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Math Help - projections of legs

  1. #1
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    Find projections of legs

    Right angeled triangle's legs relate as 3:7 and the height between the vertex (where the angle is 90 degrees) and the hypotenuse is 42 cm. How to find the projections of the legs. The answers are 18 cm and 98 cm. By using Height's theoreme according to the formula that h^2=f*g where h is height, f and g are projections (f+g=hypotenuse), the answers are right but I don't really find the way to find these answers.
    Last edited by totalnewbie; September 4th 2005 at 07:09 AM.
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  2. #2
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    Pytagoras job

    Hi,

    Lets name the triangle ABC with B the vertex at the right angle (so, AC the hypotenuse) and AB the smallest of the two remaining sides.

    As you said, we draw the height coming from the vertex B and perpendicular to AC. Lets name R the intersection of the height with AC.

    So since both the triangles ABR and BCR are right-angled (because of the height), we can use Pytagoras. We have BR = 42. So, AB^2 = AR^2 + 42^2 so we rewrite it as AR^2 = AB^2 - 42^2 (formula 1). Same with BC^2 = RC^2 + 42^2 and we get RC^2 = BC^2 - 42^2 (formula 2).

    Now, we know that BC = 7/3 * AB (here, it is what 3:7 means, in other words the proportion is 7/3). So we replace BC by 7/3*AB is the second formula and we get RC^2 = (7/3*AB)^2 - 42^2 = 49/9 AB^2 - 42^2 (formula 2b).

    It is now that we use your height theorem : BR^2 = AR * RC. We square both sides and get BR^4 = AR^2 * RC^2. So 42^4 = (AB^2 - 42^2)*(49/9 AB^2 - 42^2) = 49/9 AB^4 - 42^2 AB^2 - 42^2 * 49/9 AB^2 + 42^4.

    So we cancel the 42^4 and get 0=49/9 AB^4 - 42^2 AB^2 - 42^2 * 49/9 AB^2 = 49/9 AB^4 - 7840 11368 AB^2. So by factorization or quadratic equation you find AB^2 = 2088 or 0. We of course don't take 0 because a measure can't be 0. So AB= 45.695.

    Of course we don't need AB so we look for AR and RC. From formula 1 we get AR = sqrt(AB^2 - 42^2) = 18 and from formula 2b we get RC = sqrt(49/9 AB^2 - 42^2) = 98. Done !

    - Here my notation is not very good because I should be using "measure of" a side instead of the name of the side in my calculations but to make the text easier to read, I didn't do it -
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  3. #3
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    Thank you. Good job.
    Before I looked at your solution I haven't a chance to get on internet. So I tried again and as a surprise I was successful.
    My way:
    leg a=3*x/10
    leg b=7*x/10
    height, h=42
    So I used pythagoras.
    a^2+b^2=c^2 where c=x*sqrt(29/50)
    f and g are the projections of the legs.
    It is right angeled triangle, so f=9x/100*sqrt(29/50)
    Then f^2+h^2=a^2 because it is right angeled triangle, so [9x/100*sqrt(29/50)]^2+42^2=9*x^2/100 where X=+-sqrt(23200)
    And by substituiting the x, we get that:
    f=18 or -18 but -18 is wrong because measure can't be with minus as you said before.
    c=116 or -116 but -116 is wrong again
    and g=c-f=116-18=98
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