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Chords \(\overline{UV}, \overline{WX},\) and \(\overline{YZ}\) of a circle are parallel. The distance between chords \(\overline{UV}\) and \(\overline{WX}\) is 4, and the distance between chords \(\overline{WX} \) and \(\overline{YZ}\) is also 4. If UV = 78 and YZ = 50, then find WX.

 Aug 13, 2020
 #1
avatar+25556 
+3

Chords \(\overline{UV}\), \(\overline{WX}\), and \(\overline{YZ}\) of a circle are parallel.
The distance between chords \(\overline{UV}\) and \(\overline{WX}\) is 4,
and the distance between chords \(\overline{WX}\) and \(\overline{YZ}\) is also 4.
If \(\overline{UV}\) = 78 and \(\overline{YZ}\) = 50, then find \(\overline{WX}\).

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{r} &=& \mathbf{a+b+8} \\ \hline & \mathbf{39^2} &=& \mathbf{(b+8)(a+r)} \\ & 39^2 &=& (b+8)(a+a+b+8) \\ (2) & \mathbf{39^2} &=& \mathbf{(b+8)(2a+b+8)} \\ \hline & \mathbf{25^2} &=& \mathbf{b(8+a+r)} \\ & 25^2 &=& b(8+a+a+b+8) \\ (3) & \mathbf{25^2} &=& \mathbf{b(2a+b+16)} \\ \hline & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (b+4)(4+a+r) \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (b+4)(4+a+a+b+8) \\ (4) & \mathbf{\left(\dfrac{\overline{WX}}{2}\right)^2} &=& \mathbf{(b+4)(2a+b+12)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (3): & \mathbf{25^2} &=& \mathbf{b(2a+b+16)} \\ & 625 &=& b(2a+b+16) \\ & 625 &=& 2ab+b(b+16) \\ & 2ab &=& 625- b(b+16) \\ & \mathbf{a} &=& \mathbf{\dfrac{625- b(b+16)}{2b}} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (2): & \mathbf{39^2} &=& \mathbf{(b+8)(2a+b+8)} \\ & 1521 &=& (b+8)(2a+b+8) \quad | \quad \mathbf{a=\dfrac{625- b(b+16)}{2b}} \\ & 1521 &=& (b+8)\left(2\dfrac{\Big(625- b(b+16)\Big)}{2b}+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{ 625- b(b+16) }{b}+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{625}{b}- (b+16)+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{625}{b}-8 \right) \\ & 1521 &=& (b+8)\dfrac{(625-8b)}{b} \\ & 1521b &=& (b+8)( 625-8b) \\ & 1521b &=& 625b-8b^2+8*625-64b \\ & 8b^2 + 960b - 5000 &=& 0 \quad | \quad : 8 \\ & \mathbf{b^2 + 120b - 625} &=& \mathbf{0} \\ \hline &b &=& \dfrac{-120\pm \sqrt{120^2-4*(-625) } }{2} \\ &b &=& \dfrac{-120\pm \sqrt{16900} }{2} \\ &b &=& \dfrac{-120\pm 130 }{2} \\\\ &\mathbf{b} &=& \mathbf{\dfrac{-120+ 130 }{2}} \\ &\mathbf{b} &=& \mathbf{5} \\\\ &\mathbf{b} &=& \mathbf{\dfrac{-120- 130 }{2}} \quad | \quad \text{no solution } b < 0!\\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{\dfrac{625- b(b+16)}{2b}} \quad | \quad \mathbf{b=5} \\\\ a &=& \dfrac{625- 5(5+16)}{2*5} \\\\ a &=& \dfrac{625- 5*21}{10} \\\\ a &=& \dfrac{625- 105}{10} \\\\ a &=& \dfrac{520}{10} \\\\ \mathbf{a} &=& \mathbf{52} \\ \hline \end{array}\)

 

\(\mathbf{\overline{WX}=\ ?}\)

\(\begin{array}{|rcll|} \hline (4): & \mathbf{\left(\dfrac{\overline{WX}}{2}\right)^2} &=& \mathbf{(b+4)(2a+b+12)} \quad | \quad a=52,\ b=5 \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (5+4)(2*52+5+12) \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& 9*121 \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& 1089 \\\\ & \dfrac{\overline{WX}}{2} &=& \sqrt{1089} \\\\ & \dfrac{\overline{WX}}{2} &=& 33 \\\\ &\mathbf{\overline{WX}} &=& \mathbf{66} \\ \hline \end{array}\)

 

laugh

 Aug 13, 2020
 #2
avatar+1326 
+4

This is amazing:  r = ( UY / 2 )2  surprise

 

So far I have only this:  r = 65

 

And now I have:   WX = 2*[sqrt( 652 - 562 )] = 66  laugh

 

 Aug 13, 2020
edited by Dragan  Aug 13, 2020
edited by Dragan  Aug 20, 2020

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