Rhombus Proofs Practice

Question

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Question
Rhombus LMNO
has a diagonal line MO¯¯¯¯¯¯¯¯¯
. If the length of LO¯¯¯¯¯¯¯=5
units and MO¯¯¯¯¯¯¯¯¯=8
units, what is the unit length of LM¯¯¯¯¯¯¯¯¯
?(1 point)
LM¯¯¯¯¯¯¯¯¯=
units
Remaining Attempts : 3

Answer 1

To find the length of \( LM \) in rhombus \( LMNO \), we can use the properties of a rhombus and its diagonals.

1. In a rhombus, the diagonals bisect each other at right angles. Therefore, when we draw diagonal \( MO \), it divides the rhombus into four right triangles.

2. We know:

   • \( LO = 5 \) units (one half of diagonal \( LN \))
   • \( MO = 8 \) units (the entire length of diagonal \( MO \))

3. Since diagonal \( MO \) is bisected, each half of diagonal \( MO \) is: \[ MO' = \frac{MO}{2} = \frac{8}{2} = 4 \text{ units} \]

4. Now, considering triangle \( LOM \):

   • The lengths of the sides are:
     • \( LO = 5 \) units
     • \( MO' = 4 \) units
   • We want to find the length of \( LM \). It would be the hypotenuse of the right triangle formed, using the Pythagorean theorem: \[ LM^2 = LO^2 + MO'^2 \] \[ LM^2 = 5^2 + 4^2 \] \[ LM^2 = 25 + 16 \] \[ LM^2 = 41 \] \[ LM = \sqrt{41} \]

Thus, the length of \( LM \) is approximately \( \sqrt{41} \) units, which is about \( 6.4 \) units when calculated.

So the answer is: \[ LM = \sqrt{41} \text{ units} \] or approximately 6.4 units.