# Circle Geometry - chords in a circle

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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 -

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 -

thanks

• Circle Geometry - chords in a circle -

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=(8-k)^2+25^2 and hence k=-52. After that r = sqrt(60^2 + 25^2)=65. From there sqrt(65^2-56^2)=33

• Circle Geometry - chords in a circle -

and the middle chord is not the average of the other two chords.

• Circle Geometry - chords in a circle -

Yes Shaun you are right

• Circle Geometry - chords in a circle -

Shaun is right

• Circle Geometry - chords in a circle -

Stop Cheating!

• Circle Geometry - chords in a circle -

Please post any questions that you have on the message board.

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