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Math Help - Find Shortest Distance

  1. #1
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    Find Shortest Distance

    To get from his high school to his home, Jamal travels 5.0 miles eastand then 4.0 miles north. When Sheila goes to her home from the same high school, she travels 8.0 miles east and 2.0 miles south. What is the measure of the shortest distance, to the nearest tenth of a mile, between Jamal’s home and Sheila’s home?

    I sketched 2 right triangles. I sketched a map with east, west, north and south and was not able to find the answer.
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  2. #2
    Super Member 11rdc11's Avatar
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    \sqrt{45}
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  3. #3
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    Think of this in the coordinate plane. Assume that school is on (0,0). Then Jamal travels east from (0,0) to (5,0), and then from (5,0) to (5,4). P(5, 4) is his home. Now find the point where Sheila lives, and use the distance formula to find the distance between the two points:

    d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2
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  4. #4
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    Smile Can you.....

    Quote Originally Posted by Chop Suey View Post
    Think of this in the coordinate plane. Assume that school is on (0,0). Then Jamal travels east from (0,0) to (5,0), and then from (5,0) to (5,4). P(5, 4) is his home. Now find the point where Sheila lives, and use the distance formula to find the distance between the two points:

    d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2
    Which coordinates go into the formula?

    Can you place the right coordinates into the given formula and I'll take it from there?

    Thanks
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  5. #5
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    Yes...

    Quote Originally Posted by 11rdc11 View Post
    \sqrt{45}
    Yes, the answer is sqrt{45} but how did you get it?
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  6. #6
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    Quote Originally Posted by magentarita View Post
    Can you place the right coordinates into the given formula and I'll take it from there?
    No. You should know that a point in a 2D coordinate plane is in the form of (x, y)

    Here's a different approach.
    Attached Thumbnails Attached Thumbnails Find Shortest Distance-picture1.png  
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by magentarita View Post
    Yes, the answer is sqrt{45} but how did you get it?
    I solved it the way chop suey told you.
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  8. #8
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    Hello, magentarita!

    To get from his high school to his home, Jamal travels 5 miles east and 4 miles north.
    When Sheila goes to her home from the same high school,
    she travels 8 miles east and 2 miles south.
    What is the measure of the shortest distance, to the nearest tenth of a mile,
    between Jamal’s home and Sheila’s home?
    Code:
                    J
                    *
                    |
                    | 4
          : -  5  - |
        S *---------*-------*
          : - - - 8 - - - - |
                            | 2
                            *
                            S
    The school is at S.

    Jamal travels 5 miles east and 4 miles north at arrives at his home at J.

    Sheila travels 8 miles east and 2 miles south and arrives at her home at S.

    We want the distance from J to S (the length of line segment JS).



    Re-draw the diagram.
    Code:
                    J
                    *
                    |
                  4 | *
                    |
        S * - - - - *   *
                    |
                  2 |     *
                    |
                    * - - - *
                        3    S

    See that right triangle?

    We have: . JS^2 \:=\:3^2 + 6^2 \:=\:45

    Therefore: . JS \;=\;\sqrt{45} \;\approx\;6.7 miles.

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  9. #9
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    Thanks

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Code:
                    J
                    *
                    |
                    | 4
          : -  5  - |
        S *---------*-------*
          : - - - 8 - - - - |
                            | 2
                            *
                            S
    The school is at S.

    Jamal travels 5 miles east and 4 miles north at arrives at his home at J.

    Sheila travels 8 miles east and 2 miles south and arrives at her home at S.

    We want the distance from J to S (the length of line segment JS).



    Re-draw the diagram.
    Code:
                    J
                    *
                    |
                  4 | *
                    |
        S * - - - - *   *
                    |
                  2 |     *
                    |
                    * - - - *
                        3    S
    See that right triangle?

    We have: . JS^2 \:=\:3^2 + 6^2 \:=\:45

    Therefore: . JS \;=\;\sqrt{45} \;\approx\;6.7 miles.
    Thank you, Soroban. I would like you to reply more often to my questions.

    I just love your replies.
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