A new Youth Activity Center is being built in Alston. The perimeter of the rectangular playing field is 342 yards. The length of the field is 5 yards less than triple the width. What are the dimensions of the playing field? The width is yards. The length is yards
A new Youth Activity Center is being built in Alston. The perimeter of the rectangular playing field is 342 yards. The length of the field is 5 yards less than triple the width. What are the dimensions of the playing field? The width is yards. The length is yards
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Homework: Homework Section 2.4 - Question 4, 2.4.11
#### Problem Statement:
A new Youth Activity Center is being built in Alston. The perimeter of the rectangular playing field is 342 yards. The length of the field is 5 yards less than triple the width. What are the dimensions of the playing field?
#### Solution:
1. **Identify the Variables:**
- Let the width of the playing field be \( W \) yards.
- Let the length of the playing field be \( L \) yards.
2. **Formulate the Relationships:**
- Given that the length \( L \) is 5 yards less than triple the width \( W \):
\[
L = 3W - 5
\]
3. **Perimeter Relationship:**
- The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2L + 2W
\]
- Given the perimeter is 342 yards:
\[
2L + 2W = 342
\]
4. **Substitute the Length in the Perimeter Formula:**
\[
2(3W - 5) + 2W = 342
\]
5. **Simplify and Solve for the Width:**
\[
6W - 10 + 2W = 342
\]
\[
8W - 10 = 342
\]
\[
8W = 352
\]
\[
W = 44
\]
6. **Calculate the Length:**
\[
L = 3W - 5
\]
\[
L = 3(44) - 5
\]
\[
L = 132 - 5
\]
\[
L = 127
\]
#### Final Dimensions of the Playing Field:
- Width = 44 yards
- Length = 127 yards
#### Interactive Options:
- **Help me solve this**
- **View an example**
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#### Answer Submission:
- **Clear all**
- **Check answer**
(You can use these interactive buttons to proceed with solving or verifying other problems.)
#### HW Score & Points](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c8847fe-5e54-48a4-8290-47d27bfa1967%2F74a71bfd-92e1-42ec-92c3-2e812a6fc0c7%2Fi14tpov_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Homework: Homework Section 2.4 - Question 4, 2.4.11
#### Problem Statement:
A new Youth Activity Center is being built in Alston. The perimeter of the rectangular playing field is 342 yards. The length of the field is 5 yards less than triple the width. What are the dimensions of the playing field?
#### Solution:
1. **Identify the Variables:**
- Let the width of the playing field be \( W \) yards.
- Let the length of the playing field be \( L \) yards.
2. **Formulate the Relationships:**
- Given that the length \( L \) is 5 yards less than triple the width \( W \):
\[
L = 3W - 5
\]
3. **Perimeter Relationship:**
- The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2L + 2W
\]
- Given the perimeter is 342 yards:
\[
2L + 2W = 342
\]
4. **Substitute the Length in the Perimeter Formula:**
\[
2(3W - 5) + 2W = 342
\]
5. **Simplify and Solve for the Width:**
\[
6W - 10 + 2W = 342
\]
\[
8W - 10 = 342
\]
\[
8W = 352
\]
\[
W = 44
\]
6. **Calculate the Length:**
\[
L = 3W - 5
\]
\[
L = 3(44) - 5
\]
\[
L = 132 - 5
\]
\[
L = 127
\]
#### Final Dimensions of the Playing Field:
- Width = 44 yards
- Length = 127 yards
#### Interactive Options:
- **Help me solve this**
- **View an example**
- **Get more help**
#### Answer Submission:
- **Clear all**
- **Check answer**
(You can use these interactive buttons to proceed with solving or verifying other problems.)
#### HW Score & Points
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