Question
The sides of a square are 3 cm long. One vertex of the square is at (2,0) on a square coordinate grid marked in centimeter units. Which of the following points could
also be a vertex of the square?
F. (-4, 0)
G. ( 0
, 1)
H. ( 1,-1)
J. (4, 1)
K. (5,0)
also be a vertex of the square?
F. (-4, 0)
G. ( 0
, 1)
H. ( 1,-1)
J. (4, 1)
K. (5,0)
Answers
GPT-4o mini
To determine which points could be vertices of the square, we first note that the square has sides of length 3 cm, and one of its vertices is at (2, 0).
The possible coordinates of the other vertices can be found by considering the following points:
1. The square can be oriented in various ways. We will illustrate a few:
- If the square is aligned with the axes, the possible vertices would be:
- (2, 0) (given)
- (2 + 3, 0) = (5, 0)
- (2, 0 + 3) = (2, 3)
- (2 - 3, 0) = (-1, 0)
- (2, 0 - 3) = (2, -3)
2. If the square is rotated at a 45-degree angle, we can use the fact that a square's diagonal length is \(s\sqrt{2}\), where \(s\) is the side length. The diagonal length with a side of 3 cm is \(3\sqrt{2}\) cm.
We can calculate the other vertices of the square assuming a rotated position, but for simplicity here, we will check only the provided points against the 3 cm distance:
To find valid adjacent vertices since we know one vertex (2, 0):
For each candidate point, we calculate the distance to (2, 0) and check if it equals 3 cm:
- Distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Checking each option:
1. **F. (-4, 0)**
\(d = \sqrt{(-4 - 2)^2 + (0 - 0)^2} = \sqrt{(-6)^2} = \sqrt{36} = 6\) (not valid)
2. **G. (0, 1)**
\(d = \sqrt{(0 - 2)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\) (not valid)
3. **H. (1, -1)**
\(d = \sqrt{(1 - 2)^2 + (-1 - 0)^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\) (not valid)
4. **J. (4, 1)**
\(d = \sqrt{(4 - 2)^2 + (1 - 0)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}\) (not valid)
5. **K. (5, 0)**
\(d = \sqrt{(5 - 2)^2 + (0 - 0)^2} = \sqrt{(3)^2} = \sqrt{9} = 3\) (valid)
From this analysis, **the only valid option that could also be a vertex of the square is:**
**K. (5, 0)**.
The possible coordinates of the other vertices can be found by considering the following points:
1. The square can be oriented in various ways. We will illustrate a few:
- If the square is aligned with the axes, the possible vertices would be:
- (2, 0) (given)
- (2 + 3, 0) = (5, 0)
- (2, 0 + 3) = (2, 3)
- (2 - 3, 0) = (-1, 0)
- (2, 0 - 3) = (2, -3)
2. If the square is rotated at a 45-degree angle, we can use the fact that a square's diagonal length is \(s\sqrt{2}\), where \(s\) is the side length. The diagonal length with a side of 3 cm is \(3\sqrt{2}\) cm.
We can calculate the other vertices of the square assuming a rotated position, but for simplicity here, we will check only the provided points against the 3 cm distance:
To find valid adjacent vertices since we know one vertex (2, 0):
For each candidate point, we calculate the distance to (2, 0) and check if it equals 3 cm:
- Distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Checking each option:
1. **F. (-4, 0)**
\(d = \sqrt{(-4 - 2)^2 + (0 - 0)^2} = \sqrt{(-6)^2} = \sqrt{36} = 6\) (not valid)
2. **G. (0, 1)**
\(d = \sqrt{(0 - 2)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\) (not valid)
3. **H. (1, -1)**
\(d = \sqrt{(1 - 2)^2 + (-1 - 0)^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\) (not valid)
4. **J. (4, 1)**
\(d = \sqrt{(4 - 2)^2 + (1 - 0)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}\) (not valid)
5. **K. (5, 0)**
\(d = \sqrt{(5 - 2)^2 + (0 - 0)^2} = \sqrt{(3)^2} = \sqrt{9} = 3\) (valid)
From this analysis, **the only valid option that could also be a vertex of the square is:**
**K. (5, 0)**.