Question

Not everyone understands this but if you can help please do mi amor

Answer 1

Answers:

1. scalene
2. isosceles
3. scalene

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Part (a)

Let's find the distance from point D to point E.

In other words, this will compute the length of segment DE.

`D = (x_1,y_1) = (0,-6) \text{ and } E = (x_2, y_2) = (3,1)\\\\d = √((x_1 - x_2)^2 + (y_1 - y_2)^2)\\\\d = √((0-3)^2 + (-6-1)^2)\\\\d = √((-3)^2 + (-7)^2)\\\\d = √(9 + 49)\\\\d = √(58)\\\\d \approx 7.6158\\\\`

Segment DE is roughly 7.6158 units long.

Let's repeat these steps to find the length of segment EF.

`E = (x_1,y_1) = (3,1) \text{ and } F = (x_2, y_2) = (-2,1)\\\\d = √((x_1 - x_2)^2 + (y_1 - y_2)^2)\\\\d = √((3-(-2))^2 + (1-1)^2)\\\\d = √((3+2)^2 + (1-1)^2)\\\\d = √((5)^2 + (0)^2)\\\\d = √(25 + 0)\\\\d = √(25)\\\\d = 5\\\\`

Segment EF is exactly 5 units long.

Lastly, we'll compute the length of segment FD.

`(x_1,y_1) = (-2,1) \text{ and } (x_2, y_2) = (0,-6)\\\\d = √((x_1 - x_2)^2 + (y_1 - y_2)^2)\\\\d = √((-2-0)^2 + (1-(-6))^2)\\\\d = √((-2-0)^2 + (1+6)^2)\\\\d = √((-2)^2 + (7)^2)\\\\d = √(4 + 49)\\\\d = √(53)\\\\d \approx 7.2801\\\\`

Segment FD is about 7.2801 units long.

This is close to the length of DE, but not quite.

To recap, we found these segment lengths:

• DE = 7.6158 (approximate)
• EF = 5 (exact)
• FD = 7.2801 (approximate)

As you can see, all three sides are different lengths. Therefore, triangle DEF is scalene.

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For part (b), follow the same template as part (a).

I'll be skipping the steps.

This is what you should get for the lengths of triangle TRI

• segment TR = 6.0828 (approximate)
• segment RI = 2 (exact)
• segment IT = 6.0828 (approximate)

Segments TR and IT are the same length (both exactly
`√(37)` units long), which means triangle TRI is isosceles.

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For part (c), I'll also be skipping steps.

Here are the lengths you should find

• JK = 4.1231 (approximate)
• KL = 8.9443 (approximate)
• LJ = 10.6301 (approximate)

All three lengths are different, making this triangle scalene.