Question

in the following triangle, find length wy. round to the nearest tenth. triangle image with labels: x, y, z, w; sides: xy=7, yz=5.7, xz=4, right angles at y (between xy and yw) and at z (between yz and xw). wy is unknown, labeled ?. wy = blank show calculator

Explanation:

Step1: Identify similar triangles

In right triangle \( XYW \) with altitude \( YZ \), triangles \( XYZ \), \( YZW \), and \( XYW \) are similar. So, \( \triangle XYZ \sim \triangle YWZ \).

Step2: Use geometric mean theorem (altitude-on-hypotenuse theorem)

The theorem states that \( \frac{XY}{WY}=\frac{YZ}{WZ}=\frac{XZ}{YZ} \). Also, \( XY^2 = XZ \cdot XW \), but we can use \( \frac{XY}{WY}=\frac{XZ}{YZ} \) or directly use the proportion for similar triangles. Alternatively, in \( \triangle YZW \), we can find \( WZ \) first? Wait, no, let's use the geometric mean for the leg. Wait, actually, in a right triangle, the length of a leg is the geometric mean of the hypotenuse and the adjacent segment. Wait, maybe better to use Pythagoras in \( \triangle XYZ \) to find \( XZ \)? Wait, no, \( XZ = 4 \), \( YZ = 5.7 \), \( XY = 7 \). Wait, maybe I made a mistake. Wait, the altitude to the hypotenuse in a right triangle creates two smaller similar triangles. So, \( \triangle XYZ \sim \triangle YWZ \), so \( \frac{XY}{WY}=\frac{XZ}{YZ} \). Wait, no, let's check the sides. In \( \triangle XYZ \), right-angled at \( Z \), \( XY = 7 \), \( XZ = 4 \), \( YZ = 5.7 \). Wait, but \( 4^2 + 5.7^2 = 16 + 32.49 = 48.49 \), and \( 7^2 = 49 \), which is close (maybe rounding). So, \( \triangle XYZ \) is right-angled at \( Z \), \( \triangle XYW \) is right-angled at \( Y \), so \( YZ \) is the altitude to hypotenuse \( XW \). Then, by geometric mean, \( YZ^2 = XZ \cdot WZ \), so \( 5.7^2 = 4 \cdot WZ \), so \( WZ = \frac{5.7^2}{4} = \frac{32.49}{4} = 8.1225 \). Then, in \( \triangle YZW \), right-angled at \( Z \), \( WY = \sqrt{YZ^2 + WZ^2} = \sqrt{5.7^2 + 8.1225^2} \). Wait, no, that's not right. Wait, actually, the leg \( WY \) in \( \triangle XYW \) (right-angled at \( Y \)): the length of \( WY \) can be found by the geometric mean of \( XW \) and \( WZ \), but maybe easier to use similarity. Wait, \( \triangle XYW \sim \triangle YWZ \), so \( \frac{XY}{YW}=\frac{XW}{YW} \)? No, that's not. Wait, let's start over.

Wait, the big triangle is \( XYW \), right-angled at \( Y \). Then \( YZ \) is the altitude to hypotenuse \( XW \). So, by the geometric mean theorem:

1. \( YZ^2 = XZ \cdot WZ \)
2. \( XY^2 = XZ \cdot XW \)
3. \( WY^2 = WZ \cdot XW \)

We know \( XY = 7 \), \( XZ = 4 \), so from 2: \( 7^2 = 4 \cdot XW \implies XW = \frac{49}{4} = 12.25 \). Then \( WZ = XW - XZ = 12.25 - 4 = 8.25 \). Then, from 3: \( WY^2 = WZ \cdot XW = 8.25 \cdot 12.25 \). Wait, no, that's not the geometric mean theorem. Wait, the geometric mean theorem says that the square of a leg is equal to the product of the hypotenuse and the adjacent segment. So, \( WY^2 = WZ \cdot XW \)? No, wait, the leg is \( WY \), the adjacent segment is \( WZ \), and the hypotenuse is \( XW \). Wait, no, the hypotenuse of the big triangle is \( XW \), and the leg \( WY \) is adjacent to segment \( WZ \), so yes, \( WY^2 = WZ \cdot XW \). But we can also use Pythagoras in \( \triangle YZW \) if we know \( WZ \) and \( YZ \). Wait, \( YZ = 5.7 \), \( WZ = XW - XZ = 12.25 - 4 = 8.25 \). Then \( WY = \sqrt{YZ^2 + WZ^2} = \sqrt{5.7^2 + 8.25^2} \). Wait, but let's check the first approach. Wait, \( XY^2 = XZ \cdot XW \implies 49 = 4 \cdot XW \implies XW = 12.25 \), so \( WZ = 12.25 - 4 = 8.25 \). Then, in \( \triangle YZW \), right-angled at \( Z \), \( WY = \sqrt{YZ^2 + WZ^2} = \sqrt{5.7^2 + 8.25^2} \). Let's calculate that: \( 5.7^2 = 32.49 \), \( 8.25^2 = 68.0625 \), sum is \( 32.49 + 68.0625 = 100.5525 \), square root is \( \sqrt{100.5525} \approx 10.0 \)? Wait, no, that can't be. Wait, maybe I messed up the similar triangles. Wait, actually, the big triangle is right-angled at \( Y \), so \( \angle XYW = 90^\circ \), and \( YZ \perp XW \), so \( \triangle XYZ \sim \triangle YWZ \sim \triangle XYW \). Therefore, the ratio of corresponding sides: \( \frac{XY}{YW} = \frac{XZ}{YZ} \). So, \( XY = 7 \), \( XZ = 4 \), \( YZ = 5.7 \). So, \( \frac{7}{YW} = \frac{4}{5.7} \), so \( YW = \frac{7 \times 5.7}{4} = \frac{39.9}{4} = 9.975 \approx 10.0 \). Ah, that's better. So, cross-multiplying: \( 4 \times YW = 7 \times 5.7 \implies YW = \frac{39.9}{4} = 9.975 \approx 10.0 \) when rounded to the nearest tenth.

Step1: Set up proportion from similar triangles

Since \( \triangle XYZ \sim \triangle YWZ \), the ratio of corresponding sides is equal:
\( \frac{XY}{WY} = \frac{XZ}{YZ} \)

Step2: Substitute known values

\( XY = 7 \), \( XZ = 4 \), \( YZ = 5.7 \). Plug into the proportion:
\( \frac{7}{WY} = \frac{4}{5.7} \)

Step3: Solve for \( WY \)

Cross-multiply:
\( 4 \times WY = 7 \times 5.7 \)
\( 4 \times WY = 39.9 \)
Divide both sides by 4:
\( WY = \frac{39.9}{4} = 9.975 \)

Step4: Round to the nearest tenth

\( 9.975 \approx 10.0 \)

Answer:

\( \boxed{10.0} \)