Question

Let \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots \ldots \mathrm{S}_{n}\) be squares such that for each \(n \geq 1\), the length of a side of \(S_{n}\) equals the length of a diagonal of \(S_{n+1}\). 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. cm ? (1) 7 (2) 8 (3) 6 (4) 5

Short answer

The area of \ S_{n} \ is less than 1 sq. cm when n = 8.

Step-by-step solution

- Understanding the Problem

Given a series of squares \(\text{S}_1, \text{S}_2, \text{...}, \text{S}_n\) where the side of each square is equal to the diagonal of the next one. The goal is to find for which value of \( n \), the area of \( \text{S}_n \) becomes less than 1 sq. cm.

- Relation between side and diagonal

For a square with side length \( a \), the diagonal length \( d \) can be found using the Pythagoras theorem. Given \( d^2 = a^2 + a^2 = 2a^2 \, \ d = a\sqrt{2} \).

- Establishing the side length of next square

The side of the next square \( S_{n+1} \) is the diagonal of the current square \( S_n. \) So, \ \text{side of} \ S_{n+1} = \text{diagonal of} \ S_n \ = a_n\sqrt{2} \.

- Finding the general formula for side length

Starting with \( S_1 \) where side \( a_1 = 10 \text{ cm} \), the side of \( S_n \) can be recursively defined as \( a_{n} = a_{1} (\frac{1}{\sqrt{2})^{n-1}} \).

- Area of the n-th square

The area of the n-th square \( S_n \) can be given by \( \text{Area} = a_n^2 = (10 / \sqrt{2}^{(n-1)})^2 = \ \frac{100}{2^{(n-1)}} \).

- Solving for n such that area < 1 sq. cm

We need to solve \( \frac{100}{2^{n-1}} < 1 \). Rearrange to find: \( 2^{n-1} > 100 \). Taking log base 2 gives: \( n-1 > \ \log_2{100} \), approximately \ \log_2{100} \ = 6.64 implying \ n-1 > 6.64, thus n > 7.64. Therefore, the smallest integer n is 8.

Key concepts

Square properties

Squares are some of the most fundamental geometric shapes. They have four equal sides and four right angles. When dealing with properties of squares, remember that:

• All sides are of equal length.
• Diagonals of a square are equal in length and bisect each other at right angles.
• The diagonals of a square are \(a\sqrt{2}\) if the side length is \(a\).

In this problem, understanding the relationship between the side and the diagonal of the square is crucial. For example, if a square's side length is \(a\), its diagonal is \(a\sqrt{2}\). This is derived using the Pythagorean theorem: \[ d^2 = a^2 + a^2 = 2a^2 \ \therefore d = a\sqrt{2} \]These properties help us connect one square to the next in the given problem sequence.

Geometric series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the given exercise, the side lengths of the squares form a geometric series.
Starting from an initial side length \(a_1 = 10 \text{ cm}\), the side length of the next square is given by the diagonal of the current square, which is \(a\sqrt{2}\). So, each side length can be described as \[a_{n} = a_{1} \left(\frac{1}{\sqrt{2}}\right)^{n-1}\]Here, the initial term \(a_1\) is 10 cm and the common ratio is \(\frac{1}{\sqrt{2}}\). Each side length reduces via a multiplication of this common ratio as \(n\) increases.

Logarithms

Logarithms are the inverse operations to exponentiation. They are useful for solving equations where the unknown is an exponent. In this problem, we need to find when the area of the square is less than 1 sq. cm.
The area of the n-th square given by \(\text{Area} = a_{n}^2 = \left(\frac{10}{\sqrt{2}^{n-1}}\right)^2 = \frac{100}{2^{n-1}}\) needs to be less than 1 sq. cm.:
\