Free Math Worksheets

- Imagine unfolding a rectangular box. - Opposite faces are congruent. - Pair the three edge lengths to identify the face dimensions.

1. A rectangular prism has \(6\) faces, so its net contains \(6\) rectangles. 2. Opposite faces are congruent. The net needs two \(6\,\text{cm}\times4\,\text{cm}\) rectangles, two \(6\,\text{cm}\times3\,\text{cm}\) rectangles, and two \(4\,\text{cm}\times3\,\text{cm}\) rectangles.

a) The net contains \(6\) faces. b) It contains two \(6\,\text{cm}\times4\,\text{cm}\) rectangles, two \(6\,\text{cm}\times3\,\text{cm}\) rectangles, and two \(4\,\text{cm}\times3\,\text{cm}\) rectangles.

- Use the area formula for a square. - Multiply the area of one face by the number of faces. - The area of the net equals the surface area of the folded cube.

1. The area of one square is \(4 \times 4=16\,\text{cm}^2\). 2. The net has \(6\) congruent squares, so its total area is \(6 \times 16=96\,\text{cm}^2\). 3. Folding the net does not change its area, so the cube has a surface area of \(96\,\text{cm}^2\).

a) \(16\,\text{cm}^2\) b) \(96\,\text{cm}^2\) c) \(96\,\text{cm}^2\)

- A rectangular prism has three pairs of congruent opposite faces. - Find the area of each different rectangle in the net. - Add those three areas and double the sum.

1. Read the three edge lengths from the net: \(7\,\text{cm}\), \(4\,\text{cm}\), and \(2\,\text{cm}\). 2. Find the areas of the three different face types: \(7 \times 4 = 28\,\text{cm}^2\), \(7 \times 2 = 14\,\text{cm}^2\), and \(4 \times 2 = 8\,\text{cm}^2\). 3. Each face type appears twice, so \(S = 2(28 + 14 + 8) = 100\,\text{cm}^2\).

The surface area is \(100\,\text{cm}^2\).

- Which pairs of edge lengths form the three different face types? - Find each different rectangle's area once. - The net contains two of each face type.

1. The three different face areas are \(10 \times 5 = 50\,\text{cm}^2\), \(10 \times 3 = 30\,\text{cm}^2\), and \(5 \times 3 = 15\,\text{cm}^2\). 2. The smallest face area is \(15\,\text{cm}^2\), and the largest face area is \(50\,\text{cm}^2\). 3. Each face type appears twice in the net, so \(S = 2(50 + 30 + 15) = 190\,\text{cm}^2\).

The smallest face has area \(15\,\text{cm}^2\), the largest face has area \(50\,\text{cm}^2\), and the total surface area is \(190\,\text{cm}^2\).

- A square base means that two of the three edge lengths are equal. - Identify which faces in the net have equal areas.

1. The box dimensions are \(20\,\text{cm}\), \(20\,\text{cm}\), and \(5\,\text{cm}\). 2. The top and bottom have total area \(2 \times (20 \times 20) = 800\,\text{cm}^2\). 3. One pair of side faces has total area \(2 \times (20 \times 5) = 200\,\text{cm}^2\). 4. The other pair of side faces also has total area \(2 \times (20 \times 5) = 200\,\text{cm}^2\). 5. The total surface area is \(800 + 200 + 200 = 1200\,\text{cm}^2\).

The net requires \(1200\,\text{cm}^2\) of cardboard.

- How many faces does a cube have? - Find the area of one square in the net. - How can you use one face area to find the total?

1. A cube has \(6\) congruent square faces. 2. The area of one square is \(6 \times 6 = 36\,\text{cm}^2\). 3. The total surface area is \(S = 6 \times 36 = 216\,\text{cm}^2\).

The surface area is \(216\,\text{cm}^2\).

- Group congruent faces by their dimensions. - Find the area of one face of each type. - Add the areas of all six faces.

1. The prism has four \(1 \times 2\) rectangular faces and two \(1 \times 1\) square faces. 2. Each rectangular face has area \(1 \times 2=2\) square units, so the four rectangular faces have total area \(4 \times 2=8\) square units. 3. Each square face has area \(1 \times 1=1\) square unit. The two square faces add \(2\) square units, so the surface area is \(8+2=10\) square units.

a) Four \(1 \times 2\) rectangular faces and two \(1 \times 1\) square faces b) \(8\) square units c) \(10\) square units

- Look closely at the edge labels in the net. - Each rectangle uses two of the three edge lengths. - How many different edge-length values appear?

1. Read the three edge lengths from the net: \(5\,\text{cm}\), \(4\,\text{cm}\), and \(1\,\text{cm}\). 2. Each face type uses a different pair of edge lengths. 3. The three rectangle sizes are \(5\,\text{cm} \times 4\,\text{cm}\), \(5\,\text{cm} \times 1\,\text{cm}\), and \(4\,\text{cm} \times 1\,\text{cm}\).

