Let `A_(1), A_(2), A_(3),` ........ be squares such that for each `n ge 1`, the length of the side of `A_(n)` equals the length of diagonal of `A_(n+1)`. If the length of `A_(1)` is 12 cm, then the smallest value of n for which area of `A_(n)` is less than one, is ______.

Answer

To solve the problem, we need to find the smallest value of \( n \) such that the area of square \( A_n \) is less than 1 cm², given that the side length of square \( A_1 \) is 12 cm.

### Step-by-Step Solution:

1. **Understanding the relationship between squares:**
The side length of square \( A_n \) is equal to the diagonal of square \( A_{n+1} \). The diagonal \( d \) of a square with side length \( a \) is given by the formula:
\[ d = a \sqrt{2} \]
Therefore, if the side length of \( A_n \) is \( a_n \), we have:
\[ a_n = d_{n+1} = a_{n+1} \sqrt{2} \]

2. **Setting up the relationship:**
From the above relationship, we can express \( a_{n+1} \) in terms of \( a_n \):
\[ a_{n+1} = \frac{a_n}{\sqrt{2}} \]

3. **Finding the side lengths recursively:**
Starting with \( a_1 = 12 \) cm, we can find the subsequent side lengths:
- For \( n = 1 \):
\[ a_2 = \frac{a_1}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2} \text{ cm} \]
- For \( n = 2 \):
\[ a_3 = \frac{a_2}{\sqrt{2}} = \frac{6\sqrt{2}}{\sqrt{2}} = 6 \text{ cm} \]
- For \( n = 3 \):
\[ a_4 = \frac{a_3}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \text{ cm} \]
- For \( n = 4 \):
\[ a_5 = \frac{a_4}{\sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{2}} = 3 \text{ cm} \]
- For \( n = 5 \):
\[ a_6 = \frac{a_5}{\sqrt{2}} = \frac{3}{\sqrt{2}} = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \text{ cm} \]
- For \( n = 6 \):
\[ a_7 = \frac{a_6}{\sqrt{2}} = \frac{\frac{3\sqrt{2}}{2}}{\sqrt{2}} = \frac{3}{2} \text{ cm} \]
- For \( n = 7 \):
\[ a_8 = \frac{a_7}{\sqrt{2}} = \frac{\frac{3}{2}}{\sqrt{2}} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \text{ cm} \]
- For \( n = 8 \):
\[ a_9 = \frac{a_8}{\sqrt{2}} = \frac{\frac{3\sqrt{2}}{4}}{\sqrt{2}} = \frac{3}{4} \text{ cm} \]
- For \( n = 9 \):
\[ a_{10} = \frac{a_9}{\sqrt{2}} = \frac{\frac{3}{4}}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8} \text{ cm} \]
- For \( n = 10 \):
\[ a_{11} = \frac{a_{10}}{\sqrt{2}} = \frac{\frac{3\sqrt{2}}{8}}{\sqrt{2}} = \frac{3}{8} \text{ cm} \]
- For \( n = 11 \):
\[ a_{12} = \frac{a_{11}}{\sqrt{2}} = \frac{\frac{3}{8}}{\sqrt{2}} = \frac{3}{8\sqrt{2}} = \frac{3\sqrt{2}}{16} \text{ cm} \]
- For \( n = 12 \):
\[ a_{13} = \frac{a_{12}}{\sqrt{2}} = \frac{\frac{3\sqrt{2}}{16}}{\sqrt{2}} = \frac{3}{16} \text{ cm} \]
- For \( n = 13 \):
\[ a_{14} = \frac{a_{13}}{\sqrt{2}} = \frac{\frac{3}{16}}{\sqrt{2}} = \frac{3}{16\sqrt{2}} = \frac{3\sqrt{2}}{32} \text{ cm} \]

4. **Finding the area:**
The area of square \( A_n \) is given by \( a_n^2 \). We need to find the smallest \( n \) such that:
\[ a_n^2 < 1 \]

5. **Calculating areas:**
- For \( n = 1 \):
\[ a_1^2 = 12^2 = 144 \]
- For \( n = 2 \):
\[ a_2^2 = (6\sqrt{2})^2 = 72 \]
- For \( n = 3 \):
\[ a_3^2 = 6^2 = 36 \]
- For \( n = 4 \):
\[ a_4^2 = (3\sqrt{2})^2 = 18 \]
- For \( n = 5 \):
\[ a_5^2 = 3^2 = 9 \]
- For \( n = 6 \):
\[ a_6^2 = \left(\frac{3\sqrt{2}}{2}\right)^2 = \frac{9}{2} = 4.5 \]
- For \( n = 7 \):
\[ a_7^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \]
- For \( n = 8 \):
\[ a_8^2 = \left(\frac{3\sqrt{2}}{4}\right)^2 = \frac{9}{8} = 1.125 \]
- For \( n = 9 \):
\[ a_9^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} = 0.5625 \]

6. **Conclusion:**
The area of \( A_9 \) is the first area that is less than 1. Therefore, the smallest value of \( n \) for which the area of \( A_n \) is less than 1 is:
\[ \boxed{9} \]