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?
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
thanks
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
and the middle chord is not the average of the other two chords.
Yes Shaun you are right
Shaun is right
Stop Cheating!
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To solve this problem, let's first draw a diagram.
1. Start by drawing a circle.
2. Draw two parallel chords, PQ and TU, intersecting the circle at points P, Q, R, S, T, and U.
3. Label PQ = 78 and TU = 50 as given in the problem.
Now, we can proceed to find the length of RS.
4. Since PQ and RS are parallel chords and intersect the same diameter of the circle, we can see that the distance between PQ and RS is equal to the distance between the centers of the chords.
5. Let O be the center of the circle. Draw perpendiculars from O to the chords PQ and RS, intersecting them at points X and Y, respectively.
6. Now we have two right triangles, OPX and OQY, where OP = 39 (half of PQ = 78/2) and OQ = 25 (half of TU = 50/2).
7. Given that the distance between PQ and RS is 4, we know that XY = 4.
8. Notice that PX and QY are each half of RS (RS = 2 * PX = 2 * OQ).
9. To find PX, we can use the Pythagorean theorem in triangle OPX. PX^2 = OP^2 - OX^2.
10. Given that OX = XY/2 = 2, we can calculate PX: PX^2 = 39^2 - 2^2.
11. Solving for PX, we find PX ≈ 38.909.
12. Finally, we can calculate RS = 2 * PX ≈ 2 * 38.909 ≈ 77.818.
Therefore, RS is approximately 77.818 units long.