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Math Help - 2 geometry word problems.

  1. #1
    Ash
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    2 geometry word problems.

    #1. A building inspector is checking on an apartment building that is under construction. She finds that a window ledge, exactly 22 feet from the level ground, can be reached with a ladder that is 24 feet long.
    The bottom of the ladder is 6 feet away from the base of the wall, and the top of the ladder just reaches the window ledge is the wall perfectly vertical? Explain.



    #2
    Some community volunteers have laid out a softball field at the local park. The distance between the bases is 60 feet. The distance from home plate to second base is 90 feet. Is the infield perfectly square? If not what should be the distance from home plate to second base (to the nearest tenth)?




    Is there a common formula like ( a^2 + b^2 = c^2 ) to use for both and if so how would I go about arranging the needed information?
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  2. #2
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    Quote Originally Posted by Ash View Post
    #1. A building inspector is checking on an apartment building that is under construction. She finds that a window ledge, exactly 22 feet from the level ground, can be reached with a ladder that is 24 feet long.
    The bottom of the ladder is 6 feet away from the base of the wall, and the top of the ladder just reaches the window ledge is the wall perfectly vertical? Explain.



    #2
    Some community volunteers have laid out a softball field at the local park. The distance between the bases is 60 feet. The distance from home plate to second base is 90 feet. Is the infield perfectly square? If not what should be the distance from home plate to second base (to the nearest tenth)?




    Is there a common formula like ( a^2 + b^2 = c^2 ) to use for both and if so how would I go about arranging the needed information?
    Yes; the pythagorean theorem.

    For the first, check to see if a^2 + b^2 = c^2

    6^2 + 22^2 = c^2 ?

    522 = c^2 ?

    sqrt(522) = c ?

    Well sqrt(522) = 2*sqrt(130) = 22.8035 which is not the 24 feet, and thus it is not perfectly vertical.

    2.) Do the same thing:

    60^2 + 60^2 = c^2 ?

    sqrt(7200) = c^2 ?

    sqrt(7200) = 60*sqrt(2), and thus it is not perfectly square.

    The distance should be: 84.9 feet.
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  3. #3
    Ash
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    Quote Originally Posted by AfterShock View Post
    Yes; the pythagorean theorem.

    For the first, check to see if a^2 + b^2 = c^2

    6^2 + 22^2 = c^2 ?

    522 = c^2 ?

    sqrt(522) = c ?

    Well sqrt(522) = 2*sqrt(130) = 22.8035 which is not the 24 feet, and thus it is not perfectly vertical.

    2.) Do the same thing:

    60^2 + 60^2 = c^2 ?

    sqrt(7200) = c^2 ?

    sqrt(7200) = 60*sqrt(2), and thus it is not perfectly square.

    The distance should be: 84.9 feet.
    Thanks for your help, but I was wondering how you knew to multiply
    2* sqrt(130) and in problem 2. ) 60* sqrt(2)?
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