Question

$\overline{wx} \parallel \overline{vy}$. find $xy$.
72
w 24 v z
33
y
x
$xy = \square$
submit

Explanation:

Step1: Identify Similar Triangles

Since \( \overline{WX} \parallel \overline{VY} \), by the Basic Proportionality Theorem (Thales' theorem), triangles \( \triangle WXZ \) and \( \triangle VYZ \) are similar. So, the ratios of corresponding sides are equal: \( \frac{WV}{WZ} = \frac{XY}{XZ} \). Wait, actually, \( WZ = WV + VZ \)? Wait, no, looking at the diagram, \( WZ \) is the base? Wait, no, the top segment is \( WZ \) with \( WV = 24 \), and the total length above (the horizontal segment) is 72? Wait, maybe the horizontal line at the top is \( WZ \) extended? Wait, no, the diagram shows \( W \) to \( V \) is 24, and the top horizontal segment is 72. Wait, actually, the two parallel lines imply that \( \triangle WVY \) and \( \triangle WXZ \) are similar? Wait, no, \( \overline{VY} \parallel \overline{WX} \), so the triangles \( \triangle VYZ \) and \( \triangle WXZ \) are similar. So, the ratio of \( WV \) to \( WZ \) (wait, \( WZ \) is the entire base? Wait, no, the length from \( W \) to the end of the top horizontal line is 72? Wait, maybe \( WZ = 72 \)? Wait, the top horizontal segment is 72, and \( WV = 24 \), so \( VZ = 72 - 24 = 48 \)? Wait, no, maybe the horizontal length is \( WZ = 72 \), and \( WV = 24 \), so the ratio of similarity is \( \frac{WV}{WZ} = \frac{24}{72} = \frac{1}{3} \). Then, the vertical side \( XZ = 33 \), so \( XY \) and \( YZ \) would have a ratio of \( \frac{1}{3} \)? Wait, no, similar triangles: if \( \triangle VYZ \sim \triangle WXZ \), then \( \frac{VY}{WX} = \frac{YZ}{XZ} = \frac{WV}{WZ} \). Wait, maybe I got the sides wrong. Let's re-express:

Let \( XY = x \), then \( YZ = 33 - x \) (since \( XZ = 33 \)). The ratio of the bases \( WV \) to \( WZ \): \( WV = 24 \), \( WZ = 72 \) (the top horizontal length). So, \( \frac{WV}{WZ} = \frac{24}{72} = \frac{1}{3} \). Then, in similar triangles, \( \frac{YZ}{XZ} = \frac{WV}{WZ} \), so \( \frac{33 - x}{33} = \frac{1}{3} \)? Wait, no, that would be if \( VY \) is parallel to \( WX \), so the smaller triangle is \( \triangle VYZ \) and the larger is \( \triangle WXZ \). So, \( \frac{VZ}{WZ} = \frac{YZ}{XZ} \). Wait, \( VZ = WZ - WV = 72 - 24 = 48 \)? No, maybe \( WZ = 72 \), \( WV = 24 \), so \( VZ = 72 - 24 = 48 \). Then, the ratio \( \frac{VZ}{WZ} = \frac{48}{72} = \frac{2}{3} \). Wait, I'm confused. Let's start over.

The key is that \( \overline{VY} \parallel \overline{WX} \), so by the Basic Proportionality Theorem, \( \frac{WV}{WZ} = \frac{XY}{XZ} \). Wait, \( WZ \) is the length of the base (horizontal) which is 72, \( WV = 24 \), \( XZ = 33 \). So:

\( \frac{WV}{WZ} = \frac{XY}{XZ} \)

\( \frac{24}{72} = \frac{XY}{33} \)

Simplify \( \frac{24}{72} = \frac{1}{3} \), so:

\( \frac{1}{3} = \frac{XY}{33} \)

Multiply both sides by 33:

\( XY = \frac{33}{3} = 11 \)? Wait, no, that can't be. Wait, maybe the ratio is inverted. Because if \( VY \) is parallel to \( WX \), then the triangle \( \triangle WXY \) and \( \triangle VZY \) are similar? Wait, maybe the horizontal length is \( WX = 72 \), and \( WV = 24 \), so \( VZ = 72 - 24 = 48 \). Then, the vertical side \( XZ = 33 \), so the ratio of \( WV \) to \( WX \) is \( 24/72 = 1/3 \), so the ratio of \( XY \) to \( XZ \) is \( 1/3 \), so \( XY = 33 \times (1/3) = 11 \)? Wait, no, that would make \( YZ = 33 - 11 = 22 \), and the ratio of \( VZ \) to \( WX \) is \( 48/72 = 2/3 \), so \( YZ \) should be \( 33 \times (2/3) = 22 \), which matches. So yes, \( XY = 11 \)? Wait, no, wait, maybe I mixed up the triangles. Let's use the correct similar triangles:

Since \( \overline{VY} \parallel \overline{WX} \), \( \angle VYZ = \angle WXZ \) (corresponding angles) and \( \angle VZY = \angle WZX \) (common angle), so \( \triangle VYZ \sim \triangle WXZ \) by AA similarity. Therefore, \( \frac{VZ}{WZ} = \frac{YZ}{XZ} \).

Given \( WZ = 72 \) (the length of the top horizontal segment), \( VZ = WZ - WV = 72 - 24 = 48 \), \( XZ = 33 \). Let \( YZ = x \), then \( XY = 33 - x \).

So, \( \frac{48}{72} = \frac{x}{33} \)

Simplify \( \frac{48}{72} = \frac{2}{3} \), so:

\( \frac{2}{3} = \frac{x}{33} \)

\( x = \frac{2}{3} \times 33 = 22 \)

Then, \( XY = 33 - 22 = 11 \). Wait, that's the same as before. So \( XY = 11 \)? Wait, no, maybe the ratio is \( \frac{WV}{WZ} = \frac{XY}{XZ} \), where \( WV = 24 \), \( WZ = 72 \), so \( \frac{24}{72} = \frac{XY}{33} \), so \( XY = \frac{24 \times 33}{72} = \frac{792}{72} = 11 \). Yes, that's correct.

Step2: Calculate XY

Using the proportion from similar triangles: \( \frac{WV}{WZ} = \frac{XY}{XZ} \)

Substitute \( WV = 24 \), \( WZ = 72 \), \( XZ = 33 \):

\( XY = \frac{WV \times XZ}{WZ} = \frac{24 \times 33}{72} \)

Simplify: \( \frac{24}{72} = \frac{1}{3} \), so \( XY = \frac{1}{3} \times 33 = 11 \)

Answer:

\( \boxed{11} \)