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

• Oct 28th 2008, 06:01 PM
realintegerz
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?
• Oct 28th 2008, 07:02 PM
Soroban
Hello, realintegerz!

Quote:

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