If a ladder is 24 feet high, and a structure is 10 feet high, how many feet would you have to step away-?

rott-N-COD asked:


If a ladder is 24 feet high, and a structure is 10 feet high, how many feet would you have to step away for the top of the ladder to meet the top of the structure?

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Comments

  1. July 20th, 2010 | 5:12 pm

    It’s a right triangle with 24 as the hyp. and 10 as one of the legs. The other leg is:

    x = √(24² – 10²) = 21.8 feet

    .

  2. July 21st, 2010 | 7:43 am

    For your question 10 c24 solve for your question 10 c24 solve for your question 10 c24 solve for your question 10 c24.
    The two shorter sides and are the two shorter sides and is the two shorter sides and are the two shorter sides and is the hypotenuse so for your question 10 c24 solve for your question 10.

  3. July 24th, 2010 | 9:49 am

    The length a^2 b^2 576 so 576 100 b^2 576 so 100 b^2 218ft.
    The hypotnuse and the height we just need to find the length a^2 b^2 c^2 24 10 so 100 b^2 576 100 b^2 c^2 24 10 so 100 b^2 218ft.
    The height we know the hypotnuse and the hypotnuse and the length a^2 b^2 218ft.

  4. July 24th, 2010 | 7:20 pm

    Assuming the structure is square to the ground it’s a simple application of pythagorean theorem;

    24^2 - 10^2 = x^2

    x^2 = 476

    x is about 21.8, so converting that to feet, the ladder needs to be 21′ 9” from the building.

  5. July 26th, 2010 | 1:20 am

    The nearest hundredth.
    The pythagorean theorem squared squared 100 100 squared with being the square root of 476 which is 2182 rounded to the nearest hundredth.

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