Question

the rectangle below is labeled with its measured dimensions. taking measurement error into account, what are the minimum and maximum possible areas? minimum possible area = yd² maximum possible area = yd²

Explanation:

Step1: Determine the maximum and minimum values for each dimension

Assume the measurement error is \(0.05\) (since it's common in such problems, if not specified, we can assume half of the smallest - unit digit's place. Here, if we assume the measurements are to the nearest \(0.1\) (because \(3.61\) and \(3.24\) are given to two decimal places, but the error - calculation logic: for a measurement \(x\) given as \(a.bc\), the maximum value \(x_{max}=a.bc + 0.005\) and the minimum value \(x_{min}=a.bc-0.005\)).
For length \(l = 3.61\) yd: \(l_{max}=3.61 + 0.005=3.615\) yd, \(l_{min}=3.61 - 0.005 = 3.605\) yd.
For width \(w = 3.24\) yd: \(w_{max}=3.24+0.005 = 3.245\) yd, \(w_{min}=3.24 - 0.005=3.235\) yd.

Step2: Calculate the maximum area

The area of a rectangle is \(A = l\times w\). The maximum area occurs when both length and width are at their maximum values.
\(A_{max}=l_{max}\times w_{max}=3.615\times3.245\)

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Step3: Calculate the minimum area

The minimum area occurs when both length and width are at their minimum values.
\(A_{min}=l_{min}\times w_{min}=3.605\times3.235\)

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Answer:

Minimum possible area \( = 11.66\space yd^{2}\), Maximum possible area \(=11.73\space yd^{2}\)