Question

name
class period
hw qtr2 wk 3
problem 1
can the following net be assembled into a cube? explain how you know. label parts of the net
with letters or numbers if it helps your explanation.
problem 2
a. what polyhedron can be assembled from this net? explain how you know.
b. find the surface area of this polyhedron.

Explanation:

Problem 1

Step1: Recall cube net properties

A cube net has 6 square faces.

Step2: Check the given net

The given net has 6 square - shaped regions. When folded, opposite faces will be parallel and congruent. For example, if we consider the top - most square as the top face of the cube, we can match the adjacent squares to form the front, back, left, right and bottom faces.

Problem 2a

Step1: Identify the polyhedron

Since the net can be folded into a cube (from problem 1), the polyhedron is a cube. A cube is a type of hexahedron (a polyhedron with 6 faces).

Step2: Justify

All the faces of the net are squares. When folded, the edges will match perfectly to form a 3 - D shape with 6 square faces, 12 edges and 8 vertices, which is the definition of a cube.

Problem 2b

Step1: Let the side length of each square be \(s\)

The area of one square face is \(A_{face}=s^{2}\)

Step2: Calculate the surface area

A cube has 6 faces. The surface area \(S\) of a cube is the sum of the areas of its 6 faces. So \(S = 6\times s^{2}\)

Answer:

Problem 1

Yes, the net can be assembled into a cube. When folded, the 6 square - shaped regions of the net will form a 3 - D shape with 6 square faces (a cube).

Problem 2a

The polyhedron is a cube. Because the net has 6 square faces which can be folded to form a 3 - D shape with 6 square faces (a cube).

Problem 2b

If the side length of each square in the net is \(s\), the surface area of the cube (polyhedron) is \(6s^{2}\)