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**Rectangular Geometry Problem** **Question:** The length of a rectangle is twice its width. If the area of the rectangle is 162 cm², find its perimeter. **Solution Steps:** 1. **Determine the width and length:** - Let \( w \) be the width of the rectangle. - Then, the length \( l \) is \( 2w \). 2. **Set up the area equation:** - Area \( A \) is given by \( A = l \times w = 2w \times w = 2w^2 \). - We know the area \( A = 162 \) cm². So, \( 2w^2 = 162 \). 3. **Solve for \( w \):** - \( w^2 = \frac{162}{2} = 81 \). - \( w = \sqrt{81} = 9 \) cm. 4. **Find the length \( l \):** - \( l = 2w = 2 \times 9 = 18 \) cm. 5. **Calculate the perimeter \( P \):** - Perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). - So, \( P = 2 \times 18 + 2 \times 9 = 36 + 18 = 54 \) cm. **Answer:** The perimeter of the rectangle is 54 cm.

**Rectangular Geometry Problem** **Question:** The length of a rectangle is twice its width. If the area of the rectangle is 162 cm², find its perimeter. **Solution Steps:** 1. **Determine the width and length:** - Let \( w \) be the width of the rectangle. - Then, the length \( l \) is \( 2w \). 2. **Set up the area equation:** - Area \( A \) is given by \( A = l \times w = 2w \times w = 2w^2 \). - We know the area \( A = 162 \) cm². So, \( 2w^2 = 162 \). 3. **Solve for \( w \):** - \( w^2 = \frac{162}{2} = 81 \). - \( w = \sqrt{81} = 9 \) cm. 4. **Find the length \( l \):** - \( l = 2w = 2 \times 9 = 18 \) cm. 5. **Calculate the perimeter \( P \):** - Perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). - So, \( P = 2 \times 18 + 2 \times 9 = 36 + 18 = 54 \) cm. **Answer:** The perimeter of the rectangle is 54 cm.

Algebra and Trigonometry (6th Edition)
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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**Rectangular Geometry Problem** **Question:** The length of a rectangle is twice its width. If the area of the rectangle is 162 cm², find its perimeter. **Solution Steps:** 1. **Determine the width and length:** - Let \( w \) be the width of the rectangle. - Then, the length \( l \) is \( 2w \). 2. **Set up the area equation:** - Area \( A \) is given by \( A = l \times w = 2w \times w = 2w^2 \). - We know the area \( A = 162 \) cm². So, \( 2w^2 = 162 \). 3. **Solve for \( w \):** - \( w^2 = \frac{162}{2} = 81 \). - \( w = \sqrt{81} = 9 \) cm. 4. **Find the length \( l \):** - \( l = 2w = 2 \times 9 = 18 \) cm. 5. **Calculate the perimeter \( P \):** - Perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). - So, \( P = 2 \times 18 + 2 \times 9 = 36 + 18 = 54 \) cm. **Answer:** The perimeter of the rectangle is 54 cm.
Transcribed Image Text:**Rectangular Geometry Problem** **Question:** The length of a rectangle is twice its width. If the area of the rectangle is 162 cm², find its perimeter. **Solution Steps:** 1. **Determine the width and length:** - Let \( w \) be the width of the rectangle. - Then, the length \( l \) is \( 2w \). 2. **Set up the area equation:** - Area \( A \) is given by \( A = l \times w = 2w \times w = 2w^2 \). - We know the area \( A = 162 \) cm². So, \( 2w^2 = 162 \). 3. **Solve for \( w \):** - \( w^2 = \frac{162}{2} = 81 \). - \( w = \sqrt{81} = 9 \) cm. 4. **Find the length \( l \):** - \( l = 2w = 2 \times 9 = 18 \) cm. 5. **Calculate the perimeter \( P \):** - Perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). - So, \( P = 2 \times 18 + 2 \times 9 = 36 + 18 = 54 \) cm. **Answer:** The perimeter of the rectangle is 54 cm.
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