Rhombus LMNO has diagonal lines LN¯¯¯¯¯¯¯¯ and MO¯¯¯¯¯¯¯¯¯ that intersect at P . LN¯¯¯¯¯¯¯¯=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)

Question

MN¯¯¯¯¯¯¯¯¯¯=

Answer 1

To show that the diagonals \( LN \) and \( MO \) of rhombus \( LMNO \) are perpendicular bisectors, we will use the properties of rhombuses and the Linear Pair Perpendicular Theorem.

Step 1: Properties of a Rhombus

A rhombus has the following properties:

• The diagonals bisect each other.
• The diagonals are perpendicular to each other.

Step 2: Information Given

We are given:

• \( LN = 24 \)
• \( MO = 32 \)

Step 3: Finding the Midpoint \( P \)

Since the diagonals bisect each other, point \( P \) is the midpoint of both diagonals.

For Diagonal \( LN \):

\[ LP = \frac{LN}{2} = \frac{24}{2} = 12 \] \[ PN = \frac{LN}{2} = \frac{24}{2} = 12 \]

For Diagonal \( MO \):

\[ MP = \frac{MO}{2} = \frac{32}{2} = 16 \] \[ PO = \frac{MO}{2} = \frac{32}{2} = 16 \]

Step 4: Using the Linear Pair Perpendicular Theorem

According to the Linear Pair Perpendicular Theorem, if two lines intersect and the adjacent angles formed are complementary (sum to 90 degrees), then the lines are perpendicular.

Since \( P \) is the intersection point of diagonals \( LN \) and \( MO \), let’s consider angles formed:

• \( \angle LPM \), \( \angle MPN \), \( \angle NPO \), and \( \angle OPL \) are formed when the diagonals intersect.

Since \( LN \) and \( MO \) are diagonals of a rhombus,

• Each diagonal divides the rhombus into two congruent triangles.
• Consequently, the angles \( \angle LPM \) and \( \angle MPN \), etc., are adjacent angles that add up to \( 180^\circ \) indicating that they can be complementary.

As a result, we can conclude that: \[ LN \perp MO \] Thus, the diagonals \( LN \) and \( MO \) are perpendicular bisectors of one another.

Step 5: Finding the Length of \( MN \)

Now, we can find the length of side \( MN \) using the relationship between the sides and the diagonals of a rhombus \( LMNO \).

Using the Pythagorean theorem in triangle \( LMP \): \[ LM^2 = LP^2 + MP^2 \]

Substituting the known values: \[ LM^2 = 12^2 + 16^2 \] \[ LM^2 = 144 + 256 = 400 \] \[ LM = \sqrt{400} = 20 \]

Conclusion

The length of side \( MN \) is the same as \( LM \) because all sides of a rhombus are equal.

Thus, the length \( MN \) is: \[ \boxed{20} \]