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0
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avatar+92 

What is the length of YX¯¯¯¯¯¯?

 

Enter your answer as a decimal in the box. Round your final answer to the nearest hundredth.

CrazyDaizy  Jun 6, 2017

Best Answer 

 #2
avatar+814 
+1

This problem can be solved with a theorem known as secant-tangent product theorem. I'll illustrate this theorem with a table:
 

Theorem in WordsDiagramConclusion
If a tangent and secant intersect in the exterior of a circle, then the product of the lengths of the secant segment and its external segments equals the length of the tangent segment squared.\(BD*ED=AD^2\)
 Secant \(\overline{BD}\) and tangent \(\overline{AD}\) intersect at point D.  

 

Let's apply this theorem now:

 

\(YZ*YV=YX^2\)Valid because of the secant-tangent product theorem. Plug in the values that correspond to each length and find the unknown, YX.
\((9+19)*9=YX^2\)Simplify the left hand side of the equation
\(\sqrt{252}=\sqrt{YX^2}\)Do the square root of both sides to get rid of the exponent and isolate YX.
\(|YX|=\sqrt{252}\)Simplify the radical by finding the highest factor that is also a perfect square
\(|YX|=\sqrt{36*7}=6\sqrt{7}\)Get rid of the absolute value sign by dividing your answer into the positive and negative answer
\(YX=\pm6\sqrt{7}\)Of course, reject the negative answer as a negative sidelength is nonsensical in the context of geometry. This is your final answer in simplest radical form, too.
\(YX=6\sqrt{7}m\approx15.87m\)I also included the decimal approximation to the hundrendth place.
  
  
  
TheXSquaredFactor  Jun 6, 2017
edited by TheXSquaredFactor  Jun 6, 2017
edited by TheXSquaredFactor  Jun 7, 2017
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2+0 Answers

 #1
avatar+75316 
+2

 

We have the secant-tangent theorem

 

So

 

ZY * VY  =  XY^2

 

(19 + 9) (9)  = XY^2

 

(28) (9)  = XY^2

 

(7)(4)(9)  = XY^2

 

7(36)  = XY^2

 

(36)(7)  = XY^2       take the sq rt of both sides

 

6√7   =  XY  ≈  15.87 m

 

 

 

cool cool cool

CPhill  Jun 6, 2017
 #2
avatar+814 
+1
Best Answer

This problem can be solved with a theorem known as secant-tangent product theorem. I'll illustrate this theorem with a table:
 

Theorem in WordsDiagramConclusion
If a tangent and secant intersect in the exterior of a circle, then the product of the lengths of the secant segment and its external segments equals the length of the tangent segment squared.\(BD*ED=AD^2\)
 Secant \(\overline{BD}\) and tangent \(\overline{AD}\) intersect at point D.  

 

Let's apply this theorem now:

 

\(YZ*YV=YX^2\)Valid because of the secant-tangent product theorem. Plug in the values that correspond to each length and find the unknown, YX.
\((9+19)*9=YX^2\)Simplify the left hand side of the equation
\(\sqrt{252}=\sqrt{YX^2}\)Do the square root of both sides to get rid of the exponent and isolate YX.
\(|YX|=\sqrt{252}\)Simplify the radical by finding the highest factor that is also a perfect square
\(|YX|=\sqrt{36*7}=6\sqrt{7}\)Get rid of the absolute value sign by dividing your answer into the positive and negative answer
\(YX=\pm6\sqrt{7}\)Of course, reject the negative answer as a negative sidelength is nonsensical in the context of geometry. This is your final answer in simplest radical form, too.
\(YX=6\sqrt{7}m\approx15.87m\)I also included the decimal approximation to the hundrendth place.
  
  
  
TheXSquaredFactor  Jun 6, 2017
edited by TheXSquaredFactor  Jun 6, 2017
edited by TheXSquaredFactor  Jun 7, 2017

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