To find the area of this structure, we will find the area of the rectangle and parallelogram independently and then combine the values for the total area. The area of a rectangle is simply the length times the width.
Ar = l x w
The area of a parallelogram is found by multiplying the base time the height.
Ap = b x h
We can find the area of the parallelogram without much work. We will one of the vertical parallel lines as the base of the parallelogram. The base runs from points (3, -3) to (3, 3). The x-value does not change, therefore the length of the base is 6 units in the y-direction. The height of the parallelogram will be the length of the perpendicular line between the two sides. There is only 1 unit in the x-direction between the two sides. Now we solve for the area:
Ap = (6 units)(1 unit) = 6 units²
The area of the rectangle requires a bit more work. We must find the length of both sides of the rectangle. We can do this by drawing right triangles between the two points of a side where the sides of the rectangles are the hypotenuse. We will know the two sides of the right triangle based on the grid of the graph, and we can use pythagorean theorem to solve for the sides. The width of the rectangle can be found from points (-6, -1) and (-5, -5). The change in x is 1 unit and the change in y is 4 units.
w² = 1² + 4²
w² = 17
w = √17 units
The other side can be found using points (-5, -5) and (3, -3). The change in x is 8 units and the change in y is 2 units.We can now find the length of the rectangles.
l² = 8² + 2²
l² = 68
l = √68
l = √4*17
l = 2√17
Now we can find the area of the rectangle since we know both the length and the width.
A = 2√17 · √17
A = 2·17
A = 34 units²
And finally, since we have the area of the rectangle and the area of the parallelogram, we simply add the two values together to get the total area of the structure.
A = 34 + 6
A = 40 units²