Let `S_1 , S_2` , …. Be squares such that for each `n ge 1 ` the length of a side of `S_n` equals the lengths of a diagonal of `S_(n+1)` . If the length of a sides of `S_1` is 10 cm, then for which of the following values of n is the area of `S_n` less than 1 square cm ?

Options

A. A. 7
B. B. 8
C. C. 6
D. D. 4

Answer

To solve the problem step by step, we need to analyze the relationship between the squares \( S_n \) and their areas.

### Step 1: Understand the relationship between squares
Given that the length of a side of square \( S_n \) is equal to the length of the diagonal of square \( S_{n+1} \), we can express this relationship mathematically.

The diagonal \( d \) of a square with side length \( a \) is given by the formula:
\[ d = a \sqrt{2} \]

Thus, if the side length of \( S_n \) is \( a_n \), we have:
\[ a_n = d_{n+1} = a_{n+1} \sqrt{2} \]

This implies:
\[ a_n = a_{n+1} \sqrt{2} \]

### Step 2: Establish the relationship between successive terms
From the above equation, we can express \( a_{n+1} \) in terms of \( a_n \):
\[ a_{n+1} = \frac{a_n}{\sqrt{2}} \]

### Step 3: Determine the first term
We know that the length of the side of \( S_1 \) is given as:
\[ a_1 = 10 \text{ cm} \]

### Step 4: Find the general term
Using the recursive relationship, we can express \( a_n \):
\[ a_n = \frac{10}{(\sqrt{2})^{n-1}} = 10 \cdot 2^{-\frac{n-1}{2}} \]

### Step 5: Calculate the area of \( S_n \)
The area \( A_n \) of square \( S_n \) is given by:
\[ A_n = a_n^2 = \left(10 \cdot 2^{-\frac{n-1}{2}}\right)^2 = 100 \cdot 2^{-(n-1)} \]

### Step 6: Set up the inequality for the area
We want to find \( n \) such that:
\[ A_n < 1 \text{ square cm} \]
This leads to:
\[ 100 \cdot 2^{-(n-1)} < 1 \]

### Step 7: Solve the inequality
Dividing both sides by 100 gives:
\[ 2^{-(n-1)} < \frac{1}{100} \]

Taking the reciprocal (and flipping the inequality) gives:
\[ 2^{n-1} > 100 \]

### Step 8: Find the smallest \( n \)
To find \( n \), we can take logarithms:
\[ n - 1 > \log_2(100) \]
Calculating \( \log_2(100) \):
\[ \log_2(100) = \frac{\log_{10}(100)}{\log_{10}(2)} = \frac{2}{0.301} \approx 6.644 \]
Thus:
\[ n - 1 > 6.644 \implies n > 7.644 \]

The smallest integer \( n \) satisfying this inequality is:
\[ n = 8 \]

### Conclusion
The area of \( S_n \) is less than 1 square cm for \( n = 8 \).