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.
$\endgroup$3 Answers
$\begingroup$The perpendicular from the center on these chords will be a perpendicular bisector of the chord.
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Hope it helps!
$\endgroup$ 2 $\begingroup$$$(c+d)^2 + (a/2)^2 = r^2 = d^2 + (b/2)^2$$
Solve for $d$.
Draw a diagram
and you may get $\left| \sqrt{r^2-\left(\frac a2\right)^2} \pm \sqrt{r^2-\left(\frac b2\right)^2} \right|=c$
There is plenty of scope for spurious solutions to be introduced here if you try to solve for $r$
If you actually want to find the centre of the circle by geometric construction, take the perpendicular bisectors of the sides of the isosceles trapezium (trapezoid if you are American). This should demonstrate that there is one non-spurious solution
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