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melstutes

  • one year ago

Suppose that the endpoints of one leg of a 45°-45°-90° triangle are (0, 3) and (4, –1). What is the length of the hypotenuse?

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  1. melstutes
    • one year ago
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  2. melstutes
    • one year ago
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    HELP Please

  3. melstutes
    • one year ago
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    4 – 0 = 4 -1 -3= -4 4 sq + -4 sq = c sq 16 + 16 = c sq √32 = c The distance between the two points is √(16+16) = √32 So, the hypotenuse is √32*√2 = √64 = 8 is this right?

  4. melstutes
    • one year ago
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    I thought distance was same as hypotenuse. Why do you have to multiply by sq root of 2

  5. melstutes
    • one year ago
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    This was also given. Hint: Remember how the sides relate in a 45°-45°-90° triangle.

  6. melstutes
    • one year ago
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    I kept getting √32 as my answer.

  7. Michele_Laino
    • one year ago
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    using theformula of distance we can write: \[d = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} = \sqrt {16 + 16} = \sqrt {32} = 4\sqrt 2 \]

  8. melstutes
    • one year ago
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    Ok - I did that and that is what I thought was the hypotenuse

  9. Michele_Laino
    • one year ago
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    yes! the length of the hypotenuse is \[\sqrt {32} = 4\sqrt 2 \]

  10. Michele_Laino
    • one year ago
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    so the length of each side is: x=4

  11. melstutes
    • one year ago
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    That is not a choice. Answers include 8 units, 4 units, 10 units or 2 units. I was told to multiply by 2 and answer was 8 but without an explanation. I don't understand why

  12. Michele_Laino
    • one year ago
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    I think that your exercise asks for the value of x. So if \[x\sqrt 2 = 32 = 4\sqrt 2 \] then we get: \[x = 4\]

  13. melstutes
    • one year ago
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    Question is what is length of hypotenuse so I am lost

  14. Michele_Laino
    • one year ago
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    oops..I have made an error, hereare the right steps: \[\begin{gathered} x\sqrt 2 = \sqrt {32} = 4\sqrt 2 \hfill \\ x = 4 \hfill \\ \end{gathered} \] yes, I know, nevertheless if I look at your drawing: |dw:1434476827731:dw| I understand that we have to determine the value of x

  15. Michele_Laino
    • one year ago
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    namely in your exercise there are 2 parts: the first one asks for the hypotenuse, using the coordinates which you have provided the second part asks for the value of x, using the length of the hypotenuse computed in the first part

  16. melstutes
    • one year ago
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    Oh, ok. Question is misleading or confusing.

  17. melstutes
    • one year ago
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    Thank you

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