Easy Math Question!!?

Question

Find the perimeter of a triangle whose vertices are X(-6, 2), Y(8, 2), and Z(3, 14).

Answer 1

If its easy why are you asking us?

Answer 2

If it's so easy, why don't *YOU* find the perimeter of a triangle whose vertices are X(-6, 2), Y(8, 2), and Z(3, 14)?

This is 'homework help' not 'Help! Do my homework for me!'

Have you actually tried?! (be truthful!)

Answer 3

the length of XY:-

sqrt {[8 - (-6)]^2 + (2 - 2)^2}

= sqrt [(8 + 6)^2 + 0^2]

= sqrt (14^2)

= 14 units

the length of YZ:-

sqrt [(3 - 8)^2 + (14 - 2)^2]

= sqrt [(-5)^2 + (12)^2]

= sqrt (25 + 144)

= sqrt 169

= 13 units

the length of ZX:-

sqrt [(-6 - 3)^2 + (2 - 14)^2]

= sqrt [(-9)^2 + (-12)^2]

= sqrt (81 + 144)

= sqrt 225

= 15 units

therefore...

the perimeter of the triangle is...

XY + YZ + ZX

= 14 + 13 + 15

= 42 units

Answer 4

you should try it tell us what you got if you got it wrong then ask we wont do it for you

Answer 5

Answer is 42 units. Use the distance formula  three times for XY, YZ and XZ. Formula is = [(x2-x1)^2 + (y2-y1)]^1/2

Answer 6

If its so easy why dont you figure it out?

Answer 7

42

find distance of all sides (distance formula:

(square root): (x(2) - x(1) )^2 + (y(2) - y(1)) ^2

then square root your answer. you get  14, 13, and 15 as your side lengths and you add them up to be 42 as your perimeter. hope that helps

Answer 8

I think u add them u ,,i forget . but its ur homework

Answer 9

you can add the vertices of x,y,z.

Answer 10

i got 32 lol i think its right but im not sure i didnt have graph paper so  i wrote it on the back of a printout essay lol

Answer 11

Use the distance formula D= sqrt((a_2 - a_1) + (b_2 - b_1))

(a_1, b_1) correspond to the first point and (a_2, b_2) your second point.

Answer is 42. But you be the one to solve it. That way you learn how to do it.