+33 Chords UV,WX andYZ of a circle are parallel. The distance between chords UV and WX is 2 and the distance between chords WX and YZ is also 2. If UV=38 and YZ=22 then find WX.
+370 Let O be the center of the circle, let d be the distance from O to UV, and let r be the radius of the circle (as shown in the diagram).
By the Pythagorean Theorem applied to right triangle OUV, we have: d^2+(2UV)^2=r^2 Substituting UV = 38, we get: d^2+(19)^2=r^2
Similarly, for right triangle OYZ, we have: (d+2)^2+(2YZ)^2=r^2
Substituting YZ = 22, we get: (d+2)^2+(11)^2=r^2
Now we can equate these two expressions for r2: d^2+(19)^2=(d+2)^2+(11)^2
Expanding the right side and simplifying, we get: d^2+361=d^2+4d+145
Solving for d, we find d=1/4.
Now we can substitute this value of d back into either equation for r2.
Using the first equation, we get: (41)^2+(19)^2=r^2 Solving for r, we find r=sqrt(361/16) = 19/4.
Finally, to find the length WX, we can use the Pythagorean Theorem applied to right triangle OWX: (d+1)^2+(2WX)^2=r^2
Substituting d=41 and r=419, we get: (41+1)^2+(2WX)^2=(419)^2
Solving for WX, we find: WX= sqrt((19/4)^2 - (5/4)^2) = sqrt(336/16)
Simplifying the radical gives WX=6*sqrt(6).
