Question

question 1 of 10
what is the distance from point y to \\(\overleftrightarrow{wx}\\) in the figure below?
a. \\(16\sqrt{3}\\)
b. cannot be determined
c. 4
d. 16
e. 8
f. \\(16\sqrt{2}\\)

Explanation:

Step1: Identify the right triangle

In triangle \( WYZ \), \( WZ = 16 \), \( WY = 16\sqrt{2} \), and \( YZ \) is the distance from \( Y \) to \( \overleftrightarrow{WX} \) (let's call it \( h \)). This is a right triangle with \( \angle YZW = 90^\circ \).

Step2: Apply Pythagorean theorem

The Pythagorean theorem states that in a right triangle, \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse. Here, \( WY \) is the hypotenuse, \( WZ \) is one leg, and \( YZ \) is the other leg. So:

$$ WZ^2 + YZ^2 = WY^2 $$

Substitute \( WZ = 16 \) and \( WY = 16\sqrt{2} \):

$$ 16^2 + h^2 = (16\sqrt{2})^2 $$

Step3: Solve for \( h \)

Calculate each term:
\( 16^2 = 256 \)
\( (16\sqrt{2})^2 = 16^2 \times (\sqrt{2})^2 = 256 \times 2 = 512 \)
Now the equation becomes:

$$ 256 + h^2 = 512 $$

Subtract 256 from both sides:

$$ h^2 = 512 - 256 = 256 $$

Take the square root of both sides:

$$ h = \sqrt{256} = 16 $$

We can also verify with triangle \( YZX \): \( XZ = 30 \), \( XY = 34 \), and if \( YZ = 16 \), then \( 16^2 + 30^2 = 256 + 900 = 1156 \), and \( 34^2 = 1156 \), which confirms it's a right triangle too. So the distance from \( Y \) to \( \overleftrightarrow{WX} \) is 16.

Answer:

D. 16