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

1. freckles

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

2. anonymous

3. freckles

|dw:1437665201466:dw|

4. 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: $\text{ The distance \between } (x_1,y_1) \text{ and } (x_2,y_2) \text{ is } \\ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

5. freckles

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

6. anonymous

ok

7. freckles

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

8. anonymous

no

9. freckles

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

10. anonymous

I got 10

11. 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)

12. 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

13. anonymous

39 and 125

14. freckles

do you mean sqrt(125) for d_3 ?

15. anonymous

yes

16. freckles

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

17. freckles

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

18. freckles

use that formula above to find it

19. freckles

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

20. anonymous

I got 10

21. 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 ... $d_2=\sqrt{(4-(-3))^2+(-2-(-6))^2}$

22. anonymous

11

23. freckles

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

24. freckles

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

25. freckles

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

26. freckles

27. anonymous

i got 11

28. freckles

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

29. freckles

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

30. freckles

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

31. freckles

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

32. anonymous

29 65

33. freckles

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

34. anonymous

okay

35. freckles

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

36. anonymous

239

37. anonymous

230

38. freckles

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

39. freckles

or did you add 125 and 40 and 65?

40. freckles

because those aren't the same numbers

41. anonymous

42. 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.

43. freckles

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

44. anonymous

25.567

45. freckles

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

46. anonymous

25.57

47. freckles

yes