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3. The dimensions of a rectangle are 40 cm and 45 cm . If there is an error of \( 5 \% \) in the dimensions find the percentage error in calculating area of the rectangle. ( 3 mks )

Ask by Page Warner. in Kenya
Mar 15,2025

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Answer

The percentage error in calculating the area of the rectangle is approximately 10.25%.

Solution

To find the percentage error in calculating the area of the rectangle due to the error in its dimensions, we can follow these steps: 1. **Identify the dimensions of the rectangle:** - Length \( L = 40 \) cm - Width \( W = 45 \) cm 2. **Calculate the area of the rectangle without error:** \[ A = L \times W = 40 \times 45 \] 3. **Calculate the percentage error in the dimensions:** - Given that there is a \( 5\% \) error in both dimensions, we can express the errors in length and width as: - Error in Length \( = 0.05 \times L \) - Error in Width \( = 0.05 \times W \) 4. **Calculate the new dimensions considering the error:** - New Length \( L' = L + \text{Error in Length} = L + 0.05L = 1.05L \) - New Width \( W' = W + \text{Error in Width} = W + 0.05W = 1.05W \) 5. **Calculate the new area with the erroneous dimensions:** \[ A' = L' \times W' = (1.05L) \times (1.05W) = 1.05^2 \times (L \times W) \] 6. **Calculate the percentage error in the area:** - The percentage error in the area can be calculated using the formula: \[ \text{Percentage Error in Area} = \left( \frac{A' - A}{A} \right) \times 100\% \] Now, let's perform the calculations step by step. ### Step 2: Calculate the area of the rectangle without error \[ A = 40 \times 45 \] ### Step 4: Calculate the new dimensions \[ L' = 1.05 \times 40 \] \[ W' = 1.05 \times 45 \] ### Step 5: Calculate the new area \[ A' = L' \times W' \] ### Step 6: Calculate the percentage error in the area \[ \text{Percentage Error in Area} = \left( \frac{A' - A}{A} \right) \times 100\% \] Now, I will perform the calculations. Calculate the value by following steps: - step0: Calculate: \(40\times 45\) - step1: Multiply the numbers: \(1800\) Calculate or simplify the expression \( 1.05 * 45 \). Calculate the value by following steps: - step0: Calculate: \(1.05\times 45\) - step1: Multiply the numbers: \(47.25\) Calculate or simplify the expression \( 1.05 * 40 \). Calculate the value by following steps: - step0: Calculate: \(1.05\times 40\) - step1: Multiply the numbers: \(42\) Calculate or simplify the expression \( 42 * 47.25 \). Calculate the value by following steps: - step0: Calculate: \(42\times 47.25\) - step1: Multiply the numbers: \(1984.5\) Calculate or simplify the expression \( (1984.5 - 1800) / 1800 * 100 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(1984.5-1800\right)}{1800}\times 100\) - step1: Remove the parentheses: \(\frac{1984.5-1800}{1800}\times 100\) - step2: Subtract the numbers: \(\frac{184.5}{1800}\times 100\) - step3: Divide the terms: \(\frac{41}{400}\times 100\) - step4: Reduce the numbers: \(\frac{41}{4}\times 1\) - step5: Multiply: \(\frac{41}{4}\) Let's summarize the calculations step by step: 1. **Area of the rectangle without error:** \[ A = 40 \times 45 = 1800 \text{ cm}^2 \] 2. **New dimensions considering the \( 5\% \) error:** - New Length: \[ L' = 1.05 \times 40 = 42 \text{ cm} \] - New Width: \[ W' = 1.05 \times 45 = 47.25 \text{ cm} \] 3. **New area with the erroneous dimensions:** \[ A' = 42 \times 47.25 = 1984.5 \text{ cm}^2 \] 4. **Percentage error in the area:** \[ \text{Percentage Error in Area} = \left( \frac{1984.5 - 1800}{1800} \right) \times 100\% = \left( \frac{184.5}{1800} \right) \times 100\% \approx 10.25\% \] Thus, the percentage error in calculating the area of the rectangle is approximately **10.25%**.

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To find the percentage error in the area of the rectangle, first calculate the area using the given dimensions: Area = length × width = 40 cm × 45 cm = 1800 cm². Now, consider the \( 5\% \) error in each dimension. The maximum possible length = \( 40 \text{ cm} + 5\% \text{ of } 40 \text{ cm} = 40 \text{ cm} + 2 \text{ cm} = 42 \text{ cm} \). The minimum length = \( 40 \text{ cm} - 5\% \text{ of } 40 \text{ cm} = 40 \text{ cm} - 2 \text{ cm} = 38 \text{ cm} \). For the width, the maximum possible width = \( 45 \text{ cm} + 5\% \text{ of } 45 \text{ cm} = 45 \text{ cm} + 2.25 \text{ cm} = 47.25 \text{ cm} \) and the minimum width = \( 45 \text{ cm} - 5\% \text{ of } 45 \text{ cm} = 45 \text{ cm} - 2.25 \text{ cm} = 42.75 \text{ cm} \). Now, let's calculate the maximum and minimum areas with the adjusted dimensions. Maximum area = \( 42 \text{ cm} \times 47.25 \text{ cm} = 1983 \text{ cm}^2 \). Minimum area = \( 38 \text{ cm} \times 42.75 \text{ cm} = 1623.5 \text{ cm}^2 \). To find the percentage errors: 1. Percentage error (using max area) = \( \frac{1983 - 1800}{1800} \times 100 \approx 10.17\% \). 2. Percentage error (using min area) = \( \frac{1800 - 1623.5}{1800} \times 100 \approx 9.83\% \). Thus, the percentage error in calculating the area will be approximately \( 10\% \).

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