Derive formula for area of a circle from formula from area of rectangle I need to explain how to derive the formula for the area of a circle from the formula for the area of a rectangle. The area of a rectangle is length(width) and the formula for the area of a circle is $\pi r^2$
My original idea was to partition the circle into an even number of sections, say 6, and then cut one of the sections in half so we would have 7 sections. When you lay out the sections you will have a rectangle- but I can't seem to get the formula of a circle from that... any advice??
 A: you can deal with a semicircle then double it.
if radius is $r$ then, do as mjqxxxx suggests to get progressive approximations. the first approximation is a rectangle of base $2r$ and height $r$ so you may write:
$$
A_1 = r \times 2r = 2r^2
$$
a diagram (or a little thought) will reveal that with even subdivision of the height you will get 
$$
A_n = \sum_{k=0}^{n-1}\frac{r}{n} \times 2x_k
$$ 
where $x_k$ is the horizontal distance from the centre to the place where the $k^{th}$ horizontal meets the circle. from the fact that the equation of the circle is:
$$x^2 +y^2=r^2
$$
we know that 
$$
x_k = \sqrt{ r^2-(\frac{kr}{n})^2}
$$
to proceed further, introduce an auxiliary variable $t_k$, say, defined by
$$
t_k = \frac{kr}{n}
$$
 so that we may write 
$$\Delta t_k = t_{k+1} -t_k = \frac{r}{n}
$$
and the area formula above becomes
$$
A_n =  2 \sum_{k=0}^{n-1} \sqrt{ r^2-t_k^2} \; \Delta t_k
$$
you should now see where this is heading! and remember, you are treading in the footsteps of the great Archimedes! (look up "method of exhaustions")
A: Take these steps:
First, from the formula of area of rectangle derive the area of isosceles triangle (basically, splitting isosceles triangle in half and flipping over left half to complement the right half.
Second, partition the disk of radius $r$ into a large number $N$ of sectors by radii coming from the centre of the disk and landing at the circle at the same angle from each other. 
Next, notice that $N$ radii would create $N$ isosceles triangles with sides of length $r$ and base approximately equal $2\pi r/N$ because combining the bases of all $N$ triangles would form a polygon which fits very closely to the circle of length $2\pi$. 
Next, notice that because the triangles are very thing their hight is approximately equal to $r$, so each triangle area is $2\pi r\cdot r /2$, which makes disk area approximately $\pi r^2$.
Finally, make this rigorous by proving convergence.
