Question

Rhombus LMNO has diagonal lines IN and MO that
intersect at P. IN = 24 and MO = 32. Use the Linear Pair
Perpendicular Theorem to prove the diagonals are perpendicular bisectors. What is the length of MN?
(1 point)
MN

Answer 1

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:
1. The diagonals bisect each other.
2. 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:

- The length of \(IP\) (half of \(IN\)) is:
\[
IP = \frac{IN}{2} = \frac{24}{2} = 12
\]

- 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.