In a rhombus, the diagonals bisect each other at right angles. To find the length of side \( LM \), we can use the properties of right triangles formed by the diagonals.
Given:
- \( LO = 5 \) units (half of diagonal \( LN \))
- \( MO = 8 \) units (half of diagonal \( MO \))
Thus, the full lengths of the diagonals are:
- \( LN = 2 \times LO = 2 \times 5 = 10 \) units
- \( MO = 2 \times MO = 2 \times 8 = 16 \) units
Now, we can use the right triangle \( LOM \) to find the length of \( LM \). By applying the Pythagorean theorem:
\[ LM^2 = LO^2 + MO^2 \]
Substituting the known lengths:
\[ LM^2 = 5^2 + 8^2 \] \[ LM^2 = 25 + 64 \] \[ LM^2 = 89 \]
Taking the square root:
\[ LM = \sqrt{89} \]
Thus, the length of \( LM \) is \( \sqrt{89} \) units.
If you want to give a numerical approximation, \( \sqrt{89} \approx 9.43 \) units.
So, the answer is:
\[ LM \approx 9.43 \text{ units} \]
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