melstutes
  • melstutes
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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melstutes
  • melstutes
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melstutes
  • melstutes
HELP Please
melstutes
  • melstutes
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?

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melstutes
  • melstutes
I thought distance was same as hypotenuse. Why do you have to multiply by sq root of 2
melstutes
  • melstutes
This was also given. Hint: Remember how the sides relate in a 45°-45°-90° triangle.
melstutes
  • melstutes
I kept getting √32 as my answer.
Michele_Laino
  • Michele_Laino
using theformula of distance we can write: \[d = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} = \sqrt {16 + 16} = \sqrt {32} = 4\sqrt 2 \]
melstutes
  • melstutes
Ok - I did that and that is what I thought was the hypotenuse
Michele_Laino
  • Michele_Laino
yes! the length of the hypotenuse is \[\sqrt {32} = 4\sqrt 2 \]
Michele_Laino
  • Michele_Laino
so the length of each side is: x=4
melstutes
  • melstutes
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
Michele_Laino
  • Michele_Laino
I think that your exercise asks for the value of x. So if \[x\sqrt 2 = 32 = 4\sqrt 2 \] then we get: \[x = 4\]
melstutes
  • melstutes
Question is what is length of hypotenuse so I am lost
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
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
melstutes
  • melstutes
Oh, ok. Question is misleading or confusing.
melstutes
  • melstutes
Thank you

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