Question

use the figure and the given information to find the indicated measure. find wx when wz = 20.5, wy = 14.8, and yz = 20.5 . wx = \boxed{}

Explanation:

Step1: Identify triangle properties

In triangle \( WYZ \), \( WZ = YZ = 20.5 \), so \( \triangle WYZ \) is isosceles with \( WZ = YZ \). The segment \( ZX \) is perpendicular to \( WY \), so by the Isosceles Triangle Theorem, \( X \) is the midpoint of \( WY \)? Wait, no, wait: Wait, \( WZ = YZ = 20.5 \), and \( ZX \perp WY \). Wait, actually, in an isosceles triangle, the altitude from the apex (Z) to the base (WY) bisects the base. But wait, \( WZ = YZ \), so apex is Z, base is WY. So \( ZX \) is the altitude, so \( WX = XY \)? Wait, but we know \( WY = 14.8 \)? Wait, no, wait the given is \( WY = 14.8 \)? Wait, no, the problem says \( WZ = 20.5 \), \( WY = 14.8 \), \( YZ = 20.5 \). Wait, maybe I misread. Wait, \( WZ = 20.5 \), \( YZ = 20.5 \), so triangle \( WYZ \) has \( WZ = YZ \), so it's isosceles with legs \( WZ, YZ \) and base \( WY \). Then the altitude from Z to WY (which is \( ZX \), since \( ZX \perp WY \)) will bisect \( WY \). Wait, but \( WY = 14.8 \)? Wait, no, the problem says \( WY = 14.8 \)? Wait, no, the problem states: Find \( WX \) when \( WZ = 20.5 \), \( WY = 14.8 \), and \( YZ = 20.5 \). Wait, maybe \( WY \) is the base, and \( ZX \) is the altitude. Wait, but in triangle \( WXZ \) and \( YXZ \), we have \( WZ = YZ \), \( ZX \) is common, and \( \angle WXZ = \angle YXZ = 90^\circ \), so by HL congruence, \( \triangle WXZ \cong \triangle YXZ \). Therefore, \( WX = XY \). But \( WY = WX + XY = 2WX \)? Wait, no, \( WY = WX + XY \), and since \( WX = XY \), then \( WY = 2WX \)? Wait, but the given \( WY = 14.8 \)? Wait, that can't be, because \( WZ = 20.5 \) and \( WY = 14.8 \), but in a triangle, the sum of two sides must be greater than the third. Wait, \( WZ + YZ = 20.5 + 20.5 = 41 \), which is greater than \( WY = 14.8 \), so that's okay. Wait, maybe I made a mistake. Wait, the problem says \( WY = 14.8 \), and we need to find \( WX \). Wait, no, wait, maybe \( WY \) is not the base? Wait, the figure: \( W \) and \( Y \) are on the base, \( Z \) is the top, \( X \) is the foot of the perpendicular from \( Z \) to \( WY \). So \( WX + XY = WY \), and since \( WZ = YZ \), triangle \( WYZ \) is isosceles with \( WZ = YZ \), so the altitude \( ZX \) bisects \( WY \), so \( WX = XY \). Wait, but then \( WY = WX + XY = 2WX \), so \( WX = \frac{WY}{2} \)? Wait, but \( WY = 14.8 \), so \( WX = 14.8 / 2 = 7.4 \)? Wait, but let's check with Pythagoras. In triangle \( WXZ \), \( WZ = 20.5 \), \( WX = x \), \( ZX = h \). In triangle \( YXZ \), \( YZ = 20.5 \), \( XY = x \), \( ZX = h \). So by Pythagoras, \( WZ^2 = WX^2 + ZX^2 \) and \( YZ^2 = XY^2 + ZX^2 \). Since \( WZ = YZ \), then \( WX^2 + ZX^2 = XY^2 + ZX^2 \), so \( WX^2 = XY^2 \), so \( WX = XY \) (since lengths are positive). Therefore, \( WX = XY \), so \( WY = WX + XY = 2WX \). Therefore, \( WX = \frac{WY}{2} = \frac{14.8}{2} = 7.4 \).

Step2: Calculate WX

Given \( WY = 14.8 \), and \( WX = XY \) (from congruent triangles or isosceles triangle theorem), so \( WX = \frac{WY}{2} \).

Substitute \( WY = 14.8 \):

\( WX = \frac{14.8}{2} = 7.4 \)

Answer:

\( 7.4 \)