Question

1. find sv. 10x - 41 4x + 7

Explanation:

Step1: Set the two tangent lengths equal

According to the tangent - segment theorem, if two tangent segments are drawn to a circle from the same external point, then the two tangent segments are congruent. So, \(10x - 41=4x + 7\).

Step2: Solve the equation for \(x\)

Subtract \(4x\) from both sides: \(10x-4x - 41=4x-4x + 7\), which simplifies to \(6x-41 = 7\).
Add \(41\) to both sides: \(6x-41 + 41=7 + 41\), so \(6x=48\).
Divide both sides by \(6\): \(x=\frac{48}{6}=8\).

Step3: Find the length of \(SV\)

First, find the length of \(ST\) (or \(SU\)): Substitute \(x = 8\) into \(10x-41\) (or \(4x + 7\)). Using \(10x-41\), we get \(10\times8-41=80 - 41=39\).
Since \(FT\perp ST\) (tangent to a circle is perpendicular to the radius at the point of tangency), and \(FT = 17\).
By the Pythagorean theorem in right - triangle \(SFT\), \(SV=\sqrt{ST^{2}+FT^{2}}\).
Substitute \(ST = 39\) and \(FT = 17\): \(SV=\sqrt{39^{2}+17^{2}}=\sqrt{1521 + 289}=\sqrt{1810}\approx42.54\).

Answer:

\(SV\approx42.54\)