Posted by Knights on Thursday, March 21, 2013 at 9:01pm.
Let PQ, RS , and TU be parallel chords of a circle. The distance between chords PQ and RS is 4, and the distance between chords RS and TU is also 4. If PQ = 78 TU=50 , then find RS.
how to do this? Draw some lines?

Circle Geometry  chords in a circle  Reiny, Thursday, March 21, 2013 at 11:47pm
PTUQ is a cyclic quadrilateral, so opposite angles are supplementary
let angle P be x , let angle U be y
then x+y = 180
also since PQ  TU
angle P + angle Q = 180
so angle Q = x also ,since Q + U = 180
Extend PT and QU to meet at V
I see three similar triangles:
VTU , VRS , and VPQ
let the height of VTU be h
h/50 = (h+8)/78
78h = 50h + 400
28h = 400
h = 400/28 = 100/7
h/50 = (h+4)/RS
h(RS) =50h + 200
RS(100/7) = 50(100/7) + 200
times 7
100RS = 5000 + 1400
100RS = 6400
RS = 64
which was a long and windy way to show it was simply the average of the other two chords

Circle Geometry  chords in a circle  Knights, Friday, March 22, 2013 at 7:04am
thanks

Circle Geometry  chords in a circle  Shaun, Friday, March 29, 2013 at 8:54am
hmm I got 66. I got my radius to be 65 and from there i proceede to solve and got 2(33)=66. My method is that the radius from point O the origin perpendicular bisects the three parallel chords. Using pythagoras theorem, let the distance of perpendicular bisector from O to TU be k. Then,
k^2+36^2=(8k)^2+25^2 and hence k=52. After that r = sqrt(60^2 + 25^2)=65. From there sqrt(65^256^2)=33

Circle Geometry  chords in a circle  Shaun, Friday, March 29, 2013 at 8:59am
and the middle chord is not the average of the other two chords.

Circle Geometry  chords in a circle  Jerrrrrrry, Saturday, August 22, 2015 at 5:37pm
Yes Shaun you are right

Circle Geometry  chords in a circle  saaaaaaaaaaaa, Monday, August 24, 2015 at 7:07pm
Shaun is right
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