## 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?

1. melstutes

2. melstutes

3. 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?

4. melstutes

I thought distance was same as hypotenuse. Why do you have to multiply by sq root of 2

5. melstutes

This was also given. Hint: Remember how the sides relate in a 45°-45°-90° triangle.

6. melstutes

I kept getting √32 as my answer.

7. Michele_Laino

using theformula of distance we can write: $d = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} = \sqrt {16 + 16} = \sqrt {32} = 4\sqrt 2$

8. melstutes

Ok - I did that and that is what I thought was the hypotenuse

9. Michele_Laino

yes! the length of the hypotenuse is $\sqrt {32} = 4\sqrt 2$

10. Michele_Laino

so the length of each side is: x=4

11. 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

12. 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: $x = 4$

13. melstutes

Question is what is length of hypotenuse so I am lost

14. 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

15. 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

16. melstutes

Oh, ok. Question is misleading or confusing.

17. melstutes

Thank you