: Triangle\: =\:\frac{1}{2}\times\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|[/tex]
⇒ Area of ΔCDE = [tex]\frac{1}{2}\times\left|4(4-3)+3(3-3)+(-2)(3-4)\right|[/tex]
= [tex]\frac{1}{2}\times\left|4\times1+3\times0+(-2)\times(-1)\right|[/tex]
= [tex]\frac{1}{2}\times\left|4+0+2\right|[/tex]
= [tex]\frac{1}{2}\times6[/tex]
= 3 [tex]unit^2[/tex]
Similarly,
⇒ Area of ΔEFG = [tex]\frac{1}{2}\times\left|(-5)(3-5)+(-2)(5-3)+(-4)(3-3)\right|[/tex]
= [tex]\frac{1}{2}\times\left|(-5)\times(-2)+(-2)\times2+(-4)\times0\right|[/tex]
= [tex]\frac{1}{2}\times\left|10-4+0\right|[/tex]
= [tex]\frac{1}{2}\times6[/tex]
= 3 [tex]unit^2[/tex]
∴ Area of Figure = area of trapezium ABCG + area of ΔCDE + area of ΔEFG
= 48 + 3 + 3
= 54 [tex]unit^2[/tex]
Therefore, Area of Figure = 54 [tex]unit^2[/tex]
