Question

in the figure below, z is the center of the circle. suppose that st = 8, ru = 4, qu = 2x - 6, and uz = 10. find the following.

Explanation:

Step1: Use the property of chords

If a line from the center of a circle is perpendicular to a chord, it bisects the chord. So, \(RU = UQ\) and \(ST=2VT\). Since \(RU = 4\), then \(UQ=4\). Given \(UQ = 2x - 6\), we set up the equation \(2x-6=4\).

$$\begin{align*} 2x-6&=4\\ 2x&=4 + 6\\ 2x&=10\\ x&=5 \end{align*}$$

Step2: Use the Pythagorean theorem

In right - triangle \(RUZ\), \(RZ\) is the radius. Let \(RZ = r\), \(RU = 4\), \(UZ = 10\). In right - triangle \(VTZ\), \(ST = 8\), so \(VT=\frac{ST}{2}=4\). Since \(RZ = VZ\) (radii of the same circle).
By the Pythagorean theorem in \(\triangle RUZ\), \(r=\sqrt{RU^{2}+UZ^{2}}=\sqrt{4^{2}+10^{2}}=\sqrt{16 + 100}=\sqrt{116}\). But wait, no! Wait, actually, since \(RZ = VZ\) (radii) and using the property of perpendicular from center to chord. Wait, another approach:
Since \(ST\) and \(RQ\) are chords and \(ZT\perp ST\), \(ZW\perp RQ\). The length of \(VZ\) can be found using the fact that chords which are equidistant from the center are equal in length. Wait, no, better:
Since \(RU = 4\), \(UQ = 4\) (from step 1). In right - triangle \(RUZ\), if we consider the general formula for the distance from the center to a chord \(d\), and half - chord length \(l\), radius \(r=\sqrt{l^{2}+d^{2}}\).
For chord \(RQ\), half - chord length \(l = RU=4\), \(UZ = 10\) (distance from center to chord \(RQ\)). For chord \(ST\), half - chord length \(l'=\frac{ST}{2}=4\). Since the half - chord lengths are equal, the distance from the center to the chords \(VZ\) (distance from \(Z\) to \(ST\)) is equal to \(UZ\). So \(VZ = 10\).

Answer:

\(x = 5\), \(VZ=10\)