QUESTION IMAGE
Question
what is the area of parallelogram abcd?
13 square units
14 square units
15 square units
16 square units
Identify the coordinates of the vertices
Using the grid, we find the coordinates of the vertices of parallelogram \(ABCD\):
- \(A = (3, 6)\)
- \(B = (6, 5)\)
- \(C = (5, 1)\)
- \(D = (2, 2)\)
Calculate the area using bounding box subtraction
Using the Composite Area knowledge point:
We enclose the parallelogram in a bounding box with horizontal and vertical sides.
- The minimum \(x\)-coordinate is \(2\) (at \(D\)) and the maximum \(x\)-coordinate is \(6\) (at \(B\)). Thus, the width is \(6 - 2 = 4\).
- The minimum \(y\)-coordinate is \(1\) (at \(C\)) and the maximum \(y\)-coordinate is \(6\) (at \(A\)). Thus, the height is \(6 - 1 = 5\).
- The area of this bounding box is:
Now, we subtract the areas of the four right triangles in the corners of the bounding box:
- Top-left triangle (above \(D\) and left of \(A\)):
- Width = \(3 - 2 = 1\)
- Height = \(6 - 2 = 4\)
- \(\text{Area}_1 = \frac{1}{2} \times 1 \times 4 = 2\)
- Top-right triangle (above \(B\) and right of \(A\)):
- Width = \(6 - 3 = 3\)
- Height = \(6 - 5 = 1\)
- \(\text{Area}_2 = \frac{1}{2} \times 3 \times 1 = 1.5\)
- Bottom-right triangle (below \(B\) and right of \(C\)):
- Width = \(6 - 5 = 1\)
- Height = \(5 - 1 = 4\)
- \(\text{Area}_3 = \frac{1}{2} \times 1 \times 4 = 2\)
- Bottom-left triangle (below \(D\) and left of \(C\)):
- Width = \(5 - 2 = 3\)
- Height = \(2 - 1 = 1\)
- \(\text{Area}_4 = \frac{1}{2} \times 3 \times 1 = 1.5\)
Subtracting these corner areas from the bounding box area:
Verify using the Shoelace Formula
Using the Parallelogram Area in Coordinate Plane knowledge point:
We apply the Shoelace Formula to the vertices listed in counterclockwise order: \(A(3,6)\), \(D(2,2)\), \(C(5,1)\), \(B(6,5)\).
Both methods yield an area of \(13\) square units.
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- 13 square units (Correct answer)
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