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12. find sv. 10x - 41 4x + 7 17

Question

  1. find sv. 10x - 41 4x + 7 17

Explanation:

Step1: Use the property of tangent segments

Tangent segments from the same external point to a circle are equal. So, \(10x - 41=4x + 7\).

Step2: Solve the equation for \(x\)

Subtract \(4x\) from both sides: \(10x-4x - 41=4x-4x + 7\), which gives \(6x-41 = 7\).
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 \(ST\) or \(SU\)

Substitute \(x = 8\) into \(4x + 7\): \(4\times8+7=32 + 7=39\).

Step4: Use the Pythagorean theorem to find \(SV\)

Since \(TV = 17\) (radius) and \(ST = 39\) (tangent - segment), in right - triangle \(STV\) (a tangent to a circle is perpendicular to the radius at the point of tangency), by the Pythagorean theorem \(SV=\sqrt{ST^{2}+TV^{2}}\).
\(SV=\sqrt{39^{2}+17^{2}}=\sqrt{1521 + 289}=\sqrt{1810}\approx42.54\).

Answer:

\(SV\approx42.54\)