A triangle has a perimeter of exactly 24 units. Which of the following could be the vertices of the triangle?
> (-1,3),(2,-1),(-1,-1)
(6,0),(6,7),(0,7)
(-1,-1),(-6,-13),(-9,-9)
(-3,-4),(3,-4),(3,4)
Your choice has sides 3,4,5 whose sum is 12
If you can find vertices making sides of 6,8,10 then that will work. Luckily it is a right triangle, with a horizontal side and a vertical side. It's easiest if you plot the points on graph paper.
To determine which of the given sets of vertices could form a triangle with a perimeter of exactly 24 units, we need to calculate the distances between the vertices.
1. Calculate the distances for the first set of vertices: (-1,3), (2,-1), and (-1,-1):
- Distance between (-1,3) and (2,-1):
d1 = sqrt((2 - (-1))^2 + (-1 - 3)^2) = sqrt(3^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5
- Distance between (2,-1) and (-1,-1):
d2 = sqrt((-1 - 2)^2 + (-1 - (-1))^2) = sqrt((-3)^2 + 0^2) = sqrt(9 + 0) = sqrt(9) = 3
- Distance between (-1,-1) and (-1,3):
d3 = sqrt((-1 - (-1))^2 + (3 - (-1))^2) = sqrt(0^2 + 4^2) = sqrt(0 + 16) = sqrt(16) = 4
The perimeter of this triangle is: p = d1 + d2 + d3 = 5 + 3 + 4 = 12
Since the perimeter is not exactly 24, the first set of vertices (-1,3), (2,-1), and (-1,-1) is not a valid triangle.
2. Calculate the distances for the second set of vertices: (6,0), (6,7), and (0,7):
- Distance between (6,0) and (6,7):
d1 = sqrt((6 - 6)^2 + (7 - 0)^2) = sqrt(0^2 + 7^2) = sqrt(0 + 49) = sqrt(49) = 7
- Distance between (6,7) and (0,7):
d2 = sqrt((0 - 6)^2 + (7 - 7)^2) = sqrt((-6)^2 + 0^2) = sqrt(36 + 0) = sqrt(36) = 6
- Distance between (0,7) and (6,0):
d3 = sqrt((6 - 0)^2 + (0 - 7)^2) = sqrt(6^2 + (-7)^2) = sqrt(36 + 49) = sqrt(85)
The perimeter of this triangle is: p = d1 + d2 + d3 = 7 + 6 + sqrt(85)
It is not clear what the value of sqrt(85) is, but the sum of the known distances is 13 + sqrt(85). Therefore, it is not possible for the second set of vertices (6,0), (6,7), and (0,7) to form a triangle with a perimeter of exactly 24.
3. Calculate the distances for the third set of vertices: (-1,-1), (-6,-13), and (-9,-9):
- Distance between (-1,-1) and (-6,-13):
d1 = sqrt((-6 - (-1))^2 + (-13 - (-1))^2) = sqrt((-5)^2 + (-12)^2) = sqrt(25 + 144) = sqrt(169) = 13
- Distance between (-6,-13) and (-9,-9):
d2 = sqrt((-9 - (-6))^2 + (-9 - (-13))^2) = sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5
- Distance between (-9,-9) and (-1,-1):
d3 = sqrt((-1 - (-9))^2 + (-1 - (-9))^2) = sqrt(8^2 + 8^2) = sqrt(64 + 64) = sqrt(128)
The perimeter of this triangle is: p = d1 + d2 + d3 = 13 + 5 + sqrt(128)
It is not clear what the value of sqrt(128) is, but the sum of the known distances is 18 + sqrt(128). Therefore, it is not possible for the third set of vertices (-1,-1), (-6,-13), and (-9,-9) to form a triangle with a perimeter of exactly 24.
