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E16 inches H 20 inches SALE! 12 inches 4 inches 20 inches What is the area of the sign? A. 312 inches2 B. 336 inches? C. 384 inches² D. 400 inches?

E16 inches H 20 inches SALE! 12 inches 4 inches 20 inches What is the area of the sign? A. 312 inches2 B. 336 inches? C. 384 inches² D. 400 inches?

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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### Geometry Word Problem: Area of a Store Sign #### Problem Description A store places the sign shown below in their window to attract customers. The sign is an irregular octagon with specified dimensions. #### Diagram Explanation - The sign has a top and bottom width of 16 inches. - The height through the middle of the central rectangle is 12 inches. - The height of each trapezoidal section on the top and bottom is 4 inches. - The total width at the widest part of the sign, which includes the entire horizontal span of the trapezoidal sections, is 20 inches. #### Question What is the area of the sign? #### Multiple Choice Answers A. 312 inches² B. 336 inches² C. 384 inches² D. 400 inches² The sign can be broken down into three shapes: a rectangle and two trapezoids. - **Rectangle:** - Width = 16 inches - Height = 12 inches - Area = \(16 \text{ inches} \times 12 \text{ inches} = 192 \text{ inches}^2\) - **Trapezoids:** - Top base = 16 inches - Bottom base = 20 inches - Height = 4 inches - The area of one trapezoid can be found using the formula for the area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \) - Area of one trapezoid = \( \frac{1}{2} \times (16 \text{ inches} + 20 \text{ inches}) \times 4 \text{ inches} = \frac{1}{2} \times 36 \text{ inches} \times 4 \text{ inches} = 72 \text{ inches}^2\) - Since there are two trapezoids: Total area of trapezoids = \(2 \times 72 \text{ inches}^2 = 144 \text{ inches}^2\) - **Total Area of the Sign:** - Area of rectangle + Area of trapezoids = 192 inches² + 144 inches² = 336 inches² Thus, the
Transcribed Image Text:### Geometry Word Problem: Area of a Store Sign #### Problem Description A store places the sign shown below in their window to attract customers. The sign is an irregular octagon with specified dimensions. #### Diagram Explanation - The sign has a top and bottom width of 16 inches. - The height through the middle of the central rectangle is 12 inches. - The height of each trapezoidal section on the top and bottom is 4 inches. - The total width at the widest part of the sign, which includes the entire horizontal span of the trapezoidal sections, is 20 inches. #### Question What is the area of the sign? #### Multiple Choice Answers A. 312 inches² B. 336 inches² C. 384 inches² D. 400 inches² The sign can be broken down into three shapes: a rectangle and two trapezoids. - **Rectangle:** - Width = 16 inches - Height = 12 inches - Area = \(16 \text{ inches} \times 12 \text{ inches} = 192 \text{ inches}^2\) - **Trapezoids:** - Top base = 16 inches - Bottom base = 20 inches - Height = 4 inches - The area of one trapezoid can be found using the formula for the area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \) - Area of one trapezoid = \( \frac{1}{2} \times (16 \text{ inches} + 20 \text{ inches}) \times 4 \text{ inches} = \frac{1}{2} \times 36 \text{ inches} \times 4 \text{ inches} = 72 \text{ inches}^2\) - Since there are two trapezoids: Total area of trapezoids = \(2 \times 72 \text{ inches}^2 = 144 \text{ inches}^2\) - **Total Area of the Sign:** - Area of rectangle + Area of trapezoids = 192 inches² + 144 inches² = 336 inches² Thus, the
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