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17. 18. /60" 20 3. 寸

17. 18. /60" 20 3. 寸

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Find the area of the shaded regions, round all answers to the nearest 10th.

### Educational Resources on Geometry: Problems and Solutions #### Problem 17: This diagram illustrates a composite figure comprising two concentric circles with a shared center. - The inner circle has a radius of 3 units. - The outer circle has a radius of 4 units. To find the area of the shaded ring (the area between the two concentric circles), you can utilize the formula for the area of a circle, \(A = \pi r^2\). #### Problem 18: This diagram describes a circle with a highlighted sector. - The sector forms a 60-degree angle at the center of the circle. - The radius of the circle is 20 units. - A right triangle is also formed by the radius, where one angle is 60 degrees. From this information, you can calculate the area of the sector, the length of the arc, and the area of the triangle using trigonometric relationships and circle area formulas. Key Points for Calculation: - Use the central angle to find the fraction of the circle. - Apply trigonometric identities for the right triangle. ### Solutions and Explanations: **Problem 17 Solution:** Calculate the areas of both circles and subtract the area of the smaller circle from the area of the larger circle to find the area of the ring. - Area of inner circle: \( \pi \times 3^2 = 9\pi \) - Area of outer circle: \( \pi \times 4^2 = 16\pi \) - Shaded ring area: \( 16\pi - 9\pi = 7\pi \) **Problem 18 Solution:** Calculate the area of the sector and the right triangle. - The area of the sector is \(\frac{60}{360} \times \pi \times (20)^2 = \frac{1}{6} \pi \times 400 = \frac{400\pi}{6} = \frac{200\pi}{3}\). - For the right triangle: - The length of the side opposite the 60-degree angle (height) can be found using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Height = 20 units \(\times \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\). The educational content above will
Transcribed Image Text:### Educational Resources on Geometry: Problems and Solutions #### Problem 17: This diagram illustrates a composite figure comprising two concentric circles with a shared center. - The inner circle has a radius of 3 units. - The outer circle has a radius of 4 units. To find the area of the shaded ring (the area between the two concentric circles), you can utilize the formula for the area of a circle, \(A = \pi r^2\). #### Problem 18: This diagram describes a circle with a highlighted sector. - The sector forms a 60-degree angle at the center of the circle. - The radius of the circle is 20 units. - A right triangle is also formed by the radius, where one angle is 60 degrees. From this information, you can calculate the area of the sector, the length of the arc, and the area of the triangle using trigonometric relationships and circle area formulas. Key Points for Calculation: - Use the central angle to find the fraction of the circle. - Apply trigonometric identities for the right triangle. ### Solutions and Explanations: **Problem 17 Solution:** Calculate the areas of both circles and subtract the area of the smaller circle from the area of the larger circle to find the area of the ring. - Area of inner circle: \( \pi \times 3^2 = 9\pi \) - Area of outer circle: \( \pi \times 4^2 = 16\pi \) - Shaded ring area: \( 16\pi - 9\pi = 7\pi \) **Problem 18 Solution:** Calculate the area of the sector and the right triangle. - The area of the sector is \(\frac{60}{360} \times \pi \times (20)^2 = \frac{1}{6} \pi \times 400 = \frac{400\pi}{6} = \frac{200\pi}{3}\). - For the right triangle: - The length of the side opposite the 60-degree angle (height) can be found using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Height = 20 units \(\times \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\). The educational content above will
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