Question

find tu.
tu =

Explanation:

Step1: Use the mid - segment theorem

Since \(VT\parallel RS\), by the mid - segment theorem of a triangle (\(\frac{SU}{US + ST}=\frac{TU}{RS}\)), and because of the parallel lines and the ratio of segments. Here, \(SU = 20 - TU\) (assuming \(ST = TU\) from the parallel line and similar triangle properties, more precisely, using the basic proportionality theorem (Thales' theorem): \(\frac{SU}{ST}=\frac{RU}{VU}\). Given \(RU=42\), \(VU = 14\), so \(\frac{SU}{ST}=\frac{42}{14}=3\). Let \(ST=x\), \(SU = 20 - x\). Then \(\frac{20 - x}{x}=3\). But a simpler way is using the ratio of similar triangles. The two triangles \(\triangle VTU\) and \(\triangle VRS\) are similar. The ratio of their sides is \(\frac{VU}{RU}=\frac{14}{42}=\frac{1}{3}\). Also, \(SU = 20\), and if we consider the ratio of sides of similar triangles \(\triangle VTU\) and \(\triangle VRS\), we know that \(\frac{TU}{RS}\) is related to the ratio of the segments of the other two sides. Another approach: Since \(VT\parallel RS\), we have \(\frac{TU}{RS}=\frac{VU}{RU}\).

Step2: Calculate \(TU\)

We know that \(RS\) is not directly given, but using the property of similar triangles ( \(\triangle VTU\sim\triangle VRS\) ). The ratio of the sides of similar triangles is \(\frac{TU}{RS}=\frac{VU}{RU}\). Also, from the property of the line parallel to one side of a triangle ( \(VT\parallel RS\) ), we can use the formula \(TU=\frac{1}{3}\times20\) (wait, no, correct formula: Since \(\frac{TU}{RS}=\frac{VU}{RU}\), and \(RS\) is not needed. Let's use the formula for the length of \(TU\) based on the proportion. If we assume the big - triangle and the small - triangle. Let \(TU = x\), then \(SU=20 - x\). Since \(\frac{VU}{RU}=\frac{14}{42}=\frac{1}{3}\), and \(\frac{TU}{RS}=\frac{1}{3}\) (from similar triangles \(\triangle VTU\sim\triangle VRS\)). Also, using the formula \(TU=\frac{1}{3}\times20\) is wrong. Correct: Since \(VT\parallel RS\), we have \(\frac{TU}{20}=\frac{14}{42}\). Cross - multiply: \(42TU=14\times20\). Then \(TU=\frac{14\times20}{42}\). Simplify \(\frac{14\times20}{42}=\frac{20}{3}\approx6.67\) (wrong). Wait, correct:
Since \(VT\parallel RS\), by the basic proportionality theorem (Thales' theorem) in \(\triangle RSU\) (assuming \(R - V - U\) and \(S - T - U\)), we have \(\frac{TU}{20}=\frac{14}{42}\). Cross - multiply:

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No, wait, another way: The length of \(TU\):
Since \(VT\parallel RS\), the ratio of \(VU\) to \(RU\) is \(\frac{14}{42}=\frac{1}{3}\). And \(TU\) and \(20\) ( \(ST + TU=20\), assume \(ST = TU\) no, from the similar triangles \(\triangle VTU\sim\triangle VRS\) (incorrect similarity statement). Correct: \(\triangle VTU\) and \(\triangle VRS\) are not the right notation. Let's use the theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally.
In \(\triangle RSU\) (assuming \(R - V - U\) and \(S - T - U\)), \(\frac{VU}{RU}=\frac{TU}{SU}\). Let \(TU=x\), \(SU = 20 - x\), \(VU = 14\), \(RU=42\). Then \(\frac{14}{42}=\frac{x}{20 - x}\). Cross - multiply: \(14(20 - x)=42x\). Expand: \(280-14x = 42x\). Add \(14x\) to both sides: \(280=56x\). Solve for \(x\): \(x = 5\).

Answer:

\(5\)