Question

find uv.
uv = \boxed{}

Explanation:

Step1: Identify Similar Triangles

The triangles \( \triangle UVY \) and \( \triangle WVX \) are similar (by the Basic Proportionality Theorem or AA similarity, as the lines \( UY \) and \( VX \) are parallel, so corresponding angles are equal).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \( \frac{UV}{VW}=\frac{XY + XW}{XW} \)? Wait, no, let's check the sides. Wait, the vertical sides: \( YX = 30 \), \( XW = 25 \)? Wait, no, the vertical segment from \( Y \) to \( X \) is 30? Wait, no, looking at the diagram: \( YW \) is composed of \( YX = 30 \) and \( XW = 25 \)? Wait, no, the labels: \( Y \) to \( X \) is 30? Wait, no, the right side: \( Y \) to \( X \) is 30, \( X \) to \( W \) is 25? Wait, no, the length from \( Y \) to \( W \) is \( 30 + 25 = 55 \)? Wait, no, the two triangles: \( \triangle UVY \) and \( \triangle WVX \). The base of the smaller triangle is \( VW = 30 \), and the base of the larger triangle is \( UV + VW \)? Wait, no, \( U \) to \( V \) to \( W \), so \( UV \) is the base of the larger triangle, \( VW = 30 \) is the base of the smaller triangle. The vertical sides: the height of the larger triangle is \( 30 + 25 = 55 \)? Wait, no, the height of the larger triangle (from \( Y \) to \( U \)) and the height of the smaller triangle (from \( X \) to \( V \)): the height of the smaller triangle is 25, the height of the larger triangle is \( 25 + 30 = 55 \)? Wait, no, the diagram shows \( Y \) to \( X \) is 30, \( X \) to \( W \) is 25. So the height of \( \triangle UVY \) is \( 30 + 25 = 55 \), and the height of \( \triangle WVX \) is 25. Since the triangles are similar, the ratio of their corresponding sides is equal to the ratio of their corresponding heights. So \( \frac{UV}{VW}=\frac{55}{25} \). Wait, \( VW = 30 \), so \( \frac{UV}{30}=\frac{55}{25} \). Wait, no, that can't be. Wait, maybe I got the heights wrong. Wait, the lines \( UY \) and \( VX \) are parallel, so the triangles \( \triangle UVY \) and \( \triangle WVX \) are similar. So the ratio of the bases is equal to the ratio of the heights. The height of \( \triangle UVY \) is \( 30 + 25 = 55 \)? Wait, no, the vertical segment from \( Y \) to \( U \) (the base \( UV \)) is the height? Wait, no, the triangles are right triangles? Wait, \( UW \) is a horizontal line, \( YW \) is vertical, so \( \angle W \) is a right angle. So \( \triangle UVY \) and \( \triangle WVX \) are right triangles, with right angle at \( W \). So they are similar by AA (right angle and the angle at \( V \) is common? Wait, no, the lines \( UY \) and \( VX \) are parallel, so \( \angle UYV = \angle XVW \) (corresponding angles). So \( \triangle UVY \sim \triangle WVX \) by AA similarity (right angle and corresponding angle equal). Therefore, the ratio of corresponding sides: \( \frac{UV}{VW}=\frac{YW}{XW} \). \( YW = 30 + 25 = 55 \), \( XW = 25 \), \( VW = 30 \). So \( \frac{UV}{30}=\frac{55}{25} \). Then \( UV = 30\times\frac{55}{25} = 30\times\frac{11}{5} = 6\times11 = 66 \). Wait, that seems off. Wait, maybe the height of the larger triangle is \( 30 \) and the height of the smaller is \( 25 \)? Wait, no, the diagram: \( Y \) to \( X \) is 30, \( X \) to \( W \) is 25. So the length from \( Y \) to \( W \) is \( 30 + 25 = 55 \), and from \( X \) to \( W \) is 25. So the ratio of the heights is \( 55:25 = 11:5 \). Therefore, the ratio of the bases ( \( UV \) and \( VW \)) should also be \( 11:5 \). Since \( VW = 30 \), let \( UV = x \). Then \( \frac{x}{30}=\frac{11}{5} \), so \( x = 30\times\frac{11}{5} = 66 \). Wait, but let's check again. Wait, maybe the height of the larger triangle is \( 30 \) and the smaller is \( 25 \), but that would mean \( YW = 30 \) and \( XW = 25 \), but that doesn't make sense. Wait, maybe the diagram is \( Y \) to \( X \) is 30, and \( X \) to \( W \) is 25, so the total height from \( Y \) to \( W \) is \( 30 + 25 = 55 \), and the height from \( X \) to \( W \) is 25. So the two triangles: \( \triangle UVY \) has height \( 55 \), base \( UV \); \( \triangle WVX \) has height \( 25 \), base \( VW = 30 \). Since they are similar, \( \frac{UV}{VW}=\frac{55}{25} \), so \( UV = 30\times\frac{55}{25} = 66 \). Yes, that makes sense.

Step1 (Correct): Identify Similar Triangles

The triangles \( \triangle UVY \) and \( \triangle WVX \) are similar (by AA similarity, as \( UY \parallel VX \), so \( \angle U = \angle VXW \) and \( \angle Y = \angle XVW \), and both have a right angle at \( W \)).

Step2 (Correct): Set Up Proportion

For similar triangles, \( \frac{UV}{VW}=\frac{YW}{XW} \). Here, \( YW = 30 + 25 = 55 \), \( XW = 25 \), \( VW = 30 \). So:
\[ \frac{UV}{30}=\frac{55}{25} \]

Step3: Solve for \( UV \)

Simplify \( \frac{55}{25}=\frac{11}{5} \), so:
\[ UV = 30\times\frac{11}{5} = 6\times11 = 66 \]

Answer:

\( 66 \)