The problem with that reasoning is that when showing a rectangle's area by dividing the rectangle's length and height into units, the length and height units form squares. Rotating the radius around the circle by a set number of degrees to divide the circle does not produce squares; these slices are more like triangles. Recall that the area of a triangle is $\dfrac{bh}{2}$; the math's a bit more complex but you can draw a parallel here to the area of a circle versus a rectangle. All other things being equal, the rectangle is double the triangle's area, and so the area of a shape divided into a number of congruent triangles will be half the area of a shape divided into the same number of rectangles of the same length and height as the triangles.
The actual math to prove the area of a circle is very closely related to this, but incorporates an additional concept from calculus:
For any arbitrary $n$, draw an $n$-gon (hexagon, octagon, hectogon, etc) around a circle of radius $r$. Each side of this shape will have length $s$, and $s*n > 2\pi r$; the perimeter of the n-gon will be greater than the circumference of the circle (recall that $\pi = \dfrac{c}{d}, d=2r \therefore c=2\pi r$). However, as $n$ increases, $s$ decreases, and the perimeter of the n-gon will approach the circumference of the circle. It never quite gets there for any finite n, but it gets close enough, allowing us to define what's known in calculus as a limit: $\lim_{n\to \infty}ns = 2\pi r$.
Now, for each side of the n-gon, we can define an isoceles triangle between the vertices of the side and the center of the circle. The symmetrical sides of this triangle have length $l$ which is $>r$ (because the line connects to the vertex of the n-gon, outside the circle) but, similar to the way $ns$ approaches $2\pi r$, $l$ approaches $r$ as $n\to \infty$. This isoceles triangle of base $s$ and height $r$ can be split into two right triangles with base $s/2$ and height $r$. The area of a right triangle is $\dfrac{bh}{2}$ as previously stated, and so the area of the isoceles triangle is two of these, or $2*\dfrac{\dfrac{s}{2}r}{2}= \dfrac{sr}{2}$. There are $n$ of these triangles, one per side, so the area of the n-gon is $\dfrac{nsr}{2}$. Finally recall our limit; as $n\to\infty$, $ns\to 2\pi r$. The limit allows us to equate these two in the general case we are considering, where for our purposes $n=\infty$; $ns=2\pi r$. Plug that into the area of the n-gon, and behold: $\dfrac{nsr}{2} = \dfrac{(2\pi r)r}{2} = \dfrac{2\pi r^2}{2} = \pi r^2$.
QED.