Question

HELP!
http://gyazo.com/e7bc4de262daa0a9da272702ba995677

Answer 1

Have you done problems like this in class?

Answer 2

no homeschooled for this year
(well online school)

Answer 3

Okay. I was just wondering how you were allowed to decide the answer: whether you can use a picture, or if you have to prove it with numbers. The picture is easier, the numbers are more certain.

Answer 4

Do you know the definition of ech kind of triangle?

Answer 5

no

Answer 6

Alright! That's an excellent place to start, then.

Answer 7

ya so is openstudy.com

Answer 8

Let's start with the keyword "right". "Right" means that two of the sides meet at a right angle, a 90 degree angle.

Answer 9

ya

Answer 10

So when you see a right angle in a triangle, you're looking at a "right triangle".

Answer 11

Do you understand that part?

Answer 12

ya

Answer 13

Cool. Moving on....

There's also the keyword "obtuse". "Obtuse" is a keyword that indicates the angle too, but this time there must be an angle that is greater than a right angle, greater than a 90 degree angle.

Answer 14

yes

Answer 15

So if you see an obtuse angle, then you're looking at an obtuse triangle.

Answer 16

Now, what's left are the keywords "scalene" and "isosceles". "Equilateral" is another, but you don't need to know that right now, I guess!

Answer 17

ok

Answer 18

These tell you how the lengths compare to each other.
Scalene: no lengths are the same
Isosceles: 2 lengths are the same
Equilateral: all 3 lengths are the same

Answer 19

Do you understand those?

Answer 20

ya

Answer 21

Cool! Now you can either draw the triangle on a grid and look to see what it looks like, or find all of the triangles' sides lengths.

Answer 22

Do you know how to use two points to mathematically find the length of the line segment that connects them?

Answer 23

The formula is messy looking, but very precise and I can tell you how to get it exactly if you want. Let d be the distance between the points\[(x_1, y_1)\]and\[(x_2, y_2)\].

Then.......

Answer 24

\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Answer 25

You can pick any given point to be\[(x_1, y_1)\] and any other given point to be \[(x_2, y_2)\]

Answer 26

But lets's try graphing first. That isn't "solid evidence", you know, but we can make good guesses with our choices! Drawing!....

Answer 27

Good luck! It helps to make a grid and plot your triangles on there.
I've got to go! Take care!