Question

1. find yz.

Explanation:

Step1: Identify triangle type

Triangle \( XWZ \) is isosceles (\( XW = XZ = 40 \)), so \( XY \) is perpendicular bisector of \( WZ \), meaning \( WY = YZ \)? Wait, no—wait, \( XY \) is perpendicular to \( WZ \), so triangle \( XYW \) is right-angled at \( Y \).

Step2: Apply Pythagorean theorem

In right triangle \( XYW \), \( XW = 40 \), \( WY = 24 \). Let \( XY = h \), but we need \( YZ \). Wait, since \( XW = XZ = 40 \), triangle \( XWZ \) is isosceles with \( XY \perp WZ \), so \( WY = YZ \)? Wait, no, \( WZ = WY + YZ \), and \( XY \) is the altitude, so in isosceles triangle, altitude bisects the base. Wait, \( XW = XZ = 40 \), so \( WZ \) is the base, and \( XY \) is the altitude, so \( Y \) is the midpoint? Wait, no, \( WY = 24 \), so \( YZ \) should be such that \( WZ = WY + YZ \), but in right triangle \( XYW \), \( XY^2 + WY^2 = XW^2 \). Wait, no, we can find \( XY \) first, but actually, since \( XZ = 40 \) and \( XY \) is perpendicular to \( YZ \) (wait, \( XY \perp WZ \), so \( XY \perp YZ \) because \( YZ \) is part of \( WZ \)). So triangle \( XYZ \) is also right-angled at \( Y \). So \( XY^2 + YZ^2 = XZ^2 \). But we can find \( XY \) from triangle \( XYW \): \( XY^2 + 24^2 = 40^2 \).

Step3: Calculate \( XY^2 \)

\( XY^2 = 40^2 - 24^2 = 1600 - 576 = 1024 \), so \( XY = 32 \).

Step4: Calculate \( YZ \)

Now in triangle \( XYZ \), right-angled at \( Y \), \( XZ = 40 \), \( XY = 32 \), so \( YZ^2 = XZ^2 - XY^2 = 40^2 - 32^2 = 1600 - 1024 = 576 \), so \( YZ = \sqrt{576} = 24 \)? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, \( WY = 24 \), and since \( XW = XZ \), triangle \( XWZ \) is isosceles, so the altitude \( XY \) bisects \( WZ \), so \( WY = YZ \). Wait, that's the property of isosceles triangle: altitude from apex bisects the base. So \( WY = YZ \), so \( YZ = 24 \)? But wait, when we calculated using Pythagoras, we got \( YZ = 24 \). Wait, let's check again. In triangle \( XYW \): \( XY^2 + 24^2 = 40^2 \) → \( XY^2 = 1600 - 576 = 1024 \) → \( XY = 32 \). Then in triangle \( XYZ \): \( XY^2 + YZ^2 = XZ^2 \) → \( 32^2 + YZ^2 = 40^2 \) → \( YZ^2 = 1600 - 1024 = 576 \) → \( YZ = 24 \). Yes, that matches the isosceles triangle property (altitude bisects base), so \( WY = YZ = 24 \).

Answer:

\( 24 \)