Question

perservere jayden measures and labels the dimensions of a box.
4.92 in
15.3 in
2.28 in

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 (square) in(^{2}), two faces that have an area of (square) in(^{2}), and two faces that have an area of (square) in(^{2}).

b. determine the surface area of the box. round your answer to the nearest tenth of a square inch.
(square) in(^{2})

Explanation:

Step1: Calculate the area of each pair of faces

• For the pair of faces with dimensions \(15.3\times4.92\)

The area formula for a rectangle is \(A = l\times w\). So, \(A_1=15.3\times4.92 = 75.276\approx75.3\) \(in^2\)

• For the pair of faces with dimensions \(15.3\times2.28\)

Using the area formula \(A = l\times w\), \(A_2=15.3\times2.28=34.884\approx34.9\) \(in^2\)

• For the pair of faces with dimensions \(4.92\times2.28\)

Using the area formula \(A = l\times w\), \(A_3=4.92\times2.28 = 11.1176\approx11.1\) \(in^2\)

Step2: Calculate the surface area of the box

The surface area formula for a rectangular prism is \(SA=2(A_1 + A_2+A_3)\)
Substitute \(A_1 = 75.3\), \(A_2=34.9\), \(A_3 = 11.1\) into the formula:
\(SA=2(75.3 + 34.9+11.1)=2\times121.3=242.6\) \(in^2\)

Answer:

a. There are two faces that have an area of \(75.3\) \(in^2\), two faces that have an area of \(34.9\) \(in^2\), and two faces that have an area of \(11.1\) \(in^2\)
b. \(242.6\) \(in^2\)