Question

he segments $\overline{vw}$ and $\overline{vx}$ are tangent to the circle centered at $o$. given that $ow = 7.2$ and $vx = 13.5$, find $ov$.

Explanation:

Step1: Use the property of tangent to a circle

Since \( \overline{VW}\) and \( \overline{VX}\) are tangent to the circle, \(OW\perp VW\) (a line tangent to a circle is perpendicular to the radius at the point of tangency). So, \(\triangle OWV\) is a right - triangle. Also, \(VW = VX=13.5\) (tangents from the same external point to a circle are equal).

Step2: Apply the Pythagorean theorem

In right - triangle \(OWV\), by the Pythagorean theorem \(OV^{2}=OW^{2}+VW^{2}\). Given \(OW = 7.2\) and \(VW = 13.5\). Then \(OV^{2}=7.2^{2}+13.5^{2}\).
Calculate \(7.2^{2}=51.84\) and \(13.5^{2}=182.25\).
So \(OV^{2}=51.84 + 182.25=234.09\).
Take the square root of both sides: \(OV=\sqrt{234.09}\).
Since \(15.3\times15.3 = 234.09\), \(OV = 15.3\).

Answer:

\(15.3\)