Question

find uy.
uy = \boxed{}
submit

Explanation:

Step1: Identify Similar Triangles

Since \( UW \parallel YX \), triangles \( UVW \) and \( UVX \) (wait, actually \( \triangle UVW \sim \triangle UVX \)? No, correct: \( \triangle UVW \sim \triangle YVX \) by the Basic Proportionality Theorem (Thales' theorem), so the sides are proportional. Wait, the segment \( UW \) is parallel to \( YX \), so \( \triangle VUW \sim \triangle VYX \) by AA similarity (corresponding angles equal). So the ratio of corresponding sides should be equal. Let's denote \( VW = 21 \), \( VX = 63 \), so the ratio of \( VW \) to \( VX \) is \( \frac{21}{63}=\frac{1}{3} \). Wait, no: \( VX = VW + WX \)? Wait, the length from \( V \) to the outer segment is 63, and from \( W \) to the end is 21? Wait, maybe \( VW = 21 \), \( VX = 63 \), so the ratio of similarity is \( \frac{VW}{VX}=\frac{21}{63}=\frac{1}{3} \)? No, wait, actually, the side \( VU = 40 \), and we need to find \( UY \), so \( VY = VU + UY = 40 + UY \). Since \( \triangle VUW \sim \triangle VYX \), the ratio of \( VU \) to \( VY \) is equal to the ratio of \( VW \) to \( VX \). So \( \frac{VU}{VY}=\frac{VW}{VX} \). We know \( VU = 40 \), \( VW = 21 \), \( VX = 63 \). Wait, no, maybe \( VW = 21 \), \( WX = 63 - 21 = 42 \)? Wait, the diagram: the side from \( V \) to \( W \) is 21, and from \( W \) to the end (X) is 63? No, the label 63 is the entire length from \( V \) to the outer point, and 21 is from \( W \) to \( X \)? Wait, maybe the correct ratio is \( \frac{VW}{VX}=\frac{21}{63}=\frac{1}{3} \), so the smaller triangle \( VUW \) has sides in ratio \( \frac{1}{3} \) of the larger triangle \( VYX \). So \( VU = \frac{1}{3} VY \). So \( 40 = \frac{1}{3}(40 + UY) \). Solving for \( UY \): multiply both sides by 3: \( 120 = 40 + UY \), so \( UY = 120 - 40 = 80 \)? Wait, that can't be. Wait, maybe I got the ratio reversed. Let's check again. If \( UW \parallel YX \), then \( \frac{VU}{VY}=\frac{VW}{VX} \). Let's assume \( VW = 63 - 21 = 42 \)? No, the diagram shows 63 as the length from \( V \) to the outer line, and 21 from \( W \) to \( X \). Wait, maybe the correct segments are \( VW = 21 \), \( VX = 63 \), so the ratio of \( VW \) to \( VX \) is \( \frac{21}{63}=\frac{1}{3} \), meaning the smaller triangle is \( \frac{1}{3} \) the size of the larger one. So \( VU \) is a side of the smaller triangle, and \( VY \) is the corresponding side of the larger triangle. So \( VU = \frac{1}{3} VY \)? No, that would make \( VY = 3 \times VU = 120 \), so \( UY = VY - VU = 120 - 40 = 80 \). Wait, but let's verify. If \( VU = 40 \), \( VY = 120 \), then \( UY = 80 \). The ratio of \( VW \) (21) to \( VX \) (63) is \( \frac{21}{63}=\frac{1}{3} \), and \( \frac{VU}{VY}=\frac{40}{120}=\frac{1}{3} \), which matches. So that works.

Step2: Solve for \( UY \)

We have the similarity ratio \( \frac{VW}{VX}=\frac{21}{63}=\frac{1}{3} \). Let \( VY = VU + UY = 40 + UY \). Since \( \triangle VUW \sim \triangle VYX \), \( \frac{VU}{VY}=\frac{VW}{VX} \). Substitute the known values: \( \frac{40}{40 + UY}=\frac{21}{63} \). Simplify \( \frac{21}{63}=\frac{1}{3} \), so \( \frac{40}{40 + UY}=\frac{1}{3} \). Cross-multiplying: \( 40 \times 3 = 40 + UY \). So \( 120 = 40 + UY \). Subtract 40 from both sides: \( UY = 120 - 40 = 80 \).

Answer:

\( 80 \)