Question

question 5 of 5

drag the tiles to the boxes to form correct pairs. not all tiles will be used.

determine each segment length in right triangle \\(xyz\\).

\\(wy\\) \\(\
ightarrow\\)
\\(xy\\) \\(\
ightarrow\\)
\\(yz\\) \\(\
ightarrow\\)
\\(xw\\) \\(\
ightarrow\\)

tiles:
\\(10\sqrt{3}\\)
\\(10\\)
\\(5\sqrt{3}\\)
\\(5\sqrt{2}\\)
\\(5\\)
\\(15\\)
\\(10\sqrt{2}\\)

Explanation:

Find the length of segment XW

We are given that the hypotenuse \(ZX = 20\) and the segment \(ZW = 5\).
Since \(W\) lies on the segment \(ZX\), we can find \(XW\) by subtraction:

$$ XW = ZX - ZW = 20 - 5 = 15 $$

Find the length of segment WY

Using the Geometric Mean Theorem (specifically the altitude rule) in right triangle \(\triangle XYZ\) with altitude \(YW\):

$$ YW^2 = ZW \cdot XW $$

Substitute the known values:

$$ YW^2 = 5 \cdot 15 = 75 $$

$$ WY = \sqrt{75} = 5\sqrt{3} $$

Find the length of segment YZ

Using the Special Right Triangles properties or the Geometric Mean Theorem (leg rule) for \(\triangle XYZ\):

$$ YZ^2 = ZW \cdot ZX $$

Substitute the known values:

$$ YZ^2 = 5 \cdot 20 = 100 $$

$$ YZ = \sqrt{100} = 10 $$

Alternatively, in the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) right triangle \(\triangle XYZ\), the side opposite the \(30^\circ\) angle (\(YZ\)) is half the hypotenuse (\(ZX\)):

$$ YZ = \frac{1}{2} \cdot 20 = 10 $$

Find the length of segment XY

Using the Special Right Triangles properties for the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) right triangle \(\triangle XYZ\), the side opposite the \(60^\circ\) angle (\(XY\)) is \(\sqrt{3}\) times the shorter leg (\(YZ\)):

$$ XY = YZ \cdot \sqrt{3} = 10\sqrt{3} $$

Alternatively, using the leg rule of the Geometric Mean Theorem:

$$ XY^2 = XW \cdot ZX = 15 \cdot 20 = 300 $$

$$ XY = \sqrt{300} = 10\sqrt{3} $$

Answer:

• \(WY

ightarrow 5\sqrt{3}\)

• \(XY

ightarrow 10\sqrt{3}\)

• \(YZ

ightarrow 10\)

• \(XW

ightarrow 15\)