To solve for the length of $MN$, we need to understand that the triangles $\triangle ALMQ$ and $\triangle ALNP$ are similar. This similarity implies that the corresponding sides of the triangles are in proportion. Given that the similarity condition is satisfied, we can set up a proportion between the corresponding sides of these triangles to solve for $MN$.

Given Data:

• $MQ = 10$
• $NP = 28$
• $LN = 34.9$
• $\triangle ALMQ \sim \triangle ALNP$

Step-by-Step Solution

1. Identify the Corresponding Sides:

   Since $\triangle ALMQ$ is similar to $\triangle ALNP$, the corresponding sides are proportional:

   $\frac{MQ}{NP} = \frac{LM}{LN} = \frac{AQ}{AP}$

   We need to find the length of $MN$.

2. Proportionality Relationship:

   From the given similar triangles, the sides $MQ$ and $NP$ are corresponding sides. Therefore, we can write:

   $\frac{MQ}{NP} = \frac{LM}{LN}$

3. Using the Proportionality:

   Plug in the known values into the equation:

   $\frac{10}{28} = \frac{LM}{34.9}$

4. Solve for $LM$:

   Cross-multiply to solve for $LM$:

   $LM = \frac{10 \times 34.9}{28}$

   Calculate:

   $LM = \frac{349}{28} = 12.4643$

   Thus, $LM \approx 12.4643$.

5. Calculate $MN$:

   Since $MN = LN - LM$:

   $MN = 34.9 - 12.4643 = 22.4357$

   Therefore, the length of $MN \approx 22.44$.

Conclusion

The length of $MN$ is approximately 22.44.

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Would you like more details or have any questions?

Related Questions:

1. What are the conditions for two triangles to be similar?
2. How do you determine which sides correspond in similar triangles?
3. Can similar triangles be different in shape?
4. What is the importance of using proportions in similar triangles?
5. How does the Pythagorean theorem apply to similar triangles?
6. Can angle measurements help in determining triangle similarity?
7. How do transformations like scaling affect triangle similarity?
8. How are similarity and congruence different in geometry?

Tip:

Remember that similar triangles maintain the same angles and have proportional sides, making them a powerful tool for solving geometric problems.