Problem 23
Question
Let \(S_{1}, S_{2}, \ldots .\) be squares such that for each \(n \geq 1\), the length of a side of \(S_{n}\) equals the length of a diagonal of \(S_{s+1^{*}}\) If the length of a side of \(S_{1}\) is \(10 \mathrm{~cm}\), then for which of the following values of \(n\) is the area of \(S_{n}\) less than 1 sq. \(\mathrm{cm} ?\) a. 7 b. 8 c. 9 d. 10
Step-by-Step Solution
Verified Answer
n = 10.
1Step 1: Understand the Problem
We are given a sequence of squares \(S_1, S_2, \ldots\) such that the side length of each square \(S_n\) is equal to the diagonal of square \(S_{n-1}\). The side length of \(S_1\) is 10 cm. We need to find the smallest \(n\) for which the area of \(S_n\) is less than 1 sq. cm.
2Step 2: Calculate the Diagonal Length
Let's calculate the side length of \(S_2\). Since the side length of \(S_1\) is 10 cm, the diagonal of \(S_1\) can be calculated using the formula for the diagonal of a square, which is \( s\sqrt{2} \). Thus, the side length of \(S_2\) is \(10\sqrt{2}\) cm.
3Step 3: Derive the General Formula for Side Length
Continuing this pattern, the side length of \(S_3\) will be \((10\sqrt{2})\sqrt{2} = 10\times 2 = 20\) cm, and for \(S_4\), the side will be \(20\sqrt{2}\), and so on. Generally, the side length of \(S_n\) can be represented as \(10\cdot(\sqrt{2})^{n-1}\).
4Step 4: Determine the Area of each Square
The area of square \(S_n\) is given by the square of its side length: \((10\times (\sqrt{2})^{n-1})^2 = 100\cdot(2)^{n-1}\).
5Step 5: Solve for Area Less Than 1 sq. cm
We need \(100\cdot(2)^{n-1} < 1\). Dividing both sides by 100 gives \((2)^{n-1} < \frac{1}{100}\). Taking the base-2 logarithm, we have: \(n-1 < \log_2 \left(0.01\right)\).
6Step 6: Calculate the Value of n
\(\log_2 (0.01) = \log_2 (10^{-2}) = -2\cdot \log_2(10)\). Since \(\log_2(10) \approx 3.32\), \(\log_2(0.01) \approx -6.64\). Thus, \(n-1 < -6.64\), so \(n < -5.64\).
Key Concepts
Diagonal of a SquareSide Length CalculationArea CalculationExponentiation
Diagonal of a Square
The diagonal of a square is an important concept that helps us understand relationships within the square. The diagonal cuts across the square, connecting two opposite corners. When solving problems that involve diagonals, knowing how to calculate their length is essential.
In a square with side length \(s\), the diagonal can be calculated using the Pythagorean theorem. Since a square is a special type of rectangle with equal sides, its diagonal forms a right triangle with its sides. The formula for the diagonal becomes \(s\sqrt{2}\), where \(s\) is the side length. This equation helps us link the side length and diagonal, especially useful when sequences involve transformations from one to another.
In a square with side length \(s\), the diagonal can be calculated using the Pythagorean theorem. Since a square is a special type of rectangle with equal sides, its diagonal forms a right triangle with its sides. The formula for the diagonal becomes \(s\sqrt{2}\), where \(s\) is the side length. This equation helps us link the side length and diagonal, especially useful when sequences involve transformations from one to another.
Side Length Calculation
Calculating the side length of subsequent squares in a geometric sequence often involves using the diagonal as a stepping stone. For each square in the sequence, the side length of the current square is the diagonal of the preceding square.
Consider the initial square \(S_1\) with a 10 cm side length. Its diagonal is \(10\sqrt{2}\) cm. This diagonal becomes the side length of square \(S_2\). Continuing this process, the side length of \(S_n\) can be determined by a formula:
Consider the initial square \(S_1\) with a 10 cm side length. Its diagonal is \(10\sqrt{2}\) cm. This diagonal becomes the side length of square \(S_2\). Continuing this process, the side length of \(S_n\) can be determined by a formula:
- For \(S_2\), it is \(10\sqrt{2}\).
- For \(S_3\), it becomes \(10(\sqrt{2})^2 = 20\).
- For \(S_4\), the side is \(20\sqrt{2}\), and so forth.
Area Calculation
Calculating the area of squares in a sequence involves squaring their side lengths. For any square \(S_n\), where the side length is given by \(10\cdot(\sqrt{2})^{n-1}\), we find the area by squaring this expression.
The formula for the area then becomes:\[(10 \cdot (\sqrt{2})^{n-1})^2 = 100 \cdot (2)^{n-1}\]This expression highlights how the area grows exponentially due to the powers of two. Squaring the side length multiple \((2)^{n-1}\) amplifies the area as \(n\) increases. In some problems, like determining when an area falls below a certain threshold, understanding this exponential growth is key to finding the correct sequence position.
The formula for the area then becomes:\[(10 \cdot (\sqrt{2})^{n-1})^2 = 100 \cdot (2)^{n-1}\]This expression highlights how the area grows exponentially due to the powers of two. Squaring the side length multiple \((2)^{n-1}\) amplifies the area as \(n\) increases. In some problems, like determining when an area falls below a certain threshold, understanding this exponential growth is key to finding the correct sequence position.
Exponentiation
Exponentiation is a powerful tool in mathematics, especially in sequences involving repeated multiplication, like geometric progressions. When calculating the area of squares in our sequence, exponentiation becomes crucial.
For the formula \( (10 \cdot (\sqrt{2})^{n-1})^2 \), the term \((\sqrt{2})^{n-1}\) uses exponentiation to represent how the side length changes with each step. This is central to understanding how the series evolves.
Given that \((\sqrt{2})^{n}\) keeps multiplying the side length by \(\sqrt{2}\) repeatedly, the exponential factor \((2)^{n-1}\) in the area calculation represents this compounding effect. Understanding exponentiation helps break down how seemingly small changes in side length add up across multiple squares in the sequence. This is particularly useful when determining when an outcome like the area becomes less than a specified value.
For the formula \( (10 \cdot (\sqrt{2})^{n-1})^2 \), the term \((\sqrt{2})^{n-1}\) uses exponentiation to represent how the side length changes with each step. This is central to understanding how the series evolves.
Given that \((\sqrt{2})^{n}\) keeps multiplying the side length by \(\sqrt{2}\) repeatedly, the exponential factor \((2)^{n-1}\) in the area calculation represents this compounding effect. Understanding exponentiation helps break down how seemingly small changes in side length add up across multiple squares in the sequence. This is particularly useful when determining when an outcome like the area becomes less than a specified value.
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