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} $