Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
Section4.9: Area Problems
Problem 1P
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![### Area Calculation of a Composite Figure: Parallelogram and Semicircle
#### Question:
**Find the Area of the figure below, composed of a parallelogram and one semicircle. Rounded to the nearest tenths place.**
![Figure with dimensions](image_url)
This geometric figure consists of a parallelogram and a semicircle attached to one side of the parallelogram.
#### Given Dimensions:
- **Parallelogram:**
- Base: \(18\) units
- Height: \(9\) units
- **Semicircle:**
- Diameter: \(18\) units
- Radius (half of diameter): \(9\) units
### Steps to Find the Area:
#### Area of the Parallelogram:
The area \(A_p\) of a parallelogram can be calculated using the formula:
\[ A_p = \text{Base} \times \text{Height} \]
\[ A_p = 18 \, \text{units} \times 9 \, \text{units} \]
\[ A_p = 162 \, \text{square units} \]
#### Area of the Semicircle:
The area \(A_s\) of a semicircle can be calculated using the formula. The area of a full circle is \(\pi r^2\), so the area of the semicircle is half of that:
\[ A_s = \frac{1}{2} \pi r^2 \]
Given the radius \(r = 9\) units,
\[ A_s = \frac{1}{2} \pi (9 \, \text{units})^2 \]
\[ A_s = \frac{1}{2} \pi \times 81 \, \text{square units} \]
\[ A_s = \frac{81}{2} \pi \, \text{square units} \]
\[ A_s \approx \frac{81}{2} \times 3.14 \]
\[ A_s \approx 127.2 \, \text{square units} \]
### Total Area of the Composite Figure:
The total area \(A_t\) is the sum of the area of the parallelogram and the area of the semicircle:
\[ A_t = A_p + A_s \]
\[ A_t = 162 \, \text{square units} + 127](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb223a19b-0fd0-443c-881a-6f762159fad5%2Fd1253b24-5c5a-40e0-bfb1-75703d280e51%2F1zf3an_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Area Calculation of a Composite Figure: Parallelogram and Semicircle
#### Question:
**Find the Area of the figure below, composed of a parallelogram and one semicircle. Rounded to the nearest tenths place.**
![Figure with dimensions](image_url)
This geometric figure consists of a parallelogram and a semicircle attached to one side of the parallelogram.
#### Given Dimensions:
- **Parallelogram:**
- Base: \(18\) units
- Height: \(9\) units
- **Semicircle:**
- Diameter: \(18\) units
- Radius (half of diameter): \(9\) units
### Steps to Find the Area:
#### Area of the Parallelogram:
The area \(A_p\) of a parallelogram can be calculated using the formula:
\[ A_p = \text{Base} \times \text{Height} \]
\[ A_p = 18 \, \text{units} \times 9 \, \text{units} \]
\[ A_p = 162 \, \text{square units} \]
#### Area of the Semicircle:
The area \(A_s\) of a semicircle can be calculated using the formula. The area of a full circle is \(\pi r^2\), so the area of the semicircle is half of that:
\[ A_s = \frac{1}{2} \pi r^2 \]
Given the radius \(r = 9\) units,
\[ A_s = \frac{1}{2} \pi (9 \, \text{units})^2 \]
\[ A_s = \frac{1}{2} \pi \times 81 \, \text{square units} \]
\[ A_s = \frac{81}{2} \pi \, \text{square units} \]
\[ A_s \approx \frac{81}{2} \times 3.14 \]
\[ A_s \approx 127.2 \, \text{square units} \]
### Total Area of the Composite Figure:
The total area \(A_t\) is the sum of the area of the parallelogram and the area of the semicircle:
\[ A_t = A_p + A_s \]
\[ A_t = 162 \, \text{square units} + 127
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