Question

ms 3–5: here is a polyhedron and its matching net.

1. what is the name of this type of polyhedron?
2. use the polyhedron to label all the lengths in this net.
3. use the net to calculate the surface area in square units. show or explain your thinking.

Explanation:

Problem 3:

From the polyhedron (a rectangular prism or cuboid, as it has rectangular faces) and its net, the polyhedron is a rectangular prism (or cuboid). A rectangular prism has 6 faces, with opposite faces equal, and its net consists of rectangles (and possibly squares) arranged to form the 3D shape.

The polyhedron (rectangular prism) has dimensions, likely from the given numbers (4, 5, and let's assume the third dimension, say, if the square face is, for example, side 4 or 5? Wait, looking at the polyhedron, if one face is a rectangle with sides 4 and 5, and the square (or another rectangle) – wait, the net has rectangles and squares. Let's assume the rectangular prism has length \( l = 5 \), width \( w = 4 \), and height \( h = 4 \) (if there's a square face). So in the net, the rectangles will have lengths and widths corresponding to the prism's edges. So label the lengths: the longer rectangles (the lateral faces) will have length 5 and width 4, and the square faces (if any) will have side 4. Wait, maybe the prism has dimensions 4, 4, 5? So in the net, the rectangles: two with \( 5 \times 4 \), two with \( 4 \times 4 \), and two with \( 5 \times 4 \)? Wait, no, a rectangular prism has 6 faces: 2 of \( l \times w \), 2 of \( l \times h \), 2 of \( w \times h \). If from the polyhedron, we see edges 4 and 5, so let's say \( l = 5 \), \( w = 4 \), \( h = 4 \) (so two square faces of \( 4 \times 4 \) and four rectangular faces of \( 5 \times 4 \))? Wait, the net shown has some squares and rectangles. So label the lengths: the square faces have side 4, the rectangular faces have length 5 and width 4. So in the net, each square is labeled 4 (side), and each rectangle is labeled 5 (length) and 4 (width).

Step 1: Identify the faces of the net

The net of the rectangular prism (cuboid) has 2 square faces (side \( 4 \)) and 4 rectangular faces (length \( 5 \), width \( 4 \))? Wait, no, a rectangular prism with dimensions \( l = 5 \), \( w = 4 \), \( h = 4 \) has:

• 2 faces: \( l \times w = 5 \times 4 \)
• 2 faces: \( l \times h = 5 \times 4 \)
• 2 faces: \( w \times h = 4 \times 4 \)

Wait, actually, if \( w = h = 4 \), then it's a rectangular prism with two square faces (\( 4 \times 4 \)) and four rectangular faces (\( 5 \times 4 \)).

Step 2: Calculate area of each type of face

• Area of square face: \( 4 \times 4 = 16 \) square units. There are 2 such faces, so total area for squares: \( 2 \times 16 = 32 \).
• Area of rectangular face: \( 5 \times 4 = 20 \) square units. There are 4 such faces, so total area for rectangles: \( 4 \times 20 = 80 \).

Step 3: Sum the areas

Total surface area = Area of squares + Area of rectangles = \( 32 + 80 = 112 \) square units. Wait, alternatively, maybe the prism has dimensions \( 4 \), \( 5 \), and another side, say, if the net has 3 rectangles of \( 5 \times 4 \), and 2 squares of \( 4 \times 4 \), no, wait the standard formula for surface area of a rectangular prism is \( 2(lw + lh + wh) \). Let's use that formula. Let \( l = 5 \), \( w = 4 \), \( h = 4 \). Then:
\( 2(lw + lh + wh) = 2(5 \times 4 + 5 \times 4 + 4 \times 4) = 2(20 + 20 + 16) = 2(56) = 112 \).

Answer:

Rectangular Prism (or Cuboid)

Problem 4: