A triangle has vertices at (2, 4), (-3, -6), and (4, -2). Find its perimeter rounded to two decimal places.

- anonymous

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- freckles

I would plot the points so it is easier for me to see which three distances to calculate.

- anonymous

##### 1 Attachment

- freckles

|dw:1437665201466:dw|

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## More answers

- freckles

|dw:1437665261302:dw|
we need to find all three side measurements
to do that we can use the distance formula
so for example let's go ahead and find the distance between (2,4) and (4,-2)
you do that by doing:
\[d_1=\sqrt{(2-4)^2+(4-(-2))^2} \\ d_1=\sqrt{(-2)^2+(6)^2} \\ d_1=\sqrt{4+36} \\ d_1=\sqrt{40}\]
so we have that|dw:1437665356288:dw|
we also need to find the other side lengths
you try the other two legs
and remember:
\[\text{ The distance \between } (x_1,y_1) \text{ and } (x_2,y_2) \text{ is } \\ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

- freckles

once you find all the side lengths
you add the side lengths to find the perimeter

- anonymous

ok

- freckles

so have you computed the distance between (2,4) and (-3,-6) ?

- anonymous

no

- freckles

|dw:1437665675706:dw|
well you still need to find these two distances (these two side lengths)

- anonymous

I got 10

- freckles

that is you need to find the distance between (2,4) and (-3,-6)
you also need to find the distance between (-3,-6) and (4,-2)

- freckles

the distance between (2,4) and (-3,-6) is given by:
\[d_3=\sqrt{(2-(-3))^2+(4-(-6))^2}=\sqrt{5^2+(10)^2}\]
see if you can finish simplifying this

- anonymous

39 and 125

- freckles

do you mean sqrt(125) for d_3 ?

- anonymous

yes

- freckles

|dw:1437665932861:dw|
now you just need to find d_2

- freckles

which is the distance between (4,-2) and (-3,-6)

- freckles

use that formula above to find it

- freckles

\[\text{ The distance between } (x_1,y_1) \text{ and } (x_2,y_2) \text{ is } \\ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \]

- anonymous

I got 10

- freckles

hmm... that is what you got for the other one...
try again...
you have (4,-2) and (-3,-6)
replace x1 with 4
replace x2 with -3
replace y1 with -2
replace y2 with -6
...
\[d_2=\sqrt{(4-(-3))^2+(-2-(-6))^2}\]

- anonymous

11

- freckles

4-(-3)
is the same as
4+3
-2-(-6)
is the same as
-2+6

- freckles

\[d_2=\sqrt{(4+3)^2+(-2+6)^2}\]

- freckles

compute 4+3 and -2+6
then square both results

- freckles

then add those results

- anonymous

i got 11

- freckles

sounds like you are doing (4+3)+(-2+6)
which means you are ignoring the squares

- freckles

(4+3)^2=(7)^2
(-2+6)^2=(4)^2

- freckles

\[d_2=\sqrt{(4+3)^2+(-2+6)^2} \\ d_2=\sqrt{(7)^2+(4)^2}\]

- freckles

you square the numbers before adding them
because that is what order of operations tells us to do

- anonymous

29
65

- freckles

don't know where the 29 comes from
but sqrt(49+16)
is sqrt(65)

- anonymous

okay

- freckles

|dw:1437666447857:dw|
now add all the sides to find the perimeter

- anonymous

239

- anonymous

230

- freckles

so you added sqrt(125) and sqrt(40) and sqrt(65)?

- freckles

or did you add 125 and 40 and 65?

- freckles

because those aren't the same numbers

- anonymous

i added them all

- freckles

sqrt(125) is not the same as 125
sqrt(40) is not the same as 40
sqrt(65) is not the same as 65
so adding sqrt(125) and sqrt(40) and sqrt(65)
is not equivalent to adding 125 and 40 and 65.

- freckles

\[\sqrt{125}+\sqrt{40}+\sqrt{65}\]
just enter this into the calculator to get the approximated perimeter

- anonymous

25.567

- freckles

ok and you are suppose to round to 2 decimal places so that would actually be...

- anonymous

25.57

- freckles

yes

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