Question

$\overline{vw} \parallel \overline{ux}$. find $uv$.

$uv = \square$

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Explanation:

Step1: Identify Similar Triangles

Since \(\overline{VW} \parallel \overline{UX}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle TUX \sim \triangle TVW\) (similar triangles) because corresponding angles are equal (AA similarity: \(\angle T\) is common, and \(\angle TUX = \angle TVW\), \(\angle TXU = \angle TWV\) due to parallel lines).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \(\frac{TX}{TW} = \frac{TU}{TV}\). First, find \(TW = WX + XT = 30 + 33 = 63\), and \(TU = 22\), let \(UV = x\), so \(TV = TU + UV = 22 + x\). Wait, no, actually, the sides: \(TX = 33\), \(TW = 30 + 33 = 63\), \(TU = 22\), and \(TV = UV + 22\)? Wait, no, let's correct. The sides: in \(\triangle TUX\) and \(\triangle TVW\), \(TX\) corresponds to \(TW\), and \(TU\) corresponds to \(TV\)? Wait, no, maybe I mixed up. Wait, \(WX = 30\), \(XT = 33\), so \(WT = WX + XT = 63\). And \(UV\) is part of \(VT\), with \(UT = 22\). So the correct proportion: since \(UX \parallel VW\), then \(\frac{XT}{WT} = \frac{UT}{VT}\)? Wait, no, Thales' theorem: if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. Wait, here \(UX \parallel VW\), so in \(\triangle TVW\), line \(UX\) is parallel to \(VW\), intersecting \(TV\) at \(U\) and \(TW\) at \(X\). So by Thales' theorem, \(\frac{XW}{XT} = \frac{UV}{UT}\)? Wait, no, \(XW = 30\), \(XT = 33\), \(UT = 22\), \(UV =?\). Wait, Thales' theorem: \(\frac{XW}{XT} = \frac{UV}{UT}\)? Wait, no, the segments: \(TW\) is split into \(XW = 30\) and \(XT = 33\), and \(TV\) is split into \(UT = 22\) and \(UV =?\). So the correct proportion is \(\frac{XW}{XT} = \frac{UV}{UT}\)? Wait, no, Thales' theorem: \(\frac{XW}{WT} = \frac{UV}{VT}\)? No, let's recall: in \(\triangle TVW\), with \(UX \parallel VW\), then \(\frac{XT}{WT} = \frac{UT}{VT}\). So \(XT = 33\), \(WT = 63\), \(UT = 22\), \(VT = UV + 22\). So \(\frac{33}{63} = \frac{22}{22 + UV}\)? Wait, that can't be. Wait, maybe the other way: \(XW = 30\), \(XT = 33\), so \(XW/XT = 30/33 = 10/11\). Then, since \(UX \parallel VW\), the ratio of \(XW\) to \(XT\) should equal the ratio of \(UV\) to \(UT\)? Wait, no, let's do it correctly. Let's denote \(UV = x\). Then \(VT = x + 22\). The line \(UX\) is parallel to \(VW\), so \(\triangle TUX \sim \triangle TVW\) by AA similarity (since \(\angle T\) is common, and \(\angle TUX = \angle TVW\) because \(UX \parallel VW\), corresponding angles). Therefore, the ratio of corresponding sides is equal. So \(TX / TW = TU / TV\). \(TX = 33\), \(TW = 30 + 33 = 63\), \(TU = 22\), \(TV = x + 22\). So \(\frac{33}{63} = \frac{22}{x + 22}\). Cross-multiplying: \(33(x + 22) = 63 \times 22\). Then \(33x + 726 = 1386\). Subtract 726: \(33x = 1386 - 726 = 660\). Then \(x = 660 / 33 = 20\). Wait, that gives \(UV = 20\)? Wait, no, wait, maybe the similarity is reversed. Wait, \(\triangle TUX\) and \(\triangle TVW\): \(TX\) is 33, \(TW\) is 63; \(TU\) is 22, \(TV\) is \(UV + 22\). So ratio \(33/63 = 22/(UV + 22)\). Solving: \(33(UV + 22) = 63 \times 22\). \(33UV + 726 = 1386\). \(33UV = 660\). \(UV = 20\). Wait, but let's check with the other ratio: \(XW = 30\), \(XT = 33\), so \(XW/XT = 30/33 = 10/11\). Then \(UV/UT = x/22\). If \(x = 20\), then \(20/22 = 10/11\), which matches \(30/33 = 10/11\). Ah, there we go. So the correct proportion is \(XW/XT = UV/UT\), because \(UX \parallel VW\), so the segments on \(TW\) (XW and XT) are proportional to the segments on \(TV\) (UV and UT). So \(30/33 = UV/22\). Simplify \(30/33 = 10/11\), so \(10/11 = UV/22\). Then \(UV = (10/11) \times 22 = 20\). Yes, that's correct. So the proportion is \(XW / XT = UV / UT\), so \(30/33 = UV/22\), solving for UV: \(UV = (30 \times 22)/33 = (660)/33 = 20\).

Step1: Apply Thales' Theorem

Since \(UX \parallel VW\), by Thales' theorem (Basic Proportionality Theorem), \(\frac{WX}{XT} = \frac{UV}{UT}\).

Step2: Substitute Values

We know \(WX = 30\), \(XT = 33\), and \(UT = 22\). Substitute into the proportion:
\(\frac{30}{33} = \frac{UV}{22}\).

Step3: Solve for \(UV\)

Simplify \(\frac{30}{33} = \frac{10}{11}\), so:
\(\frac{10}{11} = \frac{UV}{22}\).
Multiply both sides by 22:
\(UV = \frac{10}{11} \times 22 = 20\).

Answer:

\(20\)