If AF = 15 mm, FC = 24 mm, and the perimeter of AAEF

is 60 mm, calculate the perimeter of ABC.

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To solve this problem, we need to calculate the perimeter of triangle \( \triangle ABC \) using the given dimensions of triangle \( \triangle AEF \).

Here's the plan:

1. **Understand the given information:**  
   - \( AF = 15 \, \text{mm} \)
   - \( FC = 24 \, \text{mm} \)
   - The perimeter of triangle \( \triangle AEF \) is \( 60 \, \text{mm} \).

2. **Find the missing side \( EF \) using the perimeter of \( \triangle AEF \):**  
   - The perimeter of \( \triangle AEF \) is calculated as:
   \[
   \text{Perimeter of } \triangle AEF = AF + EF + EA = 60 \, \text{mm}
   \]

   However, from the given information, we don't know \( EA \). Let's assume:
   \[
   EA = x
   \]
   Therefore, we have:
   \[
   15 + EF + x = 60
   \]

3. **Find \( EF \):**  
   Assuming \( EF = y \), we have:
   \[
   15 + y + x = 60
   \]

   Solving for \( y \):
   \[
   y = 60 - 15 - x
   \]
   \[
   y = 45 - x
   \]

   Thus, \( EF = 45 - x \).

4. **Consider \( \triangle ABC \):**  
   - To calculate the perimeter of \( \triangle ABC \), we'll need the dimensions \( AB \), \( BC \), and \( CA \). 
   - Without loss of generality, let's assume \( \triangle AEF \sim \triangle ABC \) (as they seem to have a proportional similarity based on the relationship between \( AF \) and \( FC \)). 
   - If \( \triangle AEF \sim \triangle ABC \), then:
   \[
   \frac{AF}{AB} = \frac{EF}{BC} = \frac{AE}{AC}
   \]

5. **Using proportionality:**

   Let's define the similarity ratio \( k \):

   \[
   k = \frac{AF}{AB} = \frac{15}{x}
   \]

   Assume that the points \( A, F, C \) and \( A, E, B \) form lines and that \( EF \parallel BC \) such that:
   \[
   k = \frac{EF}{BC} = \frac{AE}{AC}
   \]

   Solving for the perimeter of \( \triangle ABC \):
   - Since \( FC = 24 \) mm is a straight line,
   - Assume \( BC \) is parallel to \( EF \) and calculate \( BC \) based on the similarity ratio \( k \):
   \[
   BC = \frac{EF \cdot FC}{AF} = \frac{24 \cdot (45-x)}{15}
   \]

   Calculate:
   \[
   BC = \frac{24 \cdot (45-x)}{15} = \frac{24 \cdot 45 - 24 \cdot x}{15} = \frac{1080 - 24x}{15}
   \]

   Let:
   \[
   x = 15
   \]

   Hence,
   \[
   BC = \frac{1080 - 24 \cdot 15}{15} = \frac{1080 - 360}{15} = \frac{720}{15} = 48
   \]

6. **Calculate \( AB \) and \( CA \):**

   From the previous relations using similarity:
   \[
   AB = \frac{15 \cdot 15}{15} = 15
   \]

   \[
   CA = \frac{15 \cdot 24}{15} = 24
   \]

7. **Perimeter of \( \triangle ABC \):**

   The perimeter of \( \triangle ABC \) is the sum of the sides:

   \[
   \text{Perimeter of } \triangle ABC = AB + BC + CA = 15 + 48 + 24 = 87
   \]

### Conclusion

The perimeter of \( \triangle ABC \) is **87 mm**.

---

### Would you like more details or have any questions?

### Related Questions:

1. How do you calculate the perimeter of different geometric shapes?
2. What are the properties of similar triangles?
3. How can you find the perimeter of a shape when only given two sides and the total perimeter?
4. What is the importance of parallel lines in triangle similarity?
5. Can the concept of similarity be used in 3D geometry?
6. What is the relationship between similarity and scale factors in geometry?
7. How can the Pythagorean theorem be used to find the lengths of triangle sides?
8. How does the triangle inequality theorem relate to this problem?

### Tip:

When working with similar triangles, always set up a proportion between corresponding sides to find unknown lengths.

Learn how to calculate the perimeter of triangle ABC using the given dimensions of triangle AEF. This problem involves geometry and the concept of triangle similarity. Suitable for students in Grades 9-10.

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