### Geometry Word Problem: Area of a Store Sign#### Problem DescriptionA 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.#### QuestionWhat is the area of the sign?#### Multiple Choice AnswersA. 312 inches² B. 336 inches² C. 384 inches² D. 400 inches²The 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 inches²Thus, the

Answered: Image /qna-images/answer/f19aaee4-951a-42cf-a921-6a9d5ff3b05f.jpg

Answered: E16 inches H 20 inches SALE! 12 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?

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter10: Measurement, Area, And Volume

Section10.5: Surface Areas Of Prisms And Cylinders

Problem 19E

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Question

### 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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