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

**realintegerz** · October 28th 2008, 06:01 PM

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?

**Soroban** · October 28th 2008, 07:02 PM

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

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

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

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

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

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