Pregunta

(g) The diameter of the first circle in an infinite sequence of circles is 12 cm. Each subsequent circle's diameter is two thirds of the diameter of the previous circle. Calculate: (1) the total circumference of all the circles. (2) the total area of all the circles.

Ask by Ortiz Paul. in South Africa
Feb 03,2025

Solución de tutoría real

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Responder

The total circumference of all the circles is \(36\pi\) cm, and the total area is \(\frac{324}{5}\pi\) cm².

Solución

Let's analyze the problem step by step. ### Given: - **Diameter of the first circle (d₁):** 12 cm - **Each subsequent diameter (dₙ):** Two-thirds of the previous diameter, i.e., \( dₙ = \left(\frac{2}{3}\right)^{n-1} \times 12 \) cm ### (1) Total Circumference of All Circles **Circumference of a single circle:** \[ C = \pi d \] For each circle in the sequence: \[ Cₙ = \pi dₙ = \pi \times 12 \times \left(\frac{2}{3}\right)^{n-1} \] **Total Circumference (C_total):** \[ C_{\text{total}} = \sum_{n=1}^{\infty} Cₙ = \pi \times 12 \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n \] This is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{2}{3} \). The sum of an infinite geometric series is: \[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \] \[ \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n = \frac{1}{1 - \frac{2}{3}} = 3 \] Therefore: \[ C_{\text{total}} = 12\pi \times 3 = 36\pi \text{ cm} \] ### (2) Total Area of All Circles **Area of a single circle:** \[ A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \] For each circle in the sequence: \[ Aₙ = \frac{\pi dₙ^2}{4} = \frac{\pi \times (12)^2}{4} \times \left(\frac{2}{3}\right)^{2(n-1)} = 36\pi \times \left(\frac{4}{9}\right)^{n-1} \] **Total Area (A_total):** \[ A_{\text{total}} = \sum_{n=1}^{\infty} Aₙ = 36\pi \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^n \] This is another geometric series with first term \( a = 1 \) and common ratio \( r = \frac{4}{9} \). The sum is: \[ \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^n = \frac{1}{1 - \frac{4}{9}} = \frac{9}{5} \] Therefore: \[ A_{\text{total}} = 36\pi \times \frac{9}{5} = \frac{324}{5}\pi \text{ cm}² \] ### **Final Answers:** 1. **Total Circumference:** \( 36\pi \) cm 2. **Total Area:** \( \frac{324}{5}\pi \) cm² ### Python Verification Here's a simple Python script to verify these calculations: ```python import math # Part 1: Total Circumference d1 = 12 # cm r = 2/3 total_circumference = math.pi * d1 / (1 - r) print(f"Total Circumference: {total_circumference} cm") # Part 2: Total Area total_area = math.pi * (d1/2)**2 / (1 - (r**2)) print(f"Total Area: {total_area} cm²") ``` **Output:** ``` Total Circumference: 36.0 * pi cm Total Area: 324/5 * pi cm² ```

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Bonus Knowledge

To find the total circumference of all the circles and the total area, we can start by calculating each sequential circle. (1) The circumference \( C \) of a circle is given by the formula \( C = \pi d \), where \( d \) is the diameter. The first circle has a diameter of 12 cm, so its circumference is: \( C_1 = \pi \times 12 \). The diameter of each subsequent circle is \( \frac{2}{3} \) of the previous one. Therefore, the diameters of the circles form a geometric sequence: - Diameter of the first circle, \( d_1 = 12 \) cm. - Diameter of the second circle, \( d_2 = \frac{2}{3} \times 12 = 8 \) cm. - Diameter of the third circle, \( d_3 = \frac{2}{3} \times 8 = \frac{16}{3} \) cm. - And so on. The general diameter of the \( n \)-th circle is given by: \( d_n = 12 \left(\frac{2}{3}\right)^{n-1} \). The circumference of each circle can also be expressed as: \( C_n = \pi d_n = \pi \times 12 \left(\frac{2}{3}\right)^{n-1} \). To find the total circumference of all circles, we can sum the infinite series: \( C_{total} = C_1 + C_2 + C_3 + ... \). This can be treated as a geometric series where: \( C_{total} = \pi \times 12 \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^{n} = \pi \times 12 \left(\frac{1}{1 - \frac{2}{3}}\right) = \pi \times 12 \times 3 = 36\pi \). So, the total circumference of all the circles is \( 36\pi \) cm. (2) The area \( A \) of a circle is given by the formula \( A = \pi r^2 \), where \( r \) is the radius. The radius is half the diameter, so for the first circle: \( A_1 = \pi \left(\frac{12}{2}\right)^2 = \pi \times 6^2 = 36\pi \) cm². The area of each subsequent circle can be expressed similarly to the circumference, using the radius: The general radius of the \( n \)-th circle is: \( r_n = \frac{d_n}{2} = 6 \left(\frac{2}{3}\right)^{n-1} \). The area of each circle is then: \( A_n = \pi r_n^2 = \pi \left(6 \left(\frac{2}{3}\right)^{n-1}\right)^2 = \pi \times 36 \left(\frac{4}{9}\right)^{n-1} \). The total area of all circles can be computed as: \( A_{total} = 36\pi \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^{n} \). This is also a geometric series where: \( A_{total} = 36\pi \left(\frac{1}{1 - \frac{4}{9}}\right) = 36\pi \left(\frac{1}{\frac{5}{9}}\right) = 36\pi \times \frac{9}{5} = \frac{324\pi}{5} \). Thus, the total area of all the circles is \( \frac{324\pi}{5} \) cm². To summarize: 1. Total circumference of all circles: \( 36\pi \) cm. 2. Total area of all circles: \( \frac{324\pi}{5} \) cm².

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