To solve this problem, we will first confirm that the diagonals of rhombus \(LMNO\) are indeed perpendicular to each other and bisect each other.
Step 1: Understanding the Properties of a Rhombus
In a rhombus, two properties are always true:
- The diagonals bisect each other.
- The diagonals are perpendicular to each other.
Step 2: Bisecting the Diagonals
Let’s denote the diagonals:
- \(IN\) with a length of \(24\)
- \(MO\) with a length of \(32\)
Since diagonals bisect each other, point \(P\) is the midpoint of both diagonals:
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The length of \(IP\) (half of \(IN\)) is: \[ IP = \frac{IN}{2} = \frac{24}{2} = 12 \]
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The length of \(MP\) (half of \(MO\)) is: \[ MP = \frac{MO}{2} = \frac{32}{2} = 16 \]
Step 3: Proving the Diagonals are Perpendicular
According to the Linear Pair Perpendicular Theorem, if two lines intersect and the angles adjacent to one of the angles formed are supplementary (add up to \(180^\circ\)), then the two lines are perpendicular.
In rhombus (LMNO:
- Angles \(LPN\) and \(MPN\) are adjacent angles formed by diagonals \(IN\) and \(MO\).
- Since diagonals bisect each other and form two pairs of equal opposite angles in the rhombus, we can conclude that the sum of angles at \(P\) is \(180^\circ\), thus confirming the diagonals \(IN\) and \(MO\) are perpendicular.
Conclusion
Hence, the diagonals \(IN\) and \(MO\) of rhombus \(LMNO\) intersect at \(P\) such that they are perpendicular bisectors of each other.
Step 4: Calculating the Length of \(MN\)
Now, we need to calculate the length of side \(MN\).
In a rhombus, all sides are equal. We can calculate the length of side \(MN\) using the lengths of the halves of the diagonals as the two legs of a right triangle, where \(MP\) and \(IP\) are the legs.
Using the Pythagorean theorem: \[ MN = \sqrt{MP^2 + IP^2} \] Substituting the values: \[ MN = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \]
Thus, the length of \(MN\) is \(20\) units.
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