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 )
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
Beyond the Answer
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 \( 10\% \).
