Question

find wx.
wx =
work it out

Explanation:

Step1: Identify Similar Triangles

From the diagram, \( XV \parallel TU \), so \( \triangle WXV \sim \triangle WTU \) by the Basic Proportionality Theorem (Thales' theorem). The ratio of corresponding sides in similar triangles is equal. The sides \( WV = 21 \) and \( WU = 45 \), and let \( WX = x \), \( WT = 45 \) (assuming \( WT \) is proportional, wait, actually, the ratio of \( WV \) to \( WU \) should equal the ratio of \( WX \) to \( WT \)? Wait, no, maybe the sides: since \( XV \) is parallel to \( TU \), the triangles \( WXV \) and \( WTU \) are similar, so \( \frac{WX}{WT} = \frac{WV}{WU} \). Wait, but maybe \( WT \) is given as, let's see, the length from \( T \) to \( X \) is 40? Wait, the diagram shows \( TX = 40 \), so \( WT = WX + XT \)? Wait, no, maybe the sides: \( WV = 21 \), \( WU = 45 \), and \( TX = 40 \). Wait, perhaps the correct proportion is \( \frac{WX}{WT} = \frac{WV}{WU} \), where \( WT = WX + XT \), so let \( WX = x \), then \( WT = x + 40 \). So \( \frac{x}{x + 40} = \frac{21}{45} \).

Step2: Solve the Proportion

Simplify \( \frac{21}{45} = \frac{7}{15} \). So \( \frac{x}{x + 40} = \frac{7}{15} \). Cross-multiplying: \( 15x = 7(x + 40) \).

Step3: Expand and Solve

\( 15x = 7x + 280 \). Subtract \( 7x \): \( 8x = 280 \). Then \( x = \frac{280}{8} = 35 \)? Wait, no, wait, maybe I mixed up the sides. Wait, maybe \( WV = 21 \), \( WU = 45 \), and \( XV \) is parallel to \( TU \), so the ratio of \( WX \) to \( WT \) is \( \frac{WV}{WU} \), but \( WT \) is \( WX + XT \), where \( XT = 40 \). Wait, no, maybe the other way: \( \frac{WX}{WT} = \frac{WV}{WU} \), but \( WT = WX + TX \), so \( \frac{x}{x + 40} = \frac{21}{45} \). Wait, but 21 and 45, simplify to 7/15. So 15x = 7x + 280 → 8x = 280 → x = 35? Wait, but that seems off. Wait, maybe the correct proportion is \( \frac{WX}{WT} = \frac{WV}{WU} \), but \( WT \) is the entire side, and \( WX \) is the smaller side. Wait, maybe I made a mistake. Wait, let's re-examine: the length from \( W \) to \( V \) is 21, from \( W \) to \( U \) is 45, so the ratio is 21/45 = 7/15. Then, since \( XV \parallel TU \), triangle \( WXV \sim triangle WTU \), so \( \frac{WX}{WT} = \frac{WV}{WU} \). If \( TX = 40 \), then \( WT = WX + TX = WX + 40 \). Let \( WX = x \), so \( \frac{x}{x + 40} = \frac{7}{15} \). Cross multiply: 15x = 7x + 280 → 8x = 280 → x = 35. Wait, but that would mean \( WX = 35 \), but let's check again. Wait, maybe the sides are \( WV = 21 \), \( WU = 45 \), and \( XT = 40 \), so the ratio of similarity is 21/45 = 7/15. Then \( WX / WT = 7/15 \), and \( WT = WX + XT \), so \( WX = (7/15) WT \), and \( WT = WX + 40 \). So substituting, \( WX = (7/15)(WX + 40) \). Multiply both sides by 15: 15 WX = 7 WX + 280 → 8 WX = 280 → WX = 35. Yes, that works.

Answer:

35