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Thread: Isosceles triangle of fixed perimeter P

  1. #1
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    Isosceles triangle of fixed perimeter P

    Consider an isosceles triangle of fixed perimeter P.
    a) If x equals the length of one of the two equal sides, express the area A as a function of x.

    b) What is the domain of A?
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  2. #2
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    1)
    $\displaystyle
    A(x) = \frac{(P-2x)\sqrt{4Px-P^2}}{4}
    $

    (Note that A(x) above IS a function of x (and not x and P), since P is a fixed constant.)

    2)
    $\displaystyle
    \text{Domain of } A: \frac{P}{4} < x < \frac{P}{2}
    $


    Enjoy your Sunday!
    Last edited by abender; Jan 3rd 2010 at 09:13 AM. Reason: exchanging an x^2 for a p^2... I deserve this for not writing out my work.
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  3. #3
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    Quote Originally Posted by dartanian24 View Post
    Consider an isosceles triangle of fixed perimeter P.
    a) If x equals the length of one of the two equal sides, express the area A as a function of x.
    To see where you're at:
    can you calculate the height of this triangle in terms of P and x?
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  4. #4
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    Quote Originally Posted by Wilmer View Post
    To see where you're at:
    can you calculate the height of this triangle in terms of P and x?
    If he/she can calculate the height of the triangle in terms of P and x, that is solving it in terms of x, since x is the only variable in play; P is a fixed constant.

    Good day!
    -AB
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  5. #5
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    Hello, dartanian24!

    Consider an isosceles triangle of fixed perimeter $\displaystyle P$.

    a) If $\displaystyle x$ equals the length of one of the two equal sides,
    . . express the area $\displaystyle A$ as a function of $\displaystyle x\quad{\color{red}\hdots\;\text{ and }P}$
    Code:
                  A
                  *
                 /|\
                / | \
               /  |  \
            x /   |   \ x
             /    |h   \
            /     |     \
           /      |      \
        B *-------*-------* C
             Żb   D   Żb
          : - - - b - - - :
    Let $\displaystyle b$ = base of the triangle.
    Let $\displaystyle h$ = height of the triangle.

    Then: .$\displaystyle b + 2x \:=\:P \quad\Rightarrow\quad b \:=\:P-2x \;\;{\color{blue}[1]}\quad\Rightarrow\quad \frac{b}{2} \:=\:\frac{P - 2x}{2}\;\;{\color{blue}[2]}$

    In right triangle $\displaystyle ADB\!:\;\;h^2 + \left(\frac{b}{2}\right)^2 \:=\:x^2
    $

    Substitute [2]: .$\displaystyle h^2 + \left(\frac{P-2x}{2}\right)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 + \frac{P^2 - 4Px + 4x^2}{4} \:=\:x^2$

    Multiply by 4: .$\displaystyle 4h^2 + P^2 - 4Px + 4x^2 \:=\:4x^2 \quad\Rightarrow\quad h^2 \:=\:\frac{4Px-P^2}{4}$

    Hence: .$\displaystyle h \:=\:\frac{\sqrt{4Px-P^2}}{2}\;\;{\color{blue}[3]}$


    We know that: .$\displaystyle A \:=\:\tfrac{1}{2}bh$

    Subnstitute [1] and [3]: .$\displaystyle A \;=\;\tfrac{1}{2}(P-2x)\frac{\sqrt{4Px-P^2}}{2} $


    Therefore: .$\displaystyle A \;=\;\frac{1}{4}(P-2x)\sqrt{4Px-P^2}$



    b) What is the domain of $\displaystyle A$?
    You probably mean "range".


    The area has a minimum of 0 . . . when $\displaystyle b = 0.$


    The area is a maximum when the triangle is equilateral.

    . . Then: .$\displaystyle 3x \:=\:P \quad\Rightarrow\quad x \:=\:\frac{P}{3}$

    . . The area is: .$\displaystyle \frac{\sqrt{3}}{4}\left(\frac{P}{3}\right)^2 \;=\;\frac{\sqrt{3}}{36}P^2$


    The range is: .$\displaystyle \left[0,\:\frac{\sqrt{3}}{36}P^2\right]$

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