4. Calculate the distances for the fourth set of vertices: (-3,-4), (3,-4), and (3,4):
- Distance between (-3,-4) and (3,-4):
d1 = sqrt((3 - (-3))^2 + (-4 - (-4))^2) = sqrt(6^2 + 0^2) = sqrt(36 + 0) = sqrt(36) = 6
- Distance between (3,-4) and (3,4):
d2 = sqrt((3 - 3)^2 + (4 - (-4))^2) = sqrt(0^2 + 8^2) = sqrt(0 + 64) = sqrt(64) = 8
- Distance between (3,4) and (-3,-4):
d3 = sqrt((-3 - 3)^2 + (-4 - 4)^2) = sqrt((-6)^2 + (-8)^2) = sqrt(36 + 64) = sqrt(100) = 10
The perimeter of this triangle is: p = d1 + d2 + d3 = 6 + 8 + 10 = 24
Therefore, the fourth set of vertices (-3,-4), (3,-4), and (3,4) could form a triangle with a perimeter of exactly 24 units.
To determine which of the given sets of points could be the vertices of a triangle with a perimeter of 24 units, we need to calculate the distance between each pair of points and check if the sum of the distances of any three points equals 24.
Let's go through each set of points one by one:
1. (-1,3), (2,-1), (-1,-1)
To find the distances between the points, we can use the distance formula:
Distance between points (x1, y1) and (x2, y2) = √((x2 - x1)^2 + (y2 - y1)^2)
Calculating the distances:
- Distance between (-1,3) and (2,-1) = √((2 - (-1))^2 + (-1 - 3)^2) = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5
- Distance between (-1,3) and (-1,-1) = √((-1 - (-1))^2 + (-1 - 3)^2) = √(0^2 + (-4)^2) = √(0 + 16) = √16 = 4
- Distance between (2,-1) and (-1,-1) = √((-1 - 2)^2 + (-1 - (-1))^2) = √((-3)^2 + 0^2) = √(9 + 0) = √9 = 3
The sum of these distances is 5 + 4 + 3 = 12, not equal to 24.
Therefore, the set of points (-1,3), (2,-1), (-1,-1) CANNOT be the vertices of the triangle.
2. (6,0), (6,7), (0,7)
Again, calculating the distances:
- Distance between (6,0) and (6,7) = 7 units (vertically aligned points)
- Distance between (6,0) and (0,7) = √((0 - 6)^2 + (7 - 0)^2) = √((-6)^2 + 7^2) = √(36 + 49) = √85
- Distance between (6,7) and (0,7) = 6 units (horizontally aligned points)
The sum of these distances is 7 + √85 + 6 = approximately 19.33 + √85.
Therefore, the set of points (6,0), (6,7), (0,7) CANNOT be the vertices of the triangle.
3. (-1,-1), (-6,-13), (-9,-9)
Calculating the distances:
- Distance between (-1,-1) and (-6,-13) = √((-6 - (-1))^2 + (-13 - (-1))^2) = √((-5)^2 + (-12)^2) = √((25 + 144) = √169 = 13
- Distance between (-1,-1) and (-9,-9) = √((-9 - (-1))^2 + (-9 - (-1))^2) = √((-9 + 1)^2 + (-9 + 1)^2) = √((-8)^2 + (-8)^2) = √((64 + 64) = √128 = approximately 11.31
- Distance between (-6,-13) and (-9,-9) = √((-9 - (-6))^2 + (-9 - (-13))^2) = √((-9 + 6)^2 + (-9 + 13)^2) = √((-3)^2 + 4^2) = √((9 + 16) = √25 = 5
The sum of these distances is 13 + approximately 11.31 + 5 = approximately 29.31.
Therefore, the set of points (-1,-1), (-6,-13), (-9,-9) CANNOT be the vertices of the triangle.
4. (-3,-4), (3,-4), (3,4)
Calculating the distances:
- Distance between (-3,-4) and (3,-4) = 6 units (horizontally aligned points)
- Distance between (-3,-4) and (3,4) = √((3 - (-3))^2 + (4 - (-4))^2) = √((6)^2 + (8)^2) = √((36 + 64) = √100 = 10
- Distance between (3,-4) and (3,4) = 8 units (vertically aligned points)
The sum of these distances is 6 + 10 + 8 = 24 units.
Therefore, the set of points (-3,-4), (3,-4), (3,4) CAN be the vertices of the triangle with a perimeter of 24 units.
In conclusion, the only set of points that could be the vertices of the triangle with a perimeter of 24 units is (-3,-4), (3,-4), (3,4).