Since ABCD is a square with side 3 units, the diagonal AC can be found using the Pythagorean theorem: \(AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\).
Since BEMN is a square with side 5 units, the diagonal BE can be found similarly: \(BE = \sqrt{BN^2 + NE^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\).
Now, since triangles ACN and BEN are similar triangles (due to AA similarity, both having a right angle and sharing an angle at N), we can set up a proportion:
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Since ABCD is a square with side 3 units, the diagonal AC can be found using the Pythagorean theorem: \(AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\).
Since BEMN is a square with side 5 units, the diagonal BE can be found similarly: \(BE = \sqrt{BN^2 + NE^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\).
Now, since triangles ACN and BEN are similar triangles (due to AA similarity, both having a right angle and sharing an angle at N), we can set up a proportion:
\(\frac{CN}{BN} = \frac{AC}{BE}\)
Substitute the known values:
\(\frac{CN}{5} = \frac{3\sqrt{2}}{5\sqrt{2}}\)
Simplify and solve for CN:
\(CN = \frac{3\sqrt{2}}{\sqrt{2}} = 3\) units.
Therefore, the length of CN is 3 units.