In a rhombus, all sides are of equal length, and the diagonals intersect at right angles and bisect each other.
Given that \( LO = 5 \) units and \( MO = 8 \) units, we can find the lengths of the sides of the rhombus by using the properties of the right triangle formed by half the diagonals and one side of the rhombus.
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Since \( MO \) is one of the diagonals, the full length of diagonal \( MN \) (which isn't directly given but can be found using the diagonals):
- \( MO \) is 8 units, so \( MO/2 = 4 \) units.
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For diagonal \( LN \):
- Since \( LO \) is also a diagonal, the full length of diagonal \( LN = 2 \times LO = 2 \times 5 = 10 \) units.
- Therefore, \( LN/2 = 5 \) units.
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Now, we can consider triangle \( LOM \) to find the length of side \( LM \):
- We have:
- \( LO = 5 \) units
- \( MO/2 = 4 \) units (half the length of diagonal \( MO \))
- Using the Pythagorean theorem: \[ LM^2 = LO^2 + (MO/2)^2 \] \[ LM^2 = 5^2 + 4^2 \] \[ LM^2 = 25 + 16 \] \[ LM^2 = 41 \] \[ LM = \sqrt{41} \]
- We have:
So the length of \( LM \) (and hence, all sides of the rhombus) is \( \sqrt{41} \) units.
Thus, \[ LM = \sqrt{41} , \text{units}. \]