The face dimensions are \(5\,\text{cm} \times 4\,\text{cm}\), \(5\,\text{cm} \times 1\,\text{cm}\), and \(4\,\text{cm} \times 1\,\text{cm}\).

- Which two edge lengths form the smallest rectangle in the net? - How do you find the area of a rectangle? - Which product of \(6\), \(3\), and \(2\) is the least?

1. The net has three different face types with areas \(6 \times 3 = 18\,\text{cm}^2\), \(6 \times 2 = 12\,\text{cm}^2\), and \(3 \times 2 = 6\,\text{cm}^2\). 2. The least of these areas is \(6\,\text{cm}^2\).

The smallest face has area \(6\,\text{cm}^2\).

- Choose one face as the bottom and mentally fold the adjacent faces upward. - Opposite faces of a rectangular prism are congruent.

1. A rectangular prism has \(6\) faces, so its net contains \(6\) rectangles. 2. Face F is opposite face E. 3. Face C is opposite and congruent to face A. 4. A rectangular prism has \(3\) pairs of opposite, congruent faces.

a) \(6\) rectangles b) Face F c) Face C d) \(3\) pairs

- The widths of the three lateral rectangles match the three sides of the triangular base. - A prism has two congruent bases.

1. The length of the combined lateral rectangle equals the perimeter of the triangular base: \(5+12+13=30\,\text{cm}\). 2. Its height equals the prism length, \(10\,\text{cm}\). 3. The complete net also contains two congruent triangles with side lengths \(5\,\text{cm}\), \(12\,\text{cm}\), and \(13\,\text{cm}\).

a) The rectangle is \(30\,\text{cm}\) long and \(10\,\text{cm}\) high. b) The net also contains \(2\) congruent triangular faces.

- A rectangular prism has three pairs of congruent faces. - A scale of \(1{:}2\) halves every length.

1. The surface area is \(2(10\times6+10\times4+6\times4)=2(60+40+24)=248\,\text{cm}^2\). 2. At a scale of \(1{:}2\), each length is divided by \(2\). The scaled edge lengths are \(5\,\text{cm}\), \(3\,\text{cm}\), and \(2\,\text{cm}\). 3. The rectangle types are \(5\,\text{cm}\times3\,\text{cm}\), \(5\,\text{cm}\times2\,\text{cm}\), and \(3\,\text{cm}\times2\,\text{cm}\).

a) The surface area is \(248\,\text{cm}^2\). b) The rectangle types are \(5\,\text{cm}\times3\,\text{cm}\), \(5\,\text{cm}\times2\,\text{cm}\), and \(3\,\text{cm}\times2\,\text{cm}\).

- A rectangular prism has three pairs of congruent opposite faces. - Match each existing rectangle with its partner. - Add the areas of all six faces.

1. A rectangular prism has three pairs of congruent opposite faces: two \(6\,\text{cm} \times 4\,\text{cm}\) rectangles, two \(6\,\text{cm} \times 2\,\text{cm}\) rectangles, and two \(4\,\text{cm} \times 2\,\text{cm}\) rectangles. 2. The student still needs one \(6\,\text{cm} \times 2\,\text{cm}\) rectangle and one \(4\,\text{cm} \times 2\,\text{cm}\) rectangle. 3. The total area is \(2\times(6 \times 4)+2\times(6 \times 2)+2\times(4 \times 2)=48+24+16=88\,\text{cm}^2\).

a) One \(6\,\text{cm} \times 2\,\text{cm}\) rectangle and one \(4\,\text{cm} \times 2\,\text{cm}\) rectangle b) \(88\,\text{cm}^2\)

- A rectangular prism has three pairs of congruent opposite faces. - Use the area formula for each rectangle. - Add the areas of all three face pairs.

1. The net has two \(12\,\text{in.} \times 8\,\text{in.}\) rectangles, two \(12\,\text{in.} \times 4\,\text{in.}\) rectangles, and two \(8\,\text{in.} \times 4\,\text{in.}\) rectangles. 2. The areas of the three pairs are \(2 \times 12 \times 8=192\,\text{in.}^2\), \(2 \times 12 \times 4=96\,\text{in.}^2\), and \(2 \times 8 \times 4=64\,\text{in.}^2\). 3. The surface area is \(192+96+64=352\,\text{in.}^2\).

a) Two \(12\,\text{in.} \times 8\,\text{in.}\) rectangles, two \(12\,\text{in.} \times 4\,\text{in.}\) rectangles, and two \(8\,\text{in.} \times 4\,\text{in.}\) rectangles b) \(192\,\text{in.}^2\), \(96\,\text{in.}^2\), and \(64\,\text{in.}^2\) c) \(352\,\text{in.}^2\)

- What length, width, and height does the net show? - Group the six rectangles into three congruent pairs. - Add the areas of all six faces. - Multiply the number of square decimeters by the mass per square decimeter.

1. Read the rectangular prism's