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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.

 

 Jun 26, 2024
 #1
avatar+397 
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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).

 Jun 26, 2024
 #2
avatar+33 
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WX= 6sqrt6 is impossible because WX is bigger than YZ and YZ = 22

jumbean  Jun 26, 2024

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