Question

find vw.
vw =
submit
work it out
not feeling ready yet? these can he

Explanation:

Step1: Identify Similar Triangles

Since \( XY \parallel XU \) (from the arrows) and the sides are marked, triangles \( WXV \) and \( WYU \) are similar by the Basic Proportionality Theorem (Thales' theorem). The ratio of corresponding sides should be equal. The ratio of \( WX \) to \( WY \) is \( \frac{14}{42}=\frac{1}{3} \).

Step2: Apply Similar Triangles Ratio

Let \( VW = x \) and \( WU = 34 + x \). Since the triangles are similar, \( \frac{VW}{WU}=\frac{WX}{WY} \), so \( \frac{x}{34 + x}=\frac{1}{3} \). Cross - multiplying gives \( 3x=34 + x \).

Step3: Solve for \( x \)

Subtract \( x \) from both sides: \( 3x - x=34 + x - x \), which simplifies to \( 2x = 34 \). Then divide both sides by 2: \( x=\frac{34}{2}=17 \). Wait, no, wait. Wait, actually, the ratio of \( WX \) to \( WY \) is \( 14:42 = 1:3 \), so the ratio of \( VW \) to \( WU \) is also \( 1:3 \). But \( WU = 34 \)? Wait, no, maybe I misread. Wait, the side \( U V \) is 34? Wait, no, the length from \( U \) to \( V \) is 34? Wait, no, the triangle \( WYU \) has side \( WU \)? Wait, no, let's re - examine. The two sides with the same tick marks are equal, so \( WY = 42 \), \( WX = 14 \), so the ratio of similarity is \( \frac{WX}{WY}=\frac{14}{42}=\frac{1}{3} \). So the ratio of \( VW \) to \( VU \) (wait, \( VU = 34 \))? Wait, no, the line \( XV \) is parallel to \( YU \), so triangle \( WXV \sim WYU \). So \( \frac{VW}{WU}=\frac{WX}{WY} \). Wait, \( WU \) is the side from \( W \) to \( U \), and \( VW \) is from \( W \) to \( V \), \( VU \) is from \( V \) to \( U \). So \( WU=VW + VU \). Let \( VW = x \), then \( WU=x + 34 \). And \( \frac{x}{x + 34}=\frac{14}{42}=\frac{1}{3} \). Cross - multiply: \( 3x=x + 34 \), \( 3x - x=34 \), \( 2x = 34 \), \( x = 17 \)? Wait, no, that can't be. Wait, maybe the side \( U V \) is 34, and the ratio of \( WX \) to \( WY \) is \( 14:42 = 1:3 \), so \( VW=\frac{1}{3}WU \), but \( WU=VW + VU \), so \( VW=\frac{1}{3}(VW + 34) \), \( 3VW=VW + 34 \), \( 2VW = 34 \), \( VW = 17 \). Wait, but let's check again. The length of \( WX \) is 14, \( WY \) is 42, so the scale factor is \( \frac{14}{42}=\frac{1}{3} \). So the corresponding side \( VW \) should be \( \frac{1}{3} \) of \( WU \), but \( WU=VW + VU \), and \( VU = 34 \). So \( VW=\frac{1}{3}(VW + 34) \), solving gives \( VW = 17 \). Wait, but maybe the side \( U V \) is 34, and the triangle is such that \( XV \parallel YU \), so the ratio of \( WX/WY = VW/VU \). So \( 14/42=VW/34 \), then \( VW=\frac{14\times34}{42}=\frac{34}{3}\approx11.33 \)? Wait, no, I made a mistake. Let's do it correctly. The two triangles are similar, so the ratio of corresponding sides is equal. \( WX = 14 \), \( WY = 42 \), so the ratio of similarity \( k=\frac{WX}{WY}=\frac{14}{42}=\frac{1}{3} \). So the ratio of \( VW \) to \( WU \) is \( k \), but \( WU=VW + VU \), and \( VU = 34 \). Wait, no, \( VU \) is a side of the larger triangle. Wait, maybe the side \( U V \) is 34, and the side \( V W \) and \( U W \)? No, let's look at the diagram again. The triangle \( W - X - V \) and \( W - Y - U \), with \( X V \parallel Y U \). So \( \angle W \) is common, and \( \angle WXV=\angle WYU \) (corresponding angles), so \( \triangle WXV\sim\triangle WYU \) by AA similarity. Therefore, \( \frac{WX}{WY}=\frac{VX}{YU}=\frac{VW}{WU} \). We know \( WX = 14 \), \( WY = 42 \), so \( \frac{WX}{WY}=\frac{14}{42}=\frac{1}{3} \). Let \( VW = x \), \( WU=x + 34 \) (since \( U V = 34 \)). Then \( \frac{x}{x + 34}=\frac{1}{3} \), \( 3x=x + 34 \), \( 2x = 34 \), \( x = 17 \). Wait, but if \( VW = 17 \), then \( WU=17 + 34 = 51 \), and \( \frac{17}{51}=\frac{1}{3} \), and \( \frac{14}{42}=\frac{1}{3} \), so that works.

Answer:

\( 17 \)