Question

$\overline{vw} \parallel \overline{su}$. find $st$.
(image of a triangle with segment lengths: vt total 60, vu 40, and the side with length 48, needing to find st)

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{VW} \parallel \overline{SU}\), triangles \(VTW\) and \(STU\) are similar by the Basic Proportionality Theorem (Thales' theorem). So, the ratios of corresponding sides are equal.

Step2: Define Variables and Set Proportion

Let \(ST = x\). Then \(SW = 48 - x\)? Wait, no, actually, the side \(VT = 60\), \(VU = 40\), so \(UT = VT - VU = 60 - 40 = 20\)? Wait, no, looking at the diagram: \(VT\) is the top side with length 60, \(VU = 40\), so \(UT = 60 - 40 = 20\)? Wait, no, maybe the sides: the large triangle has side \(VT = 60\), and the small triangle has \(UT\)? Wait, no, let's re - examine. The two parallel lines imply \(\triangle STU \sim \triangle VTW\) (by AA similarity, since \(\angle T\) is common and \(\angle TSU=\angle TWV\) because \(SU\parallel VW\)). So, the ratio of \(UT\) to \(VT\) should equal the ratio of \(ST\) to \(WT\). Wait, \(WT = 48\)? Wait, the length from \(S\) to \(W\) is 48? Wait, the diagram shows a side with length 48, probably \(WT = 48\), and we need to find \(ST\). Let's denote \(ST = x\), then \(SW=48 - x\)? No, maybe the sides: \(VT = 60\), \(VU = 40\), so \(UT=VT - VU = 20\). Wait, no, maybe the ratio of the sides: since \(\triangle STU\sim\triangle VTW\), then \(\frac{UT}{VT}=\frac{ST}{WT}\). Wait, \(UT = 60 - 40=20\), \(VT = 60\), \(WT = 48\), and \(ST=x\). So \(\frac{20}{60}=\frac{x}{48}\)? No, that doesn't seem right. Wait, maybe I got the similar triangles wrong. Let's see: \(\overline{SU}\parallel\overline{VW}\), so \(\triangle TSU\sim\triangle TWV\) (corresponding angles are equal). So, the ratio of \(TS\) to \(TW\) is equal to the ratio of \(TU\) to \(TV\). \(TU = VT - VU=60 - 40 = 20\), \(TV = 60\), \(TW = 48\), and \(TS=x\). So \(\frac{x}{48}=\frac{20}{60}\). Wait, \(\frac{20}{60}=\frac{1}{3}\), so \(x=\frac{48}{3}=16\)? No, that can't be. Wait, maybe the other way: \(\frac{VU}{VT}=\frac{SW}{WT}\). \(VU = 40\), \(VT = 60\), \(WT = 48\), let \(ST=x\), then \(SW = 48 - x\). So \(\frac{40}{60}=\frac{48 - x}{48}\). Cross - multiply: \(40\times48=60\times(48 - x)\). \(1920 = 2880-60x\). \(60x=2880 - 1920\). \(60x = 960\). \(x = 16\)? No, that's \(SW\). Wait, we need \(ST\). Since \(WT=ST + SW\), \(WT = 48\), \(SW = 16\), then \(ST=48 - 16 = 32\)? Wait, no, let's do it correctly. If \(\triangle TSU\sim\triangle TWV\), then \(\frac{TS}{TW}=\frac{TU}{TV}\). \(TU = VT - VU=60 - 40 = 20\), \(TV = 60\), \(TW = 48\), \(TS=x\). So \(\frac{x}{48}=\frac{20}{60}\). Simplify \(\frac{20}{60}=\frac{1}{3}\), so \(x=\frac{48}{3}=16\). No, that's not correct. Wait, maybe the ratio is \(\frac{VU}{VT}=\frac{ST}{WT}\). \(VU = 40\), \(VT = 60\), \(WT = 48\), so \(\frac{40}{60}=\frac{ST}{48}\). Cross - multiply: \(60\times ST=40\times48\). \(60ST = 1920\). \(ST=\frac{1920}{60}=32\). Ah, that makes sense. Because if \(VU = 40\) and \(VT = 60\), the ratio of similarity is \(\frac{40}{60}=\frac{2}{3}\). Then \(ST\) should be \(\frac{2}{3}\) of \(WT\)? Wait, no, \(WT = 48\), \(\frac{2}{3}\times48 = 32\). Yes, that works. So the proportion is \(\frac{VU}{VT}=\frac{ST}{WT}\) because of similar triangles (since \(SU\parallel VW\), the triangles are similar and the sides are proportional).

Step1: Set Up Proportion from Similar Triangles

Since \(\overline{SU}\parallel\overline{VW}\), \(\triangle TSU\sim\triangle TWV\) (AA similarity). Thus, \(\frac{VU}{VT}=\frac{ST}{WT}\). We know \(VU = 40\), \(VT = 60\), and \(WT = 48\). Let \(ST=x\). So we have the proportion \(\frac{40}{60}=\frac{x}{48}\).

Step2: Solve the Proportion

Cross - multiply: \(60x=40\times48\).
First, calculate \(40\times48 = 1920\).
Then, \(x=\frac{1920}{60}\).
Simplify \(\frac{1920}{60}=32\).

Answer:

\(32\)