Question

$\overline{st} \parallel \overline{ru}$. find $tu$.

(image of a triangle with points t, u, q at the top, s at the bottom, r inside. lengths: tq = 30, sr = 34, sq = 51. $tu = \square$)

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{ST} \parallel \overline{RU}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle QRU \sim \triangle QST\) (similar triangles) because corresponding angles are equal (AA similarity criterion: \(\angle Q\) is common, and \(\angle QUR = \angle QTS\) as alternate interior angles due to parallel lines).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. Let \(TU = x\) and \(UQ = 30 - x\) (wait, actually, let's check the sides. The sides of the larger triangle (QST) have length 51, and the smaller triangle (QRU) has length 34? Wait, no, the sides from Q to S: QS is 51, and QR is 34? Wait, the diagram: Q to R is 34, R to S is... Wait, no, the sides: QS is 51, QR is 34, so the ratio of similarity is \(\frac{QR}{QS}=\frac{34}{51}=\frac{2}{3}\). Then, the corresponding sides \(UQ\) and \(TQ\)? Wait, no, \(TQ\) is 30? Wait, \(TQ\) is the top side, length 30. Wait, \(TU + UQ = TQ = 30\). Wait, no, actually, the similar triangles: \(\triangle QRU \sim \triangle QST\), so \(\frac{UQ}{TQ}=\frac{QR}{QS}\). Wait, no, \(TQ\) is the base of the larger triangle, \(UQ\) is the base of the smaller triangle. Wait, let's correct:

Let \(TQ = 30\) (the length from T to Q), \(UQ\) is the length from U to Q, and \(TU\) is what we need to find. The sides from Q to S: QS is 51, and QR is 34 (since R is on QS, so QR = 34, RS = QS - QR = 51 - 34 = 17? Wait, no, the diagram shows QS as 51, and QR as 34. So the ratio of similarity is \(\frac{QR}{QS}=\frac{34}{51}=\frac{2}{3}\). Therefore, the ratio of the corresponding bases (UQ and TQ) should also be \(\frac{2}{3}\). Wait, no, actually, \(\triangle QRU \sim \triangle QST\), so \(\frac{UQ}{TQ}=\frac{QR}{QS}\). Wait, \(TQ\) is 30, \(QR = 34\), \(QS = 51\). So \(\frac{UQ}{30}=\frac{34}{51}\). Let's compute \(\frac{34}{51}=\frac{2}{3}\), so \(UQ = 30 \times \frac{2}{3}=20\). Then, \(TU = TQ - UQ = 30 - 20 = 10\)? Wait, no, that can't be. Wait, maybe I mixed up the triangles. Wait, maybe \(\triangle SRU \parallel ST\), so \(\triangle SRU \sim \triangle SST\)? No, better: since \(ST \parallel RU\), \(\angle R = \angle S\) (alternate interior angles), and \(\angle U = \angle T\), so \(\triangle TRU \sim \triangle TST\)? No, let's use the correct similarity.

Wait, the correct approach: Since \(ST \parallel RU\), \(\triangle VRU \sim \triangle VST\) (where V is the common vertex at the bottom, S and T? Wait, no, the vertex is S? Wait, the diagram: T and Q are at the top, S is at the bottom. So the two triangles are \(\triangle QRU\) and \(\triangle QST\), with Q as the common vertex, RU parallel to ST. So the sides: QS is 51, QR is 34, so the ratio of similarity is \(\frac{QR}{QS}=\frac{34}{51}=\frac{2}{3}\). Therefore, the ratio of UQ to TQ is also \(\frac{2}{3}\). Wait, TQ is 30, so UQ = \(30 \times \frac{2}{3}=20\). Then, TU = TQ - UQ = 30 - 20 = 10? Wait, no, that would mean TU is 10, but let's check again. Wait, maybe the ratio is \(\frac{RS}{QS}=\frac{17}{51}=\frac{1}{3}\), but no, QR is 34, QS is 51, so QR/QS = 34/51 = 2/3. So the smaller triangle (QRU) has sides 2/3 of the larger triangle (QST). Therefore, the base UQ is 2/3 of TQ (30), so UQ = 20, so TU = TQ - UQ = 30 - 20 = 10? Wait, no, that seems off. Wait, maybe I got the triangles reversed. Maybe \(\triangle SRU \sim \triangle SST\), but no. Wait, let's use variables. Let TU = x, so UQ = 30 - x. Then, since \(\triangle QRU \sim \triangle QST\), \(\frac{UQ}{TQ}=\frac{QR}{QS}\). So \(\frac{30 - x}{30}=\frac{34}{51}\). Solve for x:

\(\frac{30 - x}{30}=\frac{2}{3}\)

Multiply both sides by 30: \(30 - x = 20\)

Subtract 20 from both sides: \(30 - 20 = x\)

So \(x = 10\)? Wait, no, that would mean TU is 10, but let's check the ratio again. Wait, QR is 34, QS is 51, so the ratio of similarity is 34/51 = 2/3, so the smaller triangle is 2/3 the size of the larger one. Therefore, the base UQ is 2/3 of TQ (30), so UQ = 20, so TU = TQ - UQ = 30 - 20 = 10. Wait, but that seems correct. Wait, maybe I made a mistake in the triangle labels. Alternatively, maybe the ratio is QR/QS = 34/51 = 2/3, so the corresponding side TU: wait, no, TU is part of TQ. Wait, TQ is 30, UQ is 20, so TU is 10. Wait, but let's verify with the ratio of the sides. If QS is 51, QR is 34, then the ratio is 34/51 = 2/3. So the height of the smaller triangle is 2/3 the height of the larger one, and the base UQ is 2/3 of TQ (30), so UQ = 20, so TU = 30 - 20 = 10. Yes, that makes sense.

Wait, no, wait: maybe the triangles are \(\triangle SRU\) and \(\triangle SST\), but no, the parallel lines are ST and RU, so the similar triangles are \(\triangle QRU\) and \(\triangle QST\) with Q as the common vertex. So the ratio of similarity is QR/QS = 34/51 = 2/3. Therefore, UQ/TQ = 2/3, so UQ = 2/3 * 30 = 20, so TU = TQ - UQ = 30 - 20 = 10. Wait, but let's check again. Wait, maybe the problem is that TU is the segment we need, and UQ is the other part. Wait, maybe I had it reversed. Let's think again: if ST is parallel to RU, then \(\triangle TRU \sim \triangle TST\)? No, better: the two triangles are similar, so the ratio of the sides from Q is 34/51 = 2/3, so the corresponding sides on the top (TQ) should have the same ratio. So UQ is 2/3 of TQ, so UQ = 20, so TU = TQ - UQ = 10. Yes, that seems correct.

Answer:

\(TU = \boxed{10}\)