Question

gent to the circle centered at o. given that ol = 12 and km = 12.6, find ok.
ok =

Explanation:

Step1: Find the length of \(LM\)

Since \(L\) is the mid - point of \(KM\) (a perpendicular from the center of a circle to a chord bisects the chord), \(LM=\frac{KM}{2}\).
Given \(KM = 12.6\), then \(LM=\frac{12.6}{2}=6.3\).

Step2: Use the Pythagorean theorem in right - triangle \(OLM\)

In right - triangle \(OLM\), by the Pythagorean theorem \(OK^{2}=OL^{2}+LM^{2}\) (where \(OK\) is the radius of the circle, \(OL\) is the distance from the center to the chord and \(LM\) is half of the chord length).
We know \(OL = 12\) and \(LM=6.3\).
Substitute the values into the formula: \(OK^{2}=12^{2}+6.3^{2}\).
First, calculate \(12^{2}=144\) and \(6.3^{2}=39.69\).
Then \(OK^{2}=144 + 39.69=183.69\).
Take the square root of both sides: \(OK=\sqrt{183.69}\).
Since \(\sqrt{183.69}=13.5\) (because \(13.5\times13.5 = 182.25\) and \(13.5^{2}=183.69\)).

Answer:

\(13.5\)