6. A rectangular garden has dimensions of 12 feet by 20 feet. The homeowner wants to increase each dimension by \( 20 \% \). a. Find the original perimeter and area of the garden. b. Find the perimeter and area of the garden with a \( 20 \% \) increase in each dimension. c. By what percent will the perimeter increase? By what percent will the area increase? d. Explain why the area and perimeter will not increase by the same percentage.

To solve the problem, let's break it down step by step. ### a. Find the original perimeter and area of the garden. 1. **Original dimensions**: - Length = 20 feet - Width = 12 feet 2. **Perimeter (P)** of a rectangle is given by the formula: \[ P = 2 \times (Length + Width) \] Substituting the values: \[ P = 2 \times (20 + 12) = 2 \times 32 = 64 \text{ feet} \] 3. **Area (A)** of a rectangle is given by the formula: \[ A = Length \times Width \] Substituting the values: \[ A = 20 \times 12 = 240 \text{ square feet} \] ### b. Find the perimeter and area of the garden with a \( 20 \% \) increase in each dimension. 1. **Increase each dimension by \( 20 \% \)**: - New Length = \( 20 + 0.2 \times 20 = 20 + 4 = 24 \) feet - New Width = \( 12 + 0.2 \times 12 = 12 + 2.4 = 14.4 \) feet 2. **New Perimeter (P')**: \[ P' = 2 \times (New Length + New Width) = 2 \times (24 + 14.4) = 2 \times 38.4 = 76.8 \text{ feet} \] 3. **New Area (A')**: \[ A' = New Length \times New Width = 24 \times 14.4 = 345.6 \text{ square feet} \] ### c. By what percent will the perimeter increase? By what percent will the area increase? 1. **Percent increase in perimeter**: \[ \text{Percent Increase in Perimeter} = \left( \frac{P' - P}{P} \right) \times 100 = \left( \frac{76.8 - 64}{64} \right) \times 100 = \left( \frac{12.8}{64} \right) \times 100 \approx 20\% \] 2. **Percent increase in area**: \[ \text{Percent Increase in Area} = \left( \frac{A' - A}{A} \right) \times 100 = \left( \frac{345.6 - 240}{240} \right) \times 100 = \left( \frac{105.6}{240} \right) \times 100 \approx 44\% \] ### d. Explain why the area and perimeter will not increase by the same percentage. The perimeter and area of a rectangle are related to its dimensions, but they are calculated differently: - The perimeter is a linear measure, which means it is directly proportional to the dimensions of the rectangle. When each dimension is increased by a certain percentage (in this case, \( 20\% \)), the perimeter also increases by that same percentage. - The area, however, is a measure of two-dimensional space and is calculated by multiplying the length and width. When both dimensions are increased by \( 20\% \), the new area is calculated as: \[ A' = (1.2 \times Length) \times (1.2 \times Width) = 1.2^2 \times (Length \times Width) = 1.44 \times A \] This means the area increases by \( 44\% \) (since \( 1.44 - 1 = 0.44 \)), which is greater than the \( 20\% \) increase in perimeter. Thus, the area increases by a greater percentage than the perimeter because the area is affected by the product of the two dimensions, while the perimeter is only affected by their sum.