Question

What is the area of this figure? Enter your answer in the box.

Answer 1

Area
=3*2/2+6*1/2+3*6/2+(6+7)*6/2
=54

Answer 2

Answer:

Area of Figure = 54 $unit^2$

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 = $\frac{1}{2}\times height\times (sum of parallel sides)$

From Figure, Height , BC = 6 unit and Parallel side, AB = 7 unit , CG = 9 units

⇒ Area of Trapezium = $\frac{1}{2}\times BC\times(AB+CG)$

                                    = $\frac{1}{2}\times 6\times(7+9)$

                                    = $3\times(16)$

                                    = 48 $unit^2$

Now for area of triangle we use Area of triangle using coordinates. Given by,

$Area\: of\: Triangle\: =\:\frac{1}{2}\times\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

⇒ Area of ΔCDE = $\frac{1}{2}\times\left|4(4-3)+3(3-3)+(-2)(3-4)\right|$

                            = $\frac{1}{2}\times\left|4\times1+3\times0+(-2)\times(-1)\right|$

                            = $\frac{1}{2}\times\left|4+0+2\right|$

                            = $\frac{1}{2}\times6$

                            = 3 $unit^2$

Similarly,

⇒ Area of ΔEFG = $\frac{1}{2}\times\left|(-5)(3-5)+(-2)(5-3)+(-4)(3-3)\right|$

                            = $\frac{1}{2}\times\left|(-5)\times(-2)+(-2)\times2+(-4)\times0\right|$

                            = $\frac{1}{2}\times\left|10-4+0\right|$

                            = $\frac{1}{2}\times6$

                            = 3 $unit^2$

∴ Area of Figure = area of trapezium ABCG + area of ΔCDE + area of ΔEFG

                           = 48 + 3 + 3

                           = 54 $unit^2$

Therefore, Area of Figure = 54 $unit^2$