# Thread: Classifying a triangle by it's vertices

1. ## Classifying a triangle by its vertices

I'm not sure my answer is correct for this question, would anyone be kind enough to look it over for me? I'll post my anwers below.

The vertices of a triangle are J(-2,2), K(-1,3), and L(5,1).

a) Is triangle JKL an equilateral, isosceles, or scalene triangle?

First I calculated the lengths of the triangles three sides.

I got: JK = square root of two, KL = square root of forty, JL = square root of fifty.

So the triangle would be scalene.

b) Determine the perimeter of triangle JKL

I added all the sides together and got approximately 14.7 units.

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Did I make any mistakes, and if so, could you please explain how I should have done this? Thanks a million.

2. Originally Posted by mathdonkey
I'm not sure my answer is correct for this question, would anyone be kind enough to look it over for me? I'll post my anwers below.

The vertices of a triangle are J(-2,2), K(-1,3), and L(5,1).

a) Is triangle JKL an equilateral, isosceles, or scalene triangle?

First I calculated the lengths of the triangles three sides.

I got: JK = square root of two, KL = square root of forty, JL = square root of fifty.

So the triangle would be scalene.

b) Determine the perimeter of triangle JKL

I added all the sides together and got approximately 14.7 units.

---

Did I make any mistakes, and if so, could you please explain how I should have done this? Thanks a million.
Nope, i don't see any mistakes. However for b) i got 14.8 units approx.

3. Originally Posted by mathdonkey
I'm not sure my answer is correct for this question, would anyone be kind enough to look it over for me? I'll post my anwers below.

The vertices of a triangle are J(-2,2), K(-1,3), and L(5,1).

a) Is triangle JKL an equilateral, isosceles, or scalene triangle?

First I calculated the lengths of the triangles three sides.

I got: JK = square root of two, KL = square root of forty, JL = square root of fifty.

So the triangle would be scalene.

b) Determine the perimeter of triangle JKL

I added all the sides together and got approximately 14.7 units.

---

Did I make any mistakes, and if so, could you please explain how I should have done this? Thanks a million.
Hello,

your results are OK but...

$\displaystyle p = \sqrt{2}+\sqrt{50}+\sqrt{40}=\sqrt{2} + 5\sqrt{2}+2\sqrt{10} \approx 14.8098...$

4. Originally Posted by earboth
Hello,

your results are OK but...

$\displaystyle p = \sqrt{2}+\sqrt{50}+\sqrt{40}=\sqrt{2} + 5\sqrt{2}+2\sqrt{10} \approx 14.8098...$
I rounded the numbers differently.

Thanks for the help, both of you!