+99 What is the length of YX¯¯¯¯¯¯?
Enter your answer as a decimal in the box. Round your final answer to the nearest hundredth.
+2440 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. |
+128407
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
+2440 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. |
