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Sagot :
To solve the problem and determine the lengths of the sides and the type of triangle [tex]\(\triangle ABC\)[/tex] with given vertices [tex]\(A(-2, 5)\)[/tex], [tex]\(B(-4, -2)\)[/tex], and [tex]\(C(3, -4)\)[/tex], we'll proceed as follows:
1. Calculate the length of [tex]\(AB\)[/tex]:
Using the distance formula:
[tex]\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ AB = \sqrt{(-4 - (-2))^2 + (-2 - 5)^2} = \sqrt{(-4 + 2)^2 + (-2 - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.28 \][/tex]
2. Calculate the length of [tex]\(AC\)[/tex]:
Using the distance formula:
[tex]\[ AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ AC = \sqrt{(3 - (-2))^2 + (-4 - 5)^2} = \sqrt{(3 + 2)^2 + (-4 - 5)^2} = \sqrt{5^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.3 \][/tex]
3. Calculate the length of [tex]\(BC\)[/tex]:
Using the distance formula:
[tex]\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(B\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ BC = \sqrt{(3 - (-4))^2 + (-4 - (-2))^2} = \sqrt{(3 + 4)^2 + (-4 + 2)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.28 \][/tex]
4. Determine the type of triangle:
Based on the lengths of the sides:
- [tex]\(AB \approx 7.28\)[/tex]
- [tex]\(AC \approx 10.3\)[/tex]
- [tex]\(BC \approx 7.28\)[/tex]
Since two sides of the triangle ([tex]\(AB\)[/tex] and [tex]\(BC\)[/tex]) are equal:
- [tex]\(AB = BC \approx 7.28\)[/tex]
- [tex]\(AC \approx 10.3\)[/tex]
The triangle [tex]\(\triangle ABC\)[/tex] is an isosceles triangle, as it has two sides of equal length.
To summarize:
- The length of [tex]\(AB\)[/tex] is 7.28.
- The length of [tex]\(AC\)[/tex] is 10.3.
- The length of [tex]\(BC\)[/tex] is 7.28.
- Therefore, the triangle is isosceles.
Hence, the completed sentences are:
The length of [tex]\(AB\)[/tex] is [tex]\( \boxed{7.28} \)[/tex]. The length of [tex]\(AC\)[/tex] is [tex]\( \boxed{10.3} \)[/tex]. The length of [tex]\(BC\)[/tex] is [tex]\( \boxed{7.28} \)[/tex]. Therefore, the triangle is [tex]\( \boxed{isosceles} \)[/tex].
1. Calculate the length of [tex]\(AB\)[/tex]:
Using the distance formula:
[tex]\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ AB = \sqrt{(-4 - (-2))^2 + (-2 - 5)^2} = \sqrt{(-4 + 2)^2 + (-2 - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.28 \][/tex]
2. Calculate the length of [tex]\(AC\)[/tex]:
Using the distance formula:
[tex]\[ AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ AC = \sqrt{(3 - (-2))^2 + (-4 - 5)^2} = \sqrt{(3 + 2)^2 + (-4 - 5)^2} = \sqrt{5^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.3 \][/tex]
3. Calculate the length of [tex]\(BC\)[/tex]:
Using the distance formula:
[tex]\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates of [tex]\(B\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ BC = \sqrt{(3 - (-4))^2 + (-4 - (-2))^2} = \sqrt{(3 + 4)^2 + (-4 + 2)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.28 \][/tex]
4. Determine the type of triangle:
Based on the lengths of the sides:
- [tex]\(AB \approx 7.28\)[/tex]
- [tex]\(AC \approx 10.3\)[/tex]
- [tex]\(BC \approx 7.28\)[/tex]
Since two sides of the triangle ([tex]\(AB\)[/tex] and [tex]\(BC\)[/tex]) are equal:
- [tex]\(AB = BC \approx 7.28\)[/tex]
- [tex]\(AC \approx 10.3\)[/tex]
The triangle [tex]\(\triangle ABC\)[/tex] is an isosceles triangle, as it has two sides of equal length.
To summarize:
- The length of [tex]\(AB\)[/tex] is 7.28.
- The length of [tex]\(AC\)[/tex] is 10.3.
- The length of [tex]\(BC\)[/tex] is 7.28.
- Therefore, the triangle is isosceles.
Hence, the completed sentences are:
The length of [tex]\(AB\)[/tex] is [tex]\( \boxed{7.28} \)[/tex]. The length of [tex]\(AC\)[/tex] is [tex]\( \boxed{10.3} \)[/tex]. The length of [tex]\(BC\)[/tex] is [tex]\( \boxed{7.28} \)[/tex]. Therefore, the triangle is [tex]\( \boxed{isosceles} \)[/tex].
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