Friday

April 25, 2014

April 25, 2014

Posted by **Knights** on Thursday, March 21, 2013 at 9:01pm.

how to do this? Draw some lines?

- Circle Geometry - chords in a circle -
**Reiny**, Thursday, March 21, 2013 at 11:47pmPTUQ 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 7:04amthanks

- Circle Geometry - chords in a circle -
**Shaun**, Friday, March 29, 2013 at 8:54amhmm 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 -
**Shaun**, Friday, March 29, 2013 at 8:54amhmm 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 -
**Shaun**, Friday, March 29, 2013 at 8:59amand the middle chord is not the average of the other two chords.

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