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Math Help - length of three sides

  1. #1
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    length of three sides

    The perimeter of a right triangle is 60 cm and the altitude perpendicular to the hypotenuse is 12 cm. What are the lengths of the three sides?
    Last edited by mr fantastic; August 7th 2009 at 11:44 PM. Reason: Restored deleted question.
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  2. #2
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    I'm assuming altitude means height. If that is so, what do we know:

    We know the perimiter of the triangle is 60cm, which means we know that B+A+H=60CM, where B is the base, A is the altitude, and H is the hypotenuse of our triangle.

    We also know that the Altitude (A) is 12CM, so:

    B+12+H=60CM

    Now we also know that B^{2}+A^{2}=H^{2} by the Pytha Theorem yes?

    Using these two formulas, we can do some substitutions and solve for these worrisome sides!

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    We also know that the Altitude (A) is 12CM, so:

    B+12+H=60CM
    No. 12 is not the length of a side; it is the length of a line inside the triangle,
    from the right angle corner to the hypotenuse, perpendicular to the hypotenuse.

    So this line forms 2 right triangles inside the given right triangle: side 12
    is common to both.
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  4. #4
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    Well good show. Even easier. We've two right triangles formed by splitting an angle of 90 degrees into two. From there its simply a matter of using properties of a 45-45-90 triangle to figure out the sides.

    My bad.
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  5. #5
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    Quote Originally Posted by ANDS! View Post
    Well good show. Even easier. We've two right triangles formed by splitting an angle of 90 degrees into two. From there its simply a matter of using properties of a 45-45-90 triangle to figure out the sides.
    No. The 90 degree angle is not split 45-45.
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  6. #6
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    OK; works like this (not as easy as it looks!):

    Let a = one leg, b = other leg, c = hypotenuse

    Since perimeter = 60, then c = 60 - a - b; so:
    a^2 + b^2 = (60 - a - b)^2 ; simplify to get:
    b = (60a - 1800) / (a - 60) [1]

    Using areas: ab = 12(60 - a - b) ; simplify to get:
    b = (720 - 12a) / (a + 12) [2]

    From [1] and [2]:
    (60a - 1800) / (a - 60) = (720 - 12a) / (a + 12) ; simplify to get:
    a^2 - 35a + 300 = 0
    (a - 20)(a - 15) = 0
    a = 20 or 15 : that's the lengths of the 2 legs.
    So hypotenuse = 60 - 15 - 20 = 25.

    So you end up with a 15-20-25 right triangle.
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  7. #7
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    Hello, Confused25!

    The perimeter of a right triangle is 60 cm and the altitude to the hypotenuse is 12 cm.
    What are the lengths of the three sides?
    Code:
                C
                *
               *|  *
            b * |     *   a
             *  |12      *
            *   |           *
           *    |              *
        A *-----+-----------------* B
          : - - - - - c - - - - - :

    Given: right triangle ABC,\;C = 90^o

    We have: . \begin{array}{cc}a + b + c \:=\:60 & {\color{blue}[1]} \\ a^2 +b^2 \:=\:c^2 & {\color{blue}[2]} \\ ab \:=\:12^2 & {\color{blue}[3]} \end{array}


    From [1], we have: . a + b \:=\:60-c

    \text{Square both sides: }\;(a+b)^2 \:=\:(60-c)^2 \quad\Rightarrow\quad \underbrace{a^2 + b^2}_{c^2} + 2\underbrace{(ab)}_{144} \:=\:3600 - 120c + c^2
    . . 120c \:=\:3312 \quad\Rightarrow\quad\boxed{ c \:=\:27.6}


    Substitute into [1]: . a + b + 27.6 \:=\:60 \quad\Rightarrow\quad a + b \:=\:32.4 \quad\Rightarrow\quad b \:=\:32.4 - a .[4]

    Substitute into [2]: . a^2 + b^2 \:=\:27.6^2 .[5]


    Substitute [4] into [5]: . a^2 + (32.4-a)^2 \:=\:27.6^2 \quad\Rightarrow\quad a^2 - 32.4a + 144 \:=\:0

    Quadratic Formula: . a \;=\;\frac{32.4 \pm \sqrt{473.76}}{2} \;=\; \frac{32.4 \pm 12\sqrt{3.29}}{2}

    Hence: . \boxed{a \;=\;16.2 \pm6\sqrt{3.29}}


    Substitute into [4]: . b \;=\;32.4 - (16.2 \pm6\sqrt{3.29}) \quad\Rightarrow\quad \boxed{b \;=\;16.2 \mp\sqrt{3.29}}

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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban View Post
    [size=3]

    We have: . \begin{array}{cc}a + b + c \:=\:60 & {\color{blue}[1]} \\ a^2 +b^2 \:=\:c^2 & {\color{blue}[2]} \\ ab \:=\:12^2 & {\color{blue}[3]} \end{array}

    I think your third equation is incorrect
    It should be:
    ab=12c
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  9. #9
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    I think Mr Soroban's having a bad day

    Right triangle 3-4-5 has height to hypotenuse = 1.2 is something
    we memorize in grade 9 or so.

    Anyhow, answer is a = 15, b = 20, c = 25.

    As a general case, given p = perimeter and h = height to hypotenuse:

    b,a = [-v +,- SQRT(v^2 - 4uw)] / (2u), where:
    u = 2(p + h)
    v = -p(p + 2h)
    w = p^2 h

    OK, Mr.Confused; now try this one:
    perimeter = 1400, height to hypotenuse = 168 ; calculate a and b
    Last edited by Wilmer; August 7th 2009 at 09:05 AM. Reason: add problem for Mr.Confused
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