Question

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find the length of \\(\overline{lw}\\).
if entering your answer as a decimal, round your final answer to the nearest hundredth.
\\(lw = \square\\) units
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Explanation:

Step1: Identify Similar Triangles

The triangles \( \triangle LAY \) and \( \triangle LYW \) are similar (by AA similarity, as both are right triangles and share \( \angle L \)). So, the ratios of corresponding sides are equal: \( \frac{LA}{LY} = \frac{LY}{LW} \). Wait, actually, using geometric mean: in a right triangle, the altitude to the hypotenuse relates the segments. Wait, \( LA = 3.5 \), \( AY = 7.0 \), and \( LY \) is the leg? Wait, no, let's see: \( \triangle LAY \) is right-angled at \( A \), \( \triangle LYW \) is right-angled at \( Y \), and \( \triangle LAW \) is right-angled at \( L \)? Wait, maybe better to use the geometric mean theorem (altitude-on-hypotenuse theorem). Wait, actually, \( LA = 3.5 \), \( AY = 7.0 \), and \( LY \) is the segment? Wait, no, \( LY \) is a leg? Wait, no, let's re-examine. The triangle \( LYW \) is right-angled at \( Y \), and \( LA = 3.5 \), \( AY = 7.0 \), with \( A \) on \( LW \) and \( Y \) on \( LW \)? Wait, no, the diagram: \( L \) to \( A \) is 3.5, \( A \) to \( Y \) is 7.0? Wait, no, the right angles: \( \angle LAY = 90^\circ \), \( \angle LYW = 90^\circ \), and \( \angle L \) is common. So \( \triangle LAY \sim \triangle LYW \) (AA similarity: \( \angle L \) common, \( \angle LAY = \angle LYW = 90^\circ \)). Therefore, the ratio of corresponding sides: \( \frac{LA}{LY} = \frac{LY}{LW} \)? Wait, no, \( LA \) corresponds to \( LY \), and \( LY \) corresponds to \( LW \)? Wait, \( LA = 3.5 \), \( LY \) is the hypotenuse of \( \triangle LAY \). Wait, in \( \triangle LAY \), \( LA = 3.5 \), \( AY = 7.0 \), so \( LY = \sqrt{3.5^2 + 7.0^2} \)? Wait, no, \( \triangle LAY \) is right-angled at \( A \), so \( LY^2 = LA^2 + AY^2 = 3.5^2 + 7.0^2 \). Let's calculate \( LY \): \( 3.5^2 = 12.25 \), \( 7.0^2 = 49 \), so \( LY^2 = 12.25 + 49 = 61.25 \), so \( LY = \sqrt{61.25} \approx 7.826 \). But then, using similarity: \( \triangle LAY \sim \triangle LYW \), so \( \frac{LA}{LY} = \frac{LY}{LW} \), so \( LW = \frac{LY^2}{LA} \). Wait, \( LA = 3.5 \), \( LY^2 = 61.25 \), so \( LW = \frac{61.25}{3.5} \). Let's compute that: \( 61.25 \div 3.5 = 17.5 \)? Wait, no, wait: \( 3.5 \times 17.5 = 61.25 \), yes. Wait, but let's check again. Alternatively, using the geometric mean: in a right triangle, the length of the hypotenuse segment (LW) can be found by \( LW = LA + AW \), but no, \( A \) is on \( LW \), and \( Y \) is on \( LW \)? Wait, maybe the correct approach is: \( LA = 3.5 \), \( AY = 7.0 \), and \( LY \) is the leg, but actually, \( \triangle LAY \) has legs 3.5 and 7.0, so \( LY = \sqrt{3.5^2 + 7^2} = \sqrt{12.25 + 49} = \sqrt{61.25} \approx 7.826 \). Then, since \( \triangle LAY \sim \triangle LYW \), \( \frac{LA}{LY} = \frac{LY}{LW} \), so \( LW = \frac{LY^2}{LA} = \frac{61.25}{3.5} = 17.5 \). Wait, that's a whole number. Let's verify: \( 3.5 \times 17.5 = 61.25 \), and \( LY^2 = 61.25 \), so yes. So \( LW = 17.5 \)? Wait, but let's check with another method. Alternatively, \( LW = LA + AW \), but \( AW \) would be \( \frac{AY^2}{LA} = \frac{7^2}{3.5} = \frac{49}{3.5} = 14 \), so \( LW = 3.5 + 14 = 17.5 \). Yes, that's the geometric mean theorem: in a right triangle, the length of a segment of the hypotenuse is the geometric mean of the adjacent segment and the hypotenuse. Wait, no, the geometric mean theorem states that \( LA^2 = LY \times LW \)? No, wait, no: if \( AY \) is the altitude to the hypotenuse \( LW \), then \( AY^2 = LA \times AW \), and \( LY^2 = LA \times LW \), \( WY^2 = AW \times LW \). Wait, yes! So \( LY^2 = LA \times LW \). So \( LW = \frac{LY^2}{LA} \). But \( LY^2 = LA^2 + AY^2 = 3.5^2 + 7^2 = 12.25 + 49 = 61.25 \). Then \( LW = \frac{61.25}{3.5} = 17.5 \). So that's the length.

Step2: Calculate LW

Using the geometric mean theorem (altitude-on-hypotenuse theorem), we have \( LY^2 = LA \times LW \). First, find \( LY \) using Pythagoras in \( \triangle LAY \): \( LY = \sqrt{LA^2 + AY^2} = \sqrt{3.5^2 + 7.0^2} = \sqrt{12.25 + 49} = \sqrt{61.25} \). Then, \( LW = \frac{LY^2}{LA} = \frac{61.25}{3.5} = 17.5 \).

Answer:

\( 17.5 \)