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Thread: Right Triangle

  1. #1
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    Right Triangle

    A right triangle has one vertex on the graph of
    y = 9 - x^2, x > 0, at (x, y), another at the origin, and the thiurd on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x.
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  2. #2
    Member Henderson's Avatar
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    The base of your triangle is your x-coordinate, and the height of the triangle is your y-coordinate. For a triangle, the area is given by

    $\displaystyle A = \frac{1}{2}(base)(height)$, so

    $\displaystyle A = \frac{1}{2}(x)(y)$

    $\displaystyle A = \frac{1}{2}(x)(9-x^2)$

    $\displaystyle
    A = \frac{9x - x^3}{2}
    $
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  3. #3
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    Hello, magentarita!

    Henderson is absolutely correct!
    Did you make a sketch? . . . It's very simple.


    A right triangle has one vertex on the graph of $\displaystyle y \:=\: 9 - x^2,\;\;x > 0,\text{ at }(x, y),$
    another at the origin, and the third on the positive x-axis at (x, 0).
    Express the area $\displaystyle A$ of the triangle as a function of $\displaystyle x.$
    Code:
            |
          9 **
            |    *   P
            |       *(x,y)
            |     *:| *
            |   *:::|  *
            | *:::::|
          - + - - - + - * - -
            O       Q   3

    The area of a triangle is: .$\displaystyle A \;=\;\tfrac{1}{2}\text{(base)(height)} $

    The base is $\displaystyle x.$
    The height is $\displaystyle y$, where $\displaystyle y \:=\:9-x^2$

    Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}x(9-x^2)$

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  4. #4
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    great work....

    Quote Originally Posted by Henderson View Post
    The base of your triangle is your x-coordinate, and the height of the triangle is your y-coordinate. For a triangle, the area is given by

    $\displaystyle A = \frac{1}{2}(base)(height)$, so

    $\displaystyle A = \frac{1}{2}(x)(y)$

    $\displaystyle A = \frac{1}{2}(x)(9-x^2)$

    $\displaystyle
    A = \frac{9x - x^3}{2}
    $
    I thank you very much.
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  5. #5
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    Soroban....

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Henderson is absolutely correct!
    Did you make a sketch? . . . It's very simple.

    Code:
            |
          9 **
            |    *   P
            |       *(x,y)
            |     *:| *
            |   *:::|  *
            | *:::::|
          - + - - - + - * - -
            O       Q   3
    The area of a triangle is: .$\displaystyle A \;=\;\tfrac{1}{2}\text{(base)(height)} $

    The base is $\displaystyle x.$
    The height is $\displaystyle y$, where $\displaystyle y \:=\:9-x^2$

    Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}x(9-x^2)$
    It's always great to receive an answer from you.
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