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Thread: Area of a triangle as a function of length of shortest side of triangle

  1. #1
    Member realintegerz's Avatar
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    Area of a triangle as a function of length of shortest side of triangle

    A triangle is incribed in a semicircle of radius 1, two of the vertices are on opposite sides of a diameter and the third vertex is somewhere on the semicircle. Find a rule that describes the area of the triangle as a function of the length of the shortest side of a triangle

    How do I do this?
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  2. #2
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    Hello, realintegerz!

    A triangle is incribed in a semicircle of radius 1,
    two of the vertices are on the ends of the diameter
    and the third vertex is somewhere on the semicircle.

    Find a rule that describes the area of the triangle as a function
    of the length of the shortest side of the triangle.
    Code:
                  * * *    C
              *           *
    . . .   *         *    \*
    . . .  *      *        a\*
    . . .     *              \*
    . . A * - - - - + - - - - * B
    . . .      1    O    1

    $\displaystyle \Delta ABC$ is inscribed in a semicircle
    . . with center $\displaystyle O$ and radius $\displaystyle OA = OB = 1.$
    Let the shortest side be $\displaystyle a.$

    Since $\displaystyle ABC$ is a right triangle: .$\displaystyle AC^2 + BC^2 \:=\:AB^2$
    . . $\displaystyle AC^2 + a^2 \:=\:2^2\quad\Rightarrow\quad AC^2 \:=\:4 - a^2 \quad\Rightarrow\quad AC \:=\:\sqrt{4-a^2}$

    The area of the triangle is: .$\displaystyle A \;=\;\tfrac{1}{2}(BC)(AC)$

    Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}a\sqrt{4-a^2} $

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