given right triangle xyz, with altitude wz, find each length.
a. if xw = 3 and wy = 12, determine 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

Question

Accepted Answer

### Part a
# 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. 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 the square root: $ WZ=\sqrt{36}=6 $.
# Answer:
$ 6 $

### Part b
# Explanation:
## Step1: Use geometric mean theorem
$ WZ^2 = XW\times WY $, so we can solve for $ WY $ as $ WY=\frac{WZ^2}{XW} $.
## Step2: Substitute values
Given $ WZ = 4 $ and $ XW = 2 $, then $ WY=\frac{4^2}{2}=\frac{16}{2}=8 $.
# Answer:
$ 8 $

### Part c
# Explanation:
## 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 $ (if we want the exact form, it's $ 3\sqrt{13} $; if decimal, approximately $ 10.82 $).
# Answer:
$ 3\sqrt{13} $ (or approximately $ 10.82 $)

### Part d
# Explanation:
## Step1: Recall geometric mean theorem for leg and segment
In a right triangle, the leg is the geometric mean of the hypotenuse and the adjacent segment. So $ XZ^2=XW\times XY $, we need to find $ XY $. Let $ XY = x $, then $ 10^2=4\times x $.
## Step2: Solve for $ x $
$ 100 = 4x $, so $ x=\frac{100}{4}=25 $.
# Answer:
$ 25 $

### Part e
# Explanation:
## Step1: Use geometric mean theorem for leg and segment
$ XZ^2=XW\times XY $, but also, from the altitude formula, we can use $ XZ^2=XW\times(XW + WY) $, but more directly, from the right triangle $ XWZ $, we can use Pythagoras? Wait, no, better to use the geometric mean relation for the leg. Wait, actually, in right triangle $ XZ W $, $ XZ^2=XW^2+WZ^2 $? Wait no, wait: Wait, $ \triangle XWZ $ is right - angled at $ W $, so by Pythagoras: $ XZ^2=XW^2 + WZ^2 $.
## Step2: Substitute values
Given $ XW = 9 $ and $ WZ = 18 $, then $ XZ^2=9^2+18^2=81 + 324=405 $.
## Step3: Solve for $ XZ $
$ XZ=\sqrt{405}=9\sqrt{5}\approx20.12 $ (exact form $ 9\sqrt{5} $, decimal approx $ 20.12 $). Wait, alternatively, using the geometric mean for the leg: $ XZ^2=XW\times XY $, but we can also use the fact that in the right triangle with altitude, the leg is the geometric mean of the hypotenuse segment and the hypotenuse, but here, since $ \triangle XWZ $ is right - angled at $ W $, Pythagoras is more direct. Wait, let's check: $ XW = 9 $, $ WZ = 18 $, so $ XZ=\sqrt{9^{2}+18^{2}}=\sqrt{81 + 324}=\sqrt{405}=9\sqrt{5}\approx20.12 $.
# Answer:
$ 9\sqrt{5} $ (or approximately $ 20.12 $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Part a

Step1: Recall geometric mean theorem

Step2: Substitute values

Step3: Solve for \( WZ \)

Step1: Use geometric mean theorem

Step2: Substitute values

Step1: Find \( WY \)

Step2: Apply geometric mean theorem

Step3: Solve for \( WZ \)

Answer:

Part b