∆ABC has the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices.
The measure of the longest side of ∆ABC is units. ∆ABC is triangle.
If ∆ABD is formed with the point D(1, 2) as its third vertex, then ∆ABD is triangle. The length of side AD is units.
To find the longest side of triangle ∆ABC, we need to calculate the lengths of all three sides and then compare them.
Using the distance formula, the lengths of the sides can be calculated as follows:
Side AB:
Length of AB = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the coordinates, we have:
Length of AB = √[(-2 - 1)^2 + (2 - 7)^2]
Length of AB = √[(-3)^2 + (-5)^2]
Length of AB = √[9 + 25]
Length of AB = √34
Side AC:
Length of AC = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the coordinates, we have:
Length of AC = √[(4 - 1)^2 + (2 - 7)^2]
Length of AC = √[(3)^2 + (-5)^2]
Length of AC = √[9 + 25]
Length of AC = √34
Side BC:
Length of BC = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the coordinates, we have:
Length of BC = √[(4 - -2)^2 + (2 - 2)^2]
Length of BC = √[(6)^2 + (0)^2]
Length of BC = √[36 + 0]
Length of BC = √36
Length of BC = 6
Comparing the lengths, we can see that the longest side of ∆ABC is BC with a length of 6 units.
Now, let's find the length of side AD in triangle ∆ABD.
Using the distance formula, the length of side AD can be calculated as follows:
Length of AD = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the coordinates, we have:
Length of AD = √[(1 - 1)^2 + (2 - 7)^2]
Length of AD = √[0^2 + (-5)^2]
Length of AD = √[0 + 25]
Length of AD = √25
Length of AD = 5 units.
Therefore, in ∆ABD, the length of side AD is 5 units.
To find the measure of the longest side of triangle ∆ABC, you can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Let's find the distances between the three points A(1, 7), B(-2, 2), and C(4, 2):
Distance AB = √((-2 - 1)^2 + (2 - 7)^2) = √((-3)^2 + (-5)^2) = √(9 + 25) = √34 units
Distance BC = √((4 - (-2))^2 + (2 - 2)^2) = √((6)^2 + (0)^2) = √(36 + 0) = √36 = 6 units
Distance AC = √((4 - 1)^2 + (2 - 7)^2) = √((3)^2 + (-5)^2) = √(9 + 25) = √34 units
Therefore, the longest side of triangle ∆ABC is √34 units.
Now, let's find the length of side AD in triangle ∆ABD using the distance formula with points A(1, 7) and D(1, 2):
Distance AD = √((1 - 1)^2 + (2 - 7)^2) = √((0)^2 + (-5)^2) = √(0 + 25) = √25 = 5 units.
Therefore, the length of side AD in triangle ∆ABD is 5 units.
5 units
∆ABC has the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices.
The measure of the longest side of ∆ABC is units. ∆ABC is triangle.
If ∆ABD is formed with the point D(1, 2) as its third vertex, then ∆ABD is triangle. The length of side AD is units.