Question 1037940: find the perimeter of triangle XYZ where X(0,2) Y(4,-1) Z(-2,-1), I took X and Y and added them together i the took X and Z and added them together and i took Y and Z and added them together fo XY i came up with -1 for XZ i came up with 5 and when i added YZ together i came up with 6 i added all three together and came up with 10 but the acellus site said i was wrong will you please tell me what i did wrong or show me the answer is p=___+the square root sign
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the perimeter of triangle XYZ where X(0,2) Y(4,-1) Z(-2,-1),
I took X and Y and added them together
i the took X and Z and added them together and
i took Y and Z and added them together
fo XY i came up with -1 **** How did you get -1 ?
----
A length can't be negative.
The length of XY is sqrt(diffx^2 + diffy^2) = sqrt(4^2 + 3^2) = 5
diffx = difference between the 2 x-values 0 & 4
------
Do the other 2 sides the same way, then add them.
========================================================
for XZ i came up with 5 and when i added YZ together i came up with 6 i added all three together and came up with 10 but the acellus site said i was wrong will you please tell me what i did wrong or show me the answer is p=___+the square root sign
|
|
|
