Question

what is the value of a?
options:
$6\frac{2}{3}$ units
5 units
$5\frac{1}{3}$ units
7 units
(image of a right triangle with segments labeled 3, 4, and vertices z, w, x, y with right angles at y and w)

Explanation:

Step1: Identify Similar Triangles

In the right triangle \( \triangle WXZ \), \( WY \perp XZ \) and \( \angle W = 90^\circ \). By the geometric mean theorem (or altitude-on-hypotenuse theorem), we know that \( WY^2 = ZY \times XY \), but more relevant here is the proportion for segments of the hypotenuse: \( \frac{ZY}{WY}=\frac{WY}{XY} \), and also \( \frac{ZY}{WX}=\frac{WX}{XZ} \)? Wait, actually, the key proportion for finding \( a \) (let's assume \( a = XY \), \( ZY = 3 \), \( WY = 4 \)): Wait, no, let's re - label. Let's see, \( \triangle ZYW \sim \triangle WYX \) (since all right triangles and share \( \angle Z \) or \( \angle X \)). So by similarity, \( \frac{ZY}{WY}=\frac{WY}{XY} \), so \( XY=\frac{WY^2}{ZY} \). Wait, \( ZY = 3 \), \( WY = 4 \), so \( XY=\frac{4^2}{3}=\frac{16}{3}=5\frac{1}{3} \)? Wait, no, maybe I mixed up. Wait, actually, in the right triangle, when an altitude is drawn to the hypotenuse, the length of each leg of the original triangle is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. Wait, let's define: Let \( XZ \) be the hypotenuse, \( ZY = 3 \), \( YX=a \), \( WY = 4 \). Then by the geometric mean theorem, \( WY^2=ZY\times YX \), so \( 4^2 = 3\times a \)? No, that's not right. Wait, no, the correct formula is that in a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. Also, each leg is the geometric mean of the hypotenuse and the adjacent segment. So \( WX^2=ZY\times XZ \), but \( XZ=ZY + YX=3 + a \). Wait, maybe I made a mistake. Wait, looking at the triangle, \( \triangle ZYW \) and \( \triangle WYX \) are similar. So \( \frac{ZY}{WY}=\frac{WY}{YX} \), so \( YX=\frac{WY^2}{ZY} \). Given \( ZY = 3 \), \( WY = 4 \), then \( YX=\frac{16}{3}=5\frac{1}{3} \)? Wait, but let's check again. Wait, the altitude - on - hypotenuse theorem states that: In a right triangle, the length of each leg is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. Wait, no, the formula for the segments of the hypotenuse: If we have a right triangle with hypotenuse \( c \), and an altitude \( h \) to the hypotenuse, dividing the hypotenuse into segments \( m \) and \( n \), then \( h^2 = m\times n \), \( leg_1^2=m\times c \), \( leg_2^2=n\times c \). Wait, in our case, let's assume the right triangle is \( \triangle WXZ \), right - angled at \( W \). The altitude from \( W \) to hypotenuse \( XZ \) is \( WY \), with \( ZY = 3 \), \( YX=a \), \( WY = 4 \). Then by \( h^2=m\times n \), we have \( 4^2=3\times a \)? No, that would be wrong because \( h^2 = m\times n \) where \( m \) and \( n \) are the two segments of the hypotenuse. Wait, yes! \( WY \) is the altitude, \( ZY = 3 \), \( YX=a \), so \( WY^{2}=ZY\times YX \), so \( 4^{2}=3\times a \)? No, that gives \( a=\frac{16}{3}=5\frac{1}{3} \). Wait, that's one of the options.

Step2: Apply the Geometric Mean Theorem

The geometric mean (altitude - on - hypotenuse) theorem states that in a right triangle, the square of the length of the altitude drawn to the hypotenuse is equal to the product of the lengths of the two segments of the hypotenuse. Also, the length of each leg of the right triangle is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. But in our case, for the two similar triangles \( \triangle ZYW \) and \( \triangle WYX \) (both right - angled and share \( \angle Z \) or \( \angle X \)), we have the proportion:

\( \frac{ZY}{WY}=\frac{WY}{YX} \)

We know that \( ZY = 3 \) and \( WY = 4 \). Substituting these values into the proportion:

\( \frac{3}{4}=\frac{4}{a} \)

Cross - multiplying gives us \( 3a=16 \), so \( a = \frac{16}{3}=5\frac{1}{3} \)

Answer:

\( 5\frac{1}{3} \) units