Question

if vw = vy = 18 and xy = 26, what is wx?
wx = \boxed{}

Explanation:

Step1: Identify Right Triangles

We have right triangles \( \triangle XVY \) and \( \triangle XVW \) since \( \angle XVY = \angle XVW = 90^\circ \), \( VY = VW = 18 \), and \( XV \) is common.

Step2: Apply Pythagorean Theorem to \( \triangle XVY \)

For \( \triangle XVY \), \( XY = 26 \), \( VY = 18 \). Let \( XV = x \). By Pythagoras:
\( XY^2 = XV^2 + VY^2 \)
\( 26^2 = x^2 + 18^2 \)
\( 676 = x^2 + 324 \)
\( x^2 = 676 - 324 = 352 \)

Step3: Apply Pythagorean Theorem to \( \triangle XVW \)

For \( \triangle XVW \), \( VW = 18 \), \( XV^2 = 352 \). Let \( WX = y \). By Pythagoras:
\( WX^2 = XV^2 + VW^2 \)
\( y^2 = 352 + 18^2 \)
\( y^2 = 352 + 324 = 676 \)
\( y = \sqrt{676} = 26 \)? Wait, no—wait, \( XY = 26 \), but \( WX \): Wait, no, correction: Wait, \( VY = VW = 18 \), \( XV \) is common, right angles at \( V \). So \( \triangle XVY \cong \triangle XVW \) (HL), so \( WX = XY \)? No, wait, no—wait, \( XY = 26 \), \( VW = VY = 18 \), so \( WX \): Wait, no, let's recalculate. Wait, \( XV^2 = 26^2 - 18^2 = 676 - 324 = 352 \). Then \( WX^2 = XV^2 + VW^2 = 352 + 324 = 676 \), so \( WX = 26 \)? But that can't be—wait, no, maybe I mixed up. Wait, no, \( XY \) and \( WX \): Wait, actually, since \( VY = VW \) and \( XV \) is common, and right angles, triangles \( XVY \) and \( XVW \) are congruent (HL), so \( WX = XY = 26 \)? Wait, no, that seems off. Wait, no, wait, \( XY = 26 \), \( WX \): Wait, no, let's check again. Wait, \( XV^2 = 26^2 - 18^2 = 352 \), then \( WX^2 = 352 + 18^2 = 352 + 324 = 676 \), so \( WX = 26 \). But that's same as \( XY \). Alternatively, maybe the triangles are congruent, so \( WX = XY = 26 \). Wait, but let's verify: If \( VY = VW = 18 \), \( XV \) common, right angles, then \( \triangle XVY \) and \( \triangle XVW \) are congruent (hypotenuse-leg), so \( WX = XY = 26 \). Wait, but that seems correct? Wait, no, maybe I made a mistake. Wait, \( XY = 26 \), \( WX \): Wait, no, \( XY \) is 26, \( WX \) should be equal? Wait, yes, because HL congruence: \( XV \) is hypotenuse? No, \( XV \) is a leg. Wait, \( VY \) and \( VW \) are legs (18), \( XV \) is another leg, and \( XY \) and \( WX \) are hypotenuses. So \( XY^2 = XV^2 + VY^2 \), \( WX^2 = XV^2 + VW^2 \). Since \( VY = VW \), then \( XY^2 = WX^2 \), so \( XY = WX \). Thus, \( WX = 26 \). Wait, but that seems too straightforward. Wait, the diagram: \( X \) connected to \( Y \) and \( W \), \( V \) is the foot with right angle. So \( XV \perp WY \), \( VY = VW \), so \( XV \) is the perpendicular bisector, so \( WX = XY \). Yes, that makes sense. So \( WX = 26 \). Wait, but let's recalculate the Pythagorean steps again. Step2: \( XV^2 = 26^2 - 18^2 = 676 - 324 = 352 \). Step3: \( WX^2 = 352 + 18^2 = 352 + 324 = 676 \), so \( WX = \sqrt{676} = 26 \). Yes, that's correct.

Answer:

\( \boxed{26} \)