+0

# i cant remember how to figure this out ugghh help?

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198
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+98

What is the length of YX¯¯¯¯¯¯?

CrazyDaizy  Jun 6, 2017

#2
+1602
+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 Words Diagram Conclusion 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
Sort:

#1
+81027
+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

CPhill  Jun 6, 2017
#2
+1602
+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 Words Diagram Conclusion 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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