anonymous
  • anonymous
A triangle has vertices at (2, 4), (-3, -6), and (4, -2). Find its perimeter rounded to two decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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freckles
  • freckles
I would plot the points so it is easier for me to see which three distances to calculate.
anonymous
  • anonymous
1 Attachment
freckles
  • freckles
|dw:1437665201466:dw|

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freckles
  • 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
  • freckles
once you find all the side lengths you add the side lengths to find the perimeter
anonymous
  • anonymous
ok
freckles
  • freckles
so have you computed the distance between (2,4) and (-3,-6) ?
anonymous
  • anonymous
no
freckles
  • freckles
|dw:1437665675706:dw| well you still need to find these two distances (these two side lengths)
anonymous
  • anonymous
I got 10
freckles
  • 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
  • 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
  • anonymous
39 and 125
freckles
  • freckles
do you mean sqrt(125) for d_3 ?
anonymous
  • anonymous
yes
freckles
  • freckles
|dw:1437665932861:dw| now you just need to find d_2
freckles
  • freckles
which is the distance between (4,-2) and (-3,-6)
freckles
  • freckles
use that formula above to find it
freckles
  • 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
  • anonymous
I got 10
freckles
  • 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
  • anonymous
11
freckles
  • freckles
4-(-3) is the same as 4+3 -2-(-6) is the same as -2+6
freckles
  • freckles
\[d_2=\sqrt{(4+3)^2+(-2+6)^2}\]
freckles
  • freckles
compute 4+3 and -2+6 then square both results
freckles
  • freckles
then add those results
anonymous
  • anonymous
i got 11
freckles
  • freckles
sounds like you are doing (4+3)+(-2+6) which means you are ignoring the squares
freckles
  • freckles
(4+3)^2=(7)^2 (-2+6)^2=(4)^2
freckles
  • freckles
\[d_2=\sqrt{(4+3)^2+(-2+6)^2} \\ d_2=\sqrt{(7)^2+(4)^2}\]
freckles
  • freckles
you square the numbers before adding them because that is what order of operations tells us to do
anonymous
  • anonymous
29 65
freckles
  • freckles
don't know where the 29 comes from but sqrt(49+16) is sqrt(65)
anonymous
  • anonymous
okay
freckles
  • freckles
|dw:1437666447857:dw| now add all the sides to find the perimeter
anonymous
  • anonymous
239
anonymous
  • anonymous
230
freckles
  • freckles
so you added sqrt(125) and sqrt(40) and sqrt(65)?
freckles
  • freckles
or did you add 125 and 40 and 65?
freckles
  • freckles
because those aren't the same numbers
anonymous
  • anonymous
i added them all
freckles
  • 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
  • freckles
\[\sqrt{125}+\sqrt{40}+\sqrt{65}\] just enter this into the calculator to get the approximated perimeter
anonymous
  • anonymous
25.567
freckles
  • freckles
ok and you are suppose to round to 2 decimal places so that would actually be...
anonymous
  • anonymous
25.57
freckles
  • freckles
yes

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