Question

$\overline{uy} \parallel \overline{vx}$. find $uv$.
(image of a triangle with points u, y, w, x, v; yx = 10, yw = 18, and a segment with length 27 parallel to uy and vx; uv is to be found)

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{UY} \parallel \overline{VX}\), triangles \(UYW\) and \(VXW\) are similar by the Basic Proportionality Theorem (Thales' theorem). So, the ratios of corresponding sides are equal. Let \(UV = x\), then \(VW=27 - x\). The ratio of \(YX\) to \(YW\) should equal the ratio of \(UV\) to \(UW\). Wait, actually, \(YX = 10\), \(XW=18 - 10 = 8\)? Wait, no, \(YW = 18\), \(YX = 10\), \(XW=18 - 10 = 8\)? Wait, no, looking at the diagram, \(YW\) is 18, \(YX = 10\), \(XW = 18 - 10 = 8\)? Wait, no, maybe the sides: \(YX = 10\), \(YW = 18\), so \(XW=18 - 10 = 8\)? Wait, no, the triangles: \(\triangle UYW\) and \(\triangle VXW\) are similar. So \(\frac{VX}{UY}=\frac{XW}{YW}=\frac{VW}{UW}\). Wait, maybe better: Let \(UV = x\), \(UW = x + 27\)? No, wait, the side with length 27 is \(UW\)? Wait, the diagram shows a segment with length 27, and \(UV\) is part of it, so \(UW = UV + VW\), \(VW = 27 - UV\) if \(UV\) is \(x\), then \(VW = 27 - x\). The base \(YW = 18\), \(YX = 10\), \(XW = 18 - 10 = 8\). So by similar triangles, \(\frac{UV}{UW}=\frac{YX}{YW}\)? Wait, no, \(\triangle VXW \sim \triangle UYW\), so \(\frac{VW}{UW}=\frac{XW}{YW}\). Let \(UV = x\), then \(UW = x + (27 - x)=27\)? No, that can't be. Wait, maybe I misread. Wait, the segment with length 27 is \(UW\), and \(UV\) is \(x\), \(VW = 27 - x\). The base \(YW = 18\), \(YX = 10\), \(XW = 18 - 10 = 8\). So \(\frac{VW}{UW}=\frac{XW}{YW}\), so \(\frac{27 - x}{27}=\frac{8}{18}\)? No, that doesn't make sense. Wait, maybe \(YX = 10\), \(YW = 18\), so \(\frac{UV}{UW}=\frac{YX}{YW}\), and \(UW = UV + VW\), but \(VW\) is related. Wait, no, let's correct: \(\triangle UY W\) and \(\triangle VX W\) are similar. So corresponding sides: \(UY\) corresponds to \(VX\), \(YW\) corresponds to \(XW\), and \(UW\) corresponds to \(VW\). Wait, no, \(YW\) is the base of the larger triangle, \(XW\) is the base of the smaller triangle. So \(\frac{XW}{YW}=\frac{VW}{UW}\). \(XW = 18 - 10 = 8\)? No, \(YW = 18\), \(YX = 10\), so \(XW = 18 - 10 = 8\)? Wait, \(YX\) is 10, \(XW\) is 8? Then \(YW = 10 + 8 = 18\), correct. So \(\frac{XW}{YW}=\frac{8}{18}=\frac{4}{9}\). Then \(\frac{VW}{UW}=\frac{4}{9}\). Let \(UV = x\), then \(UW = x + VW\), but \(VW = 27 - x\) (since \(UW = 27\)? Wait, the segment with length 27 is \(UW\)? So \(UW = 27\), then \(VW = 27 - x\). So \(\frac{27 - x}{27}=\frac{8}{18}\). Solving: \(27 - x = 27\times\frac{8}{18}=12\), so \(x = 27 - 12 = 15\). Wait, that works. Let's check again. If \(UW = 27\), \(VW = 12\), \(UV = 15\). Then \(\frac{XW}{YW}=\frac{8}{18}=\frac{4}{9}\), \(\frac{VW}{UW}=\frac{12}{27}=\frac{4}{9}\). Yes, that's correct. So the ratio of similarity is \(\frac{4}{9}\), and the sides correspond. So \(UV = 15\).

Step2: Solve for \(UV\)

Using the similarity of \(\triangle VXW\) and \(\triangle UYW\), we have \(\frac{VW}{UW}=\frac{XW}{YW}\). Let \(UV = x\), then \(VW = 27 - x\), \(UW = 27\) (since the total length is 27), \(XW = 18 - 10 = 8\), \(YW = 18\). So:

$$ \frac{27 - x}{27}=\frac{8}{18} $$

Simplify \(\frac{8}{18}=\frac{4}{9}\):

$$ 27 - x = 27\times\frac{4}{9} $$

$$ 27 - x = 12 $$

Subtract 12 from 27:

$$ x = 27 - 12 = 15 $$

Answer:

\(15\)