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Sagot :
To determine the type of the given quadrilateral with vertices [tex]\((2,4)\)[/tex], [tex]\((-4,-2)\)[/tex], [tex]\((-2,4)\)[/tex], and [tex]\((4,-2)\)[/tex], we need to check the lengths of sides and diagonals. Here's a step-by-step method:
1. Calculate the lengths of all sides:
The distance between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let's label the points for clarity:
- [tex]\(A = (2, 4)\)[/tex]
- [tex]\(B = (-4, -2)\)[/tex]
- [tex]\(C = (-2, 4)\)[/tex]
- [tex]\(D = (4, -2)\)[/tex]
Calculate [tex]\(AB\)[/tex], [tex]\(BC\)[/tex], [tex]\(CD\)[/tex], and [tex]\(DA\)[/tex]:
[tex]\[ AB = \sqrt{(2 - (-4))^2 + (4 - (-2))^2} = \sqrt{(2 + 4)^2 + (4 + 2)^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
[tex]\[ BC = \sqrt{(-4 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4 + 2)^2 + (-2 - 4)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
[tex]\[ CD = \sqrt{(-2 - 4)^2 + (4 - (-2))^2} = \sqrt{(-2 - 4)^2 + (4 + 2)^2} = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
[tex]\[ DA = \sqrt{(4 - 2)^2 + (-2 - 4)^2} = \sqrt{(4 - 2)^2 + (-2 - 4)^2} = \sqrt{(2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
So, we have the side lengths [tex]\(AB = 6\sqrt{2}\)[/tex], [tex]\(BC = 2\sqrt{10}\)[/tex], [tex]\(CD = 6\sqrt{2}\)[/tex], and [tex]\(DA = 2\sqrt{10}\)[/tex].
2. Calculate the lengths of the diagonals:
The diagonals are [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex].
[tex]\[ AC = \sqrt{(2 - (-2))^2 + (4 - 4)^2} = \sqrt{(2 + 2)^2 + 0^2} = \sqrt{4^2} = \sqrt{16} = 4 \][/tex]
[tex]\[ BD = \sqrt{(-4 - 4)^2 + (-2 - (-2))^2} = \sqrt{(-4 - 4)^2 + 0^2} = \sqrt{(-8)^2} = \sqrt{64} = 8 \][/tex]
3. Determine the type of quadrilateral:
- For a parallelogram, both pairs of opposite sides must be equal: We have [tex]\(AB = CD\)[/tex] and [tex]\(BC = DA\)[/tex], so it's a parallelogram.
- For a rectangle, the diagonals must be equal: Here, [tex]\(AC \neq BD\)[/tex], so it's not a rectangle.
- For a square, all sides and diagonals must be equal: This is not the case here, so it's not a square.
- For a trapezoid, only one pair of opposite sides need to be parallel.
Since the figure satisfies the conditions for a parallelogram but not for a rectangle or a square, the answer is:
D. parallelogram
1. Calculate the lengths of all sides:
The distance between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let's label the points for clarity:
- [tex]\(A = (2, 4)\)[/tex]
- [tex]\(B = (-4, -2)\)[/tex]
- [tex]\(C = (-2, 4)\)[/tex]
- [tex]\(D = (4, -2)\)[/tex]
Calculate [tex]\(AB\)[/tex], [tex]\(BC\)[/tex], [tex]\(CD\)[/tex], and [tex]\(DA\)[/tex]:
[tex]\[ AB = \sqrt{(2 - (-4))^2 + (4 - (-2))^2} = \sqrt{(2 + 4)^2 + (4 + 2)^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
[tex]\[ BC = \sqrt{(-4 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4 + 2)^2 + (-2 - 4)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
[tex]\[ CD = \sqrt{(-2 - 4)^2 + (4 - (-2))^2} = \sqrt{(-2 - 4)^2 + (4 + 2)^2} = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
[tex]\[ DA = \sqrt{(4 - 2)^2 + (-2 - 4)^2} = \sqrt{(4 - 2)^2 + (-2 - 4)^2} = \sqrt{(2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
So, we have the side lengths [tex]\(AB = 6\sqrt{2}\)[/tex], [tex]\(BC = 2\sqrt{10}\)[/tex], [tex]\(CD = 6\sqrt{2}\)[/tex], and [tex]\(DA = 2\sqrt{10}\)[/tex].
2. Calculate the lengths of the diagonals:
The diagonals are [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex].
[tex]\[ AC = \sqrt{(2 - (-2))^2 + (4 - 4)^2} = \sqrt{(2 + 2)^2 + 0^2} = \sqrt{4^2} = \sqrt{16} = 4 \][/tex]
[tex]\[ BD = \sqrt{(-4 - 4)^2 + (-2 - (-2))^2} = \sqrt{(-4 - 4)^2 + 0^2} = \sqrt{(-8)^2} = \sqrt{64} = 8 \][/tex]
3. Determine the type of quadrilateral:
- For a parallelogram, both pairs of opposite sides must be equal: We have [tex]\(AB = CD\)[/tex] and [tex]\(BC = DA\)[/tex], so it's a parallelogram.
- For a rectangle, the diagonals must be equal: Here, [tex]\(AC \neq BD\)[/tex], so it's not a rectangle.
- For a square, all sides and diagonals must be equal: This is not the case here, so it's not a square.
- For a trapezoid, only one pair of opposite sides need to be parallel.
Since the figure satisfies the conditions for a parallelogram but not for a rectangle or a square, the answer is:
D. parallelogram
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