QUESTION IMAGE
Question
given right triangle xyz, with altitude wz, find each length.
a. if xw = 3 and wy = 12, determine wz.
wz
b. if wz = 4 and xw = 2, determine wy.
c. if xy = 22 and xw = 9, determine wz.
d. if xz = 10 and xw = 4, determine xy.
e. if wz = 18 and xw = 9, determine xz
Part a
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. So \( WZ^2=XW\times WY \).
Step2: Substitute values
Given \( XW = 3 \) and \( WY=12 \), then \( WZ^2=3\times12 = 36 \).
Step3: Solve for \( WZ \)
Take square root: \( WZ=\sqrt{36}=6 \).
Step1: Use geometric mean theorem
\( WZ^2=XW\times WY \), so \( WY=\frac{WZ^2}{XW} \).
Step2: Substitute values
Given \( WZ = 4 \) and \( XW = 2 \), then \( WY=\frac{4^2}{2}=\frac{16}{2}=8 \).
Step1: Find \( WY \)
First, \( WY=XY - XW \). Given \( XY = 22 \) and \( XW = 9 \), so \( WY=22 - 9=13 \).
Step2: Apply geometric mean theorem
\( WZ^2=XW\times WY \), so \( WZ^2=9\times13 = 117 \).
Step3: Solve for \( WZ \)
\( WZ=\sqrt{117}=3\sqrt{13}\approx10.82 \).
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\( WZ = 6 \)