Hello, frankinaround!

Here is yet another approach . . .

$\displaystyle \text{The equal sides of an isosceles triangle are 2.}$

$\displaystyle \text{If }x\text{ is the base, express the area as a function of }x.$

Code:

*
/ \
/ @ \
/ \
2 / \ 2
/ \
/ \
/ \
* - - - - - - - *
x

Let $\displaystyle \,\theta$ = the vertex angle.

Law of Cosines: .$\displaystyle \cos\theta \:=\:\dfrac{2^2+2^2 - x^2}{2(2)(2)} \;=\;\dfrac{8-x^2}{8}$

Using $\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1$, we find that: .$\displaystyle \sin\theta \:=\:\dfrac{x\sqrt{16-x^2}}{8}$

The area of a triangle is: .$\displaystyle A \;=\;\frac{1}{2}ab\sin C$

. . One-half the product of two sides times the sine of the included angle.

Therefore: .$\displaystyle A \;=\;\frac{1}{2}(2)(2)\dfrac{x\sqrt{16-x^2}}{8} \;=\;\frac{1}{4}x\sqrt{16-x^2}$