Question

what is the value of c? 6 units 4 units 5 units 7 units

Explanation:

Step1: Find length of \( a \)

In right triangle \( WYX \), \( WY = 4 \), \( YZ = 3 \)? Wait, no, triangle \( WYZ \)? Wait, triangle \( WYX \) is right-angled at \( Y \)? Wait, \( \triangle WYZ \) and \( \triangle WYX \)? Wait, actually, in right triangle \( WYX \), \( WY = 4 \), \( YZ = 3 \)? No, wait, \( \triangle WYZ \) is right-angled at \( W \)? Wait, no, the right angle at \( Y \) for \( \triangle WYX \)? Wait, no, let's look at the diagram. \( \angle WYZ \) is right? Wait, no, \( \angle WYX \) is right? Wait, \( WY = 4 \), \( YZ = 3 \), and \( \angle WYZ \) is right? Wait, no, \( \triangle WYX \) has \( WY = 4 \), \( YX = a \), and \( \triangle WYZ \) has \( WZ = c \), \( YZ = 3 \), and \( \angle WYZ \) is right? Wait, actually, in a right triangle, the altitude to the hypotenuse relates the segments. Wait, \( WZ \) is \( c \), \( WY = 4 \), \( YZ = 3 \). Wait, in a right triangle, the length of the leg is the geometric mean of the hypotenuse segments? Wait, no, \( c \) is a leg, \( WY \) is the altitude, so \( c^2 = YZ \times ZX \)? Wait, no, \( ZX = YZ + YX = 3 + a \)? Wait, no, maybe \( \triangle WYZ \sim \triangle XWZ \)? Wait, let's use the geometric mean theorem (altitude-on-hypotenuse theorem). In right triangle \( WXZ \) (right-angled at \( W \)), \( WY \) is the altitude to hypotenuse \( XZ \). So, \( WY^2 = YZ \times YX \), but wait, \( WY = 4 \), \( YZ = 3 \), so \( 4^2 = 3 \times YX \)? No, that would be \( 16 = 3 \times YX \), which is not integer. Wait, maybe \( c \) is a leg, and \( WZ = c \), \( WY = 4 \), \( YZ = 3 \), and \( \triangle WYZ \) is right-angled at \( Y \)? Wait, no, \( \angle W \) is right? Wait, the diagram shows \( \angle W \) is right (between \( WZ \) and \( WX \)). So \( WZ = c \), \( WX = b \), \( XZ = a + 3 \)? Wait, no, \( Y \) is on \( XZ \), with \( YZ = 3 \), \( YX = a \), \( WY = 4 \), and \( \angle WYX = 90^\circ \), \( \angle WZY = 90^\circ \)? Wait, maybe I made a mistake. Let's use Pythagoras on \( \triangle WYZ \): \( WZ = c \), \( YZ = 3 \), \( WY = 4 \). Wait, no, \( \triangle WYZ \) is right-angled at \( Y \)? Then \( WZ^2 = WY^2 + YZ^2 \)? Wait, \( WY = 4 \), \( YZ = 3 \), so \( c^2 = 4^2 + 3^2 = 16 + 9 = 25 \), so \( c = 5 \). Wait, that makes sense. So \( c = 5 \) units.

Step2: Verify

Using Pythagoras in \( \triangle WYZ \) (right-angled at \( Y \)): \( c^2 = WY^2 + YZ^2 = 4^2 + 3^2 = 16 + 9 = 25 \), so \( c = \sqrt{25} = 5 \).

Answer:

5 units (the option with "5 units")