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

Ask by Colon Rogers. in South Africa
Feb 03,2025

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Tutor-Verified Answer

Answer

1. **Total Circumference:** \( 36\pi \) cm 2. **Total Area:** \( \frac{324}{5}\pi \) cm²

Solution

Let's analyze the problem step by step. ### Given: 1. **First Circle:** - **Diameter (\(d_1\))**: 12 cm - **Radius (\(r_1\))**: \( \frac{d_1}{2} = 6 \) cm 2. **Subsequent Circles:** - Each circle's **diameter** is \(\frac{2}{3}\) of the previous circle's diameter. - This creates a **geometric sequence** for diameters with a common ratio (\(r\)) of \(\frac{2}{3}\). ### 1. Total Circumference of All Circles **Circumference of a Single Circle:** \[ C_n = \pi \cdot d_n \] **Given the sequence of diameters:** \[ d_n = 12 \left(\frac{2}{3}\right)^{n-1} \] **Total Circumference (\(C_{\text{total}}\)):** \[ C_{\text{total}} = \sum_{n=1}^{\infty} C_n = \sum_{n=1}^{\infty} \pi \cdot 12 \left(\frac{2}{3}\right)^{n-1} \] Factor out \(\pi \cdot 12\): \[ C_{\text{total}} = 12\pi \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n \] This is a geometric series with sum: \[ \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n = \frac{1}{1 - \frac{2}{3}} = 3 \] Thus: \[ C_{\text{total}} = 12\pi \cdot 3 = 36\pi \text{ cm} \] ### 2. Total Area of All Circles **Area of a Single Circle:** \[ A_n = \pi r_n^2 = \pi \left(\frac{d_n}{2}\right)^2 = \frac{\pi d_n^2}{4} \] **Given the sequence of diameters:** \[ d_n = 12 \left(\frac{2}{3}\right)^{n-1} \] **Total Area (\(A_{\text{total}}\)):** \[ A_{\text{total}} = \sum_{n=1}^{\infty} A_n = \sum_{n=1}^{\infty} \frac{\pi}{4} \cdot \left[12 \left(\frac{2}{3}\right)^{n-1}\right]^2 \] Simplify inside the sum: \[ A_{\text{total}} = \frac{\pi}{4} \cdot 144 \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^n \] This is a geometric series with sum: \[ \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^n = \frac{1}{1 - \frac{4}{9}} = \frac{9}{5} \] Thus: \[ A_{\text{total}} = \frac{\pi}{4} \cdot 144 \cdot \frac{9}{5} = \frac{324}{5}\pi \text{ cm}^2 \] ### **Final Answers:** 1. **Total Circumference:** \( 36\pi \) cm 2. **Total Area:** \( \frac{324}{5}\pi \) cm² **Numerically:** - \( 36\pi \approx 113.10 \) cm - \( \frac{324}{5}\pi \approx 203.20 \) cm² --- **Summary:** 1. **Total Circumference:** \( 36\pi \) cm 2. **Total Area:** \( \frac{324}{5}\pi \) cm²

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To calculate the total circumference of all the circles, we start by understanding the diameter pattern. The sequence of diameters is 12 cm, 8 cm (12 * 2/3), 5.33 cm (8 * 2/3), and so on. The circumference of the first circle is \( C_1 = \pi \times 12 \) cm. The total circumference of all circles is found using the formula for the sum of an infinite geometric series: \[ C = \sum_{n=0}^{\infty} C_1 \left( \frac{2}{3} \right)^n = \frac{C_1}{1-\frac{2}{3}} = 12 \pi \text{ cm} \] For the total area, the area of the first circle is \( A_1 = \pi \left( \frac{12}{2} \right)^2 = 36\pi \) cm². Again, using the infinite geometric series formula to find the total area: \[ A = \sum_{n=0}^{\infty} A_1 \left( \frac{2}{3} \right)^{2n} = \frac{A_1}{1 - \left( \frac{2}{3} \right)^2} = \frac{36\pi}{1 - \frac{4}{9}} = 36\pi \div \frac{5}{9} = 64.8\pi \text{ cm}² \] So, the total circumference is \( 12\pi \) cm and the total area is \( 64.8\pi \) cm².

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