Question

a. calculate the areas of the faces of the box. round your answers to the nearest tenth of a square inch and list them in descending order.
there are two faces that have an area of in², two faces that have an area of in², and two faces that have an area of in².
b. determine the surface area of the box. round your answer to the nearest tenth of a square inch.

c. determine the range of values that should contain the actual (true) measure of the surface area of the box. explain your reasoning.
the calculation of surface area is accurate to the nearest tenth. the true surface area falls between in² and in².
d. suppose that jayden had incorrectly measured the first dimension as 15.1 inches. find the surface area of the box using this measure. round your answer to the nearest tenth of a square inch.

Explanation:

Step1: Calculate the areas of the faces

• For the face with dimensions \(15.3\times7.28\):

\(A_1 = 15.3\times7.28=111.384\approx111.4\)

• For the face with dimensions \(15.3\times4.92\):

\(A_2=15.3\times4.92 = 75.276\approx75.3\)

• For the face with dimensions \(7.28\times4.92\):

\(A_3=7.28\times4.92=35.8176\approx35.8\)

Step2: Calculate the surface area

The surface area formula for a rectangular prism is \(S = 2(A_1+A_2+A_3)\)
\(S=2(111.4 + 75.3+35.8)\)
\(S=2\times222.5\)
\(S = 445.0\)

Step3: Determine the range of values for the surface area

Since each measurement is accurate to the nearest tenth, the maximum error for each measurement is \(0.05\)

• Minimum dimensions: \(15.25\times7.23\times4.87\)

\(A_{1min}=15.25\times7.23 = 110.2575\)
\(A_{2min}=15.25\times4.87=74.3675\)
\(A_{3min}=7.23\times4.87 = 35.2101\)
\(S_{min}=2(110.2575+74.3675 + 35.2101)=2\times219.8351=439.6702\approx439.7\)

• Maximum dimensions: \(15.35\times7.33\times4.97\)

\(A_{1max}=15.35\times7.33=112.5155\)
\(A_{2max}=15.35\times4.97 = 76.2895\)
\(A_{3max}=7.33\times4.97=36.4301\)
\(S_{max}=2(112.5155+76.2895+36.4301)=2\times225.2351 = 450.4702\approx450.5\)

Step4: Calculate the surface area with incorrect measurement (\(l = 15.1\))

\(A_1=15.1\times7.28 = 109.928\)
\(A_2=15.1\times4.92=74.292\)
\(A_3=7.28\times4.92 = 35.8176\)
\(S=2(109.928+74.292+35.8176)=2\times220.0376=440.0752\approx440.1\)

Answer:

a. Two faces have an area of \(111.4\) \(in^2\), two faces have an area of \(75.3\) \(in^2\), and two faces have an area of \(35.8\) \(in^2\)
b. \(445.0\) \(in^2\)
c. The true surface area falls between \(439.7\) \(in^2\) and \(450.5\) \(in^2\)
d. \(440.1\) \(in^2\)