Question

what is the value of a?
○ 5 units
○ ( 5 \frac{1}{3} ) units
○ ( 6 \frac{2}{3} ) units
○ 7 units

Explanation:

Step1: Recall Geometric Mean Theorem

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. Here, \( WZ \perp WX \), \( WY \perp ZX \), so \( WY^2 = ZY \times XY \). Wait, actually, the leg \( WX \) (length \( b \)) and \( WZ \) (length \( c \)), and hypotenuse \( ZX = ZY + XY = 3 + a \)? Wait, no, \( ZY = 3 \), \( XY = a \)? Wait, no, the right triangle is \( WZX \), right-angled at \( W \), and \( WY \) is the altitude to hypotenuse \( ZX \). So by geometric mean theorem, \( WY^2 = ZY \times XY \)? Wait, no, \( WY \) is the altitude, so \( WY^2 = ZY \times YX \). Wait, \( WY = 4 \), \( ZY = 3 \), \( YX = a \)? Wait, no, maybe \( ZX = ZY + YX = 3 + a \), and \( WY \) is the altitude, so \( WY^2 = ZY \times YX \)? Wait, no, the formula is: in right triangle \( WZX \), right-angled at \( W \), altitude \( WY \) to hypotenuse \( ZX \), then \( WY^2 = ZY \times YX \), and \( WZ^2 = ZY \times ZX \), \( WX^2 = YX \times ZX \). Wait, but here we have \( WY = 4 \), \( ZY = 3 \), \( YX = a \)? Wait, no, maybe \( ZX = ZY + YX = 3 + a \), but actually, the segments are \( ZY = 3 \), \( YX = a \), and altitude \( WY = 4 \). So by geometric mean theorem, \( WY^2 = ZY \times YX \)? Wait, no, that would be \( 4^2 = 3 \times a \)? No, that can't be. Wait, maybe I got the segments wrong. Wait, the right triangle is \( WZX \), right angle at \( W \), and \( WY \) is perpendicular to \( ZX \), so \( \triangle WZY \sim \triangle XWZ \sim \triangle WYX \). So the ratio of corresponding sides: \( \frac{ZY}{WY} = \frac{WY}{YX} \), so \( WY^2 = ZY \times YX \). Wait, \( WY = 4 \), \( ZY = 3 \), so \( 4^2 = 3 \times YX \), so \( YX = \frac{16}{3} \)? No, that's not one of the options. Wait, maybe the hypotenuse segments are \( ZY = 3 \) and \( ZX = a \), and \( YX = a - 3 \)? Wait, the options include \( 5\frac{1}{3} \), \( 6\frac{2}{3} \), etc. Wait, maybe the formula is \( WZ^2 = ZY \times ZX \), but \( WZ \) is not given. Wait, no, the altitude is \( WY = 4 \), \( ZY = 3 \), and we need to find \( a \), which is \( YX \)? Wait, no, the options are \( 5\frac{1}{3} \) (which is \( \frac{16}{3} \)? No, \( 5\frac{1}{3} = \frac{16}{3} \)? Wait, \( 5 \times 3 + 1 = 16 \), so \( \frac{16}{3} \approx 5.333 \). Wait, but \( 4^2 = 3 \times a \) would be \( a = \frac{16}{3} = 5\frac{1}{3} \). Oh! Wait, maybe \( a \) is \( YX \), so \( WY^2 = ZY \times YX \), so \( 4^2 = 3 \times a \), so \( a = \frac{16}{3} = 5\frac{1}{3} \) units.

Step2: Apply Geometric Mean Theorem

The geometric mean theorem (altitude-on-hypotenuse theorem) states that in a right triangle, the square of the altitude to the hypotenuse is equal to the product of the lengths of the two segments of the hypotenuse. Here, \( WY = 4 \) (altitude), \( ZY = 3 \) (one segment), and \( YX = a \) (the other segment). So:

\[
WY^2 = ZY \times YX
\]

Substitute \( WY = 4 \), \( ZY = 3 \):

\[
4^2 = 3 \times a
\]

\[
16 = 3a
\]

Solve for \( a \):

\[
a = \frac{16}{3} = 5\frac{1}{3}
\]

Answer:

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