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

#2**+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

#1**+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**+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 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