# Classifying a triangle by it's vertices

• Jan 19th 2008, 10:53 AM
mathdonkey
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. :)
• Jan 19th 2008, 12:22 PM
Sean12345
Quote:

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.
• Jan 19th 2008, 12:24 PM
earboth
Quote:

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,

$p = \sqrt{2}+\sqrt{50}+\sqrt{40}=\sqrt{2} + 5\sqrt{2}+2\sqrt{10} \approx 14.8098...$
• Jan 19th 2008, 02:33 PM
mathdonkey
Quote:

Originally Posted by earboth
Hello,

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