Question

$\overline{tu} \parallel \overline{sv}$. find $tw$.
u
6
v
12
w
s
14
t
$tw = \square$

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{TU} \parallel \overline{SV}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle TUV \sim \triangle SWV\) (similar triangles). So the ratios of corresponding sides are equal. Let \(TW = x\), then \(TS = 14\), so \(WS = x - 14\)? Wait, no, wait. Wait, the segments on the side: \(UV = 6\), \(VW = 12\), so \(UW = UV + VW = 6 + 12 = 18\)? Wait, no, actually, the sides are \(VW = 12\), \(UV = 6\), and on the base, \(TS = 14\), \(TW = x\), so \(WS = x - 14\)? Wait, no, maybe I got the segments wrong. Wait, the triangle: vertex \(U\), \(V\) is on \(UW\), \(S\) is on \(TW\). So \(\overline{SV} \parallel \overline{TU}\), so \(\triangle VWS \sim \triangle UWT\) (by AA similarity, since \(\angle VWS = \angle UWT\) (common angle) and \(\angle VSW = \angle UTW\) (corresponding angles because \(SV \parallel TU\))). So the ratio of \(VW\) to \(UW\) should equal the ratio of \(WS\) to \(WT\).

Wait, \(VW = 12\), \(UW = UV + VW = 6 + 12 = 18\)? No, wait, \(UV = 6\), \(VW = 12\), so \(UW = UV + VW = 18\)? Wait, no, \(U\) to \(V\) is 6, \(V\) to \(W\) is 12, so \(U\) to \(W\) is 18. Then \(TS = 14\), let \(TW = x\), so \(WS = x - 14\)? Wait, no, \(TW\) is the entire base, \(TS = 14\), so \(WS = TW - TS = x - 14\). Then by similar triangles, \(\frac{VW}{UW} = \frac{WS}{TW}\)? Wait, no, \(\triangle VWS \sim \triangle UWT\), so corresponding sides: \(VW\) corresponds to \(UW\), \(WS\) corresponds to \(WT\), \(SV\) corresponds to \(TU\). So \(\frac{VW}{UW} = \frac{WS}{WT}\). Wait, \(VW = 12\), \(UW = UV + VW = 6 + 12 = 18\)? No, \(UV\) is from \(U\) to \(V\), \(VW\) is from \(V\) to \(W\), so \(UW\) is from \(U\) to \(W\), so \(UW = UV + VW = 6 + 12 = 18\). Then \(WS = TW - TS = x - 14\), and \(WT = x\). So \(\frac{VW}{UW} = \frac{WS}{WT}\) → \(\frac{12}{18} = \frac{x - 14}{x}\)? Wait, that doesn't seem right. Wait, maybe the ratio is \(\frac{VW}{UV} = \frac{WS}{TS}\)? Wait, no, let's re-examine.

Wait, \(\overline{SV} \parallel \overline{TU}\), so the triangles \(\triangle TUV\) and \(\triangle SVW\) are similar? Wait, \(V\) is on \(UW\), \(S\) is on \(TW\). So \(\angle T = \angle S\) (corresponding angles), \(\angle U = \angle V\) (corresponding angles). So \(\triangle TUV \sim \triangle SVW\). Then the ratio of \(UV\) to \(VW\) should equal the ratio of \(TU\) to \(SV\), but we need the ratio of the sides on the base. Wait, \(UV = 6\), \(VW = 12\), so \(UV/VW = 6/12 = 1/2\). Then \(TS/WS = 1/2\)? Wait, \(TS = 14\), so \(14/WS = 1/2\) → \(WS = 28\). Then \(TW = TS + WS = 14 + 28 = 42\)? Wait, that makes sense. Wait, because if \(SV \parallel TU\), then the ratio of the segments on \(UW\) (which is \(UV:VW = 6:12 = 1:2\)) should be equal to the ratio of the segments on \(TW\) (which is \(TS:WS = 1:2\)). So \(TS = 14\), so \(WS = 2 \times TS = 28\), so \(TW = TS + WS = 14 + 28 = 42\). Let's check with the ratio. \(\frac{UV}{VW} = \frac{6}{12} = \frac{1}{2}\), and \(\frac{TS}{WS} = \frac{14}{28} = \frac{1}{2}\), so that's consistent. So the ratio of the sides \(UV:VW = TS:WS\), so \(6:12 = 14:WS\), solving for \(WS\): \(6 \times WS = 12 \times 14\) → \(WS = \frac{12 \times 14}{6} = 28\). Then \(TW = TS + WS = 14 + 28 = 42\).

Step2: Solve for TW

Using the similarity of triangles, the ratio of \(UV\) to \(VW\) is equal to the ratio of \(TS\) to \(WS\).

Given \(UV = 6\), \(VW = 12\), \(TS = 14\). Let \(WS = y\). Then:

$$ \frac{UV}{VW} = \frac{TS}{WS} $$

$$ \frac{6}{12} = \frac{14}{y} $$

Cross-multiplying:

$$ 6y = 12 \times 14 $$

$$ 6y = 168 $$

$$ y = \frac{168}{6} = 28 $$

Now,…

Answer:

\(42\)