Question

$\overline{tu} \parallel \overline{vx}$. find $tw$.

(image of a triangle with points t, w, u; x on tw, v on wu. tw segment from x to w is 30, wv is 38, vu is 19. $tw = \square$)

Explanation:

Step1: Identify Similar Triangles

Since \( \overline{TU} \parallel \overline{VX} \), by the Basic Proportionality Theorem (Thales' theorem), triangles \( \triangle TWU \) and \( \triangle XWV \) are similar. So, the ratios of corresponding sides are equal: \( \frac{TW}{XW} = \frac{WU}{WV} \).

Step2: Calculate \( WU \)

\( WU = WV + VU = 38 + 19 = 57 \). Let \( TW = x \), then \( XW = x - 30 \) (wait, no—wait, \( XW \) is 30? Wait, looking at the diagram: \( X \) is on \( TW \), so \( TW = TX + XW \)? Wait, no, the diagram shows \( XW = 30 \)? Wait, no, the vertical segment: \( W \) to \( X \) is 30? Wait, no, the label is 30 next to \( XW \)? Wait, maybe I misread. Let's re-express: Let \( TW = x \), then \( XW = x - TX \)? No, the diagram: \( T \) at top, \( W \) at bottom left, \( U \) at bottom right. \( X \) is on \( TW \), \( V \) is on \( WU \). So \( TW \) is the left side, \( XW = 30 \), \( WV = 38 \), \( VU = 19 \). So \( WU = WV + VU = 38 + 19 = 57 \). Since \( TU \parallel VX \), \( \triangle VXW \sim \triangle TUW \) (by AA similarity, as \( \angle XWV = \angle UWT \) (common angle) and \( \angle VXW = \angle TUW \) (corresponding angles)). Therefore, \( \frac{XW}{TW} = \frac{WV}{WU} \). Wait, \( XW = 30 \), \( WV = 38 \), \( WU = 57 \). Wait, no: \( \frac{TW}{XW} = \frac{WU}{WV} \)? Wait, no, corresponding sides: \( TW \) corresponds to \( XW \)? No, \( TW \) is the entire side, \( XW \) is part of it. Wait, actually, \( \triangle XWV \) and \( \triangle TWU \): \( XW \) corresponds to \( TW \), \( WV \) corresponds to \( WU \). So \( \frac{XW}{TW} = \frac{WV}{WU} \). Wait, \( XW = 30 \), \( WV = 38 \), \( WU = 38 + 19 = 57 \). So \( \frac{30}{TW} = \frac{38}{57} \)? No, that can't be. Wait, maybe \( \frac{TW}{XW} = \frac{WU}{WV} \). Let's check: \( TW \) is the big side, \( XW \) is the small side on \( TW \); \( WU \) is the big side on the base, \( WV \) is the small side on the base. So \( \frac{TW}{XW} = \frac{WU}{WV} \). So \( TW = XW \times \frac{WU}{WV} \). Wait, \( XW \) is 30? Wait, the diagram: the vertical segment from \( W \) to \( X \) is 30, so \( XW = 30 \), and \( TW \) is the length from \( T \) to \( W \), so \( TX + XW = TW \), but \( TX \) is unknown. Wait, no—wait, the correct proportion: Since \( VX \parallel TU \), the triangles \( \triangle VXW \) and \( \triangle TUW \) are similar. Therefore, \( \frac{VX}{TU} = \frac{XW}{TW} = \frac{WV}{WU} \). So \( \frac{XW}{TW} = \frac{WV}{WU} \). \( XW = 30 \), \( WV = 38 \), \( WU = 38 + 19 = 57 \). So \( \frac{30}{TW} = \frac{38}{57} \)? No, that would give \( TW = 30 \times \frac{57}{38} = 30 \times \frac{3}{2} = 45 \)? Wait, no, \( \frac{WV}{WU} = \frac{38}{57} = \frac{2}{3} \). So \( \frac{XW}{TW} = \frac{2}{3} \), so \( TW = XW \times \frac{3}{2} = 30 \times \frac{3}{2} = 45 \)? Wait, but that would mean \( TW = 45 \), so \( TX = 45 - 30 = 15 \). Alternatively, maybe the proportion is \( \frac{TW}{XW} = \frac{WU}{WV} \). Let's try that: \( \frac{TW}{30} = \frac{57}{38} \). Simplify \( \frac{57}{38} = \frac{3}{2} \). So \( TW = 30 \times \frac{3}{2} = 45 \). Wait, that makes sense. So:

Step1: Calculate \( WU \)

\( WU = WV + VU = 38 + 19 = 57 \).

Step2: Set Up Proportion

Since \( TU \parallel VX \), by the Basic Proportionality Theorem, \( \frac{TW}{XW} = \frac{WU}{WV} \). Here, \( XW = 30 \), \( WV = 38 \), \( WU = 57 \).

Step3: Solve for \( TW \)

\( \frac{TW}{30} = \frac{57}{38} \). Simplify \( \frac{57}{38} = \frac{3}{2} \). So \( TW = 30 \times \frac{3}{2} = 45 \).

Answer:

\( 45 \)