1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

The perpendicular from the center on these chords will be a perpendicular bisector of the chord.

Below is an image with a roadmap.

enter image description here

Hope it helps!

$\endgroup$
  • 1
    $\begingroup$ Note that the chords can also be on the same "side" w.r.t the center. $\endgroup$ – Jean-Claude Arbaut May 19 at 22:35
  • $\begingroup$ Yes Jean, you're right. That didn't occur to me. $\endgroup$ – Vizag May 19 at 22:37
0
$\begingroup$

$$(c+d)^2 + (a/2)^2 = r^2 = d^2 + (b/2)^2$$

Solve for $d$.

enter image description here

$\endgroup$
0
$\begingroup$

Draw a diagram

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.