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Thread: find the hyp.

  1. #1
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    Question find the hyp.

    rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?
    what is area covered by a,d,c?
    402 +1402 =
    1600+19600=
    21200
    √21200=
    20√53
    how do i work with this?
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  2. #2
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    Quote Originally Posted by inneedofhelp!ariie View Post
    rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?
    what is area covered by a,d,c?
    402 +1402 =
    1600+19600=
    21200
    √21200=
    20√53
    how do i work with this?
    If you draw reactangle ABCD, AC=hypotenuse of triangle ADC=140 km
    and CD=AB=40 km.

    In triangle ADC,
    $\displaystyle
    AD^2 + CD^2 = AC^2$

    $\displaystyle AD^2 + 40^2 = 140^2$

    $\displaystyle AD^2 = 140^2 - 40^2$

    $\displaystyle AD = \sqrt{19600 - 1600}$

    $\displaystyle AD = \sqrt{18000}$

    $\displaystyle AD = \sqrt{5\times 3^2 \times 2^2 \times 10^2}$

    $\displaystyle AD = 60 \sqrt{5}$

    $\displaystyle AD = 60 \times 2.236$

    $\displaystyle AD = 134.164 \;km$

    Area covered by triangle ADC;

    $\displaystyle =\frac{base \times height}{2}$

    $\displaystyle =\frac{40 \times 134.164}{2}$

    $\displaystyle =2683.28\; km^2$
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  3. #3
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    no the diagonal (hypotenuse) is A to D
    A to B is 40km
    A to C is 140km

    the diagonal is always longest so cannot be 134
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  4. #4
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    If your question is not clear you can't expect to get a good answer.

    Let ad = x. Then x^2 + 40^2 = 140^2 => .....

    By the way, don't ask for help in Live Chat. It clearly says that Live Chat is for small talk.

    Quote Originally Posted by inneedofhelp!ariie View Post
    rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?

    what is area covered by a,d,c?

    402 +1402 =

    1600+19600=

    21200

    √21200=

    20√53

    how do i work with this?
    Quote Originally Posted by Shyam View Post
    If you draw reactangle ABCD, AC=hypotenuse of triangle ADC=140 km

    and CD=AB=40 km.



    In triangle ADC,

    $\displaystyle

    AD^2 + CD^2 = AC^2$



    $\displaystyle AD^2 + 40^2 = 140^2$



    $\displaystyle AD^2 = 140^2 - 40^2$



    $\displaystyle AD = \sqrt{19600 - 1600}$



    $\displaystyle AD = \sqrt{18000}$



    $\displaystyle AD = \sqrt{5\times 3^2 \times 2^2 \times 10^2}$



    $\displaystyle AD = 60 \sqrt{5}$



    $\displaystyle AD = 60 \times 2.236$



    $\displaystyle AD = 134.164 \;km$



    Area covered by triangle ADC;



    $\displaystyle =\frac{base \times height}{2}$



    $\displaystyle =\frac{40 \times 134.164}{2}$



    $\displaystyle =2683.28\; km^2$
    Quote Originally Posted by inneedofhelp!ariie View Post
    no the diagonal (hypotenuse) is A to D
    A to B is 40km
    A to C is 140km

    the diagonal is always longest so cannot be 134
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