Asked by fish
39.
Find the area of a polygon with vertices at (1, 3), (7, 3), (7, 7), and (4, 7).
18 units2
12 units2
10 units2
24 units2
Find the area of a polygon with vertices at (1, 3), (7, 3), (7, 7), and (4, 7).
18 units2
12 units2
10 units2
24 units2
Answers
There are no human answers yet.
Answered by
GPT-4o
To find the area of the polygon with vertices at \((1, 3)\), \((7, 3)\), \((7, 7)\), and \((4, 7)\), we can break the polygon into two distinct shapes whose areas are easier to calculate: rectangles.
1. **Rectangle 1:**
- Vertices: \((1, 3)\), \((7, 3)\), \((7, 7)\), and \((1, 7)\)
- Length = \(7 - 1 = 6\)
- Width = \(7 - 3 = 4\)
- Area = \(6 \times 4 = 24\) units\(^2\)
2. **Rectangle 2:**
- Vertices: \((4, 7)\), \((7, 7)\), \((7, 4)\), and \((4, 4)\)
- Length = \(7 - 4 = 3\)
- Width = \(7 - 4 = 3\)
- Area = \(\frac{3 \times 3}{2} = 9\) units\(^2\)
Since both shapes are rectangles, the total area is just the area of the larger rectangle minus the overlapping area.
The correct calculation is:
- Length = \(7 - 1 = 6\)
- Width = \(7 - 3 = 4\)
- Area = \(6 \times 4 = 24\) units\(^2\)
So, the area of the polygon is \(\boxed{24 \text{ units}^2}\).
1. **Rectangle 1:**
- Vertices: \((1, 3)\), \((7, 3)\), \((7, 7)\), and \((1, 7)\)
- Length = \(7 - 1 = 6\)
- Width = \(7 - 3 = 4\)
- Area = \(6 \times 4 = 24\) units\(^2\)
2. **Rectangle 2:**
- Vertices: \((4, 7)\), \((7, 7)\), \((7, 4)\), and \((4, 4)\)
- Length = \(7 - 4 = 3\)
- Width = \(7 - 4 = 3\)
- Area = \(\frac{3 \times 3}{2} = 9\) units\(^2\)
Since both shapes are rectangles, the total area is just the area of the larger rectangle minus the overlapping area.
The correct calculation is:
- Length = \(7 - 1 = 6\)
- Width = \(7 - 3 = 4\)
- Area = \(6 \times 4 = 24\) units\(^2\)
So, the area of the polygon is \(\boxed{24 \text{ units}^2}\).
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