Ab=x BC=y CA=z A1B1=18 см B1C1=21 A1C1=24 A1B1/AB=2

To solve the problem, let's define what you're given and what we need to find.

You have:

1. \( AB = x \)
2. \( BC = y \)
3. \( CA = z \)
4. \( A_1B_1 = 18 \, \text{cm} \)
5. \( B_1C_1 = 21 \, \text{cm} \)
6. \( A_1C_1 = 24 \, \text{cm} \)
7. \( \frac{A_1B_1}{AB} = 2 \)

From the ratio \( \frac{A_1B_1}{AB} = 2 \), we can express \( AB \) in terms of \( A_1B_1 \):

\[ AB = \frac{A_1B_1}{2} = \frac{18 \, \text{cm}}{2} = 9 \, \text{cm} \]

Now, \( AB = x = 9 \, \text{cm} \).

Next, we can use the lengths \( A_1B_1 \), \( B_1C_1 \), and \( A_1C_1 \) to find \( BC \) and \( CA \) using a similar relationship. First, let’s find an overall ratio for the smaller triangle \( A_1B_1C_1 \) to the larger triangle \( ABC \).

The sides of triangle \( A_1B_1C_1 \) are:

• \( A_1B_1 = 18 \, \text{cm} \)
• \( B_1C_1 = 21 \, \text{cm} \)
• \( A_1C_1 = 24 \, \text{cm} \)

The ratio of similarity between triangles \( A_1B_1C_1 \) and \( ABC \) can be denoted as \( k \). Since we have the length of \( AB \):

Using the known side \( A_1B_1 \):

\[ k = \frac{A_1B_1}{AB} = 2 \]

Now, we can find lengths \( BC \) and \( CA \) using \( k \):

1. For \( BC \): \[ BC = \frac{B_1C_1}{k} = \frac{21 \, \text{cm}}{2} = 10.5 \, \text{cm} \]

2. For \( CA \): \[ CA = \frac{A_1C_1}{k} = \frac{24 \, \text{cm}}{2} = 12 \, \text{cm} \]

So now we have all the lengths for triangle \( ABC \):

• \( AB = 9 \, \text{cm} \)
• \( BC = 10.5 \, \text{cm} \)
• \( CA = 12 \, \text{cm} \)

Thus, we conclude:

\[ AB = 9 \, \text{cm}, \quad BC = 10.5 \, \text{cm}, \quad CA = 12 \, \text{cm} \]