Suppose a circle has two parallel chords of lengths $a$ and $b$, and the chords are separated by a distance of $c$. Using only the usual high school geometry theorems (i.e. no trig or calculus), can we derive a formula for the radius?
I've tried drawing radii in several places without making useful progress. I can't see how to find the radius intersecting the circle and a chord. I can draw the segment from one chord-circle intersection to another, but it need not pass through the center so I can't leverage this to get the radius.
Best Answer
I just answered this question in chat, so I thought it would be worth posting the answer here.
The horizontal position of a vertical chord of length $c$ in a circle of radius $r$ would be $\pm\sqrt{r^2-\frac{c^2}4}$.
Thus, the distance, $d$, between chords of length $c_1$ and $c_2$ (with $c_1\ge c_2$) would be $$ d=\sqrt{r^2-\tfrac{c_2^2}4}\pm\sqrt{r^2-\tfrac{c_1^2}4}\tag1 $$ The "$\pm$" depends on whether or not the chords appear on the same side of the center of the circle. In either case, we have $$ \frac{c_1^2-c_2^2}{4d}=\sqrt{r^2-\tfrac{c_2^2}4}\mp\sqrt{r^2-\tfrac{c_1^2}4}\tag2 $$ To verify $(2)$, simply multiply $(1)$ and $(2)$.
Average $(1)$ and $(2)$ to get $$ \frac{c_1^2-c_2^2+4d^2}{8d}=\sqrt{r^2-\tfrac{c_2^2}4}\tag3 $$ Thus, $$ \begin{align} r^2 &=\frac{\left(c_1^2-c_2^2+4d^2\right)^2+16c_2^2d^2}{64d^2}\tag{4a}\\ &=\frac{c_1^4+c_2^4+16d^4-2c_1^2c_2^2+8c_1^2d^2+8c_2^2d^2}{64d^2}\tag{4b}\\ &=\frac{\left(c_1^2+c_2^2\right)^2+8\left(c_2^2+c_1^2\right)d^2+16d^4-4c_1^2c_2^2}{64d^2}\tag{4c}\\ &=\frac{\left(c_1^2+c_2^2+4d^2\right)^2-4c_1^2c_2^2}{64d^2}\tag{4d}\\ &=\frac{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}{64d^2}\tag{4e} \end{align} $$ Explanation:
$\text{(4a):}$ solve $(3)$ for $r^2$
$\text{(4b):}$ expand
$\text{(4c):}$ collect
$\text{(4d):}$ collect
$\text{(4e):}$ apply $a^2-b^2=(a+b)(a-b)$
Therefore, $$ r=\frac{\sqrt{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}}{8d}\tag5 $$ In $(5)$, it doesn't matter whether $c_1\ge c_2$ or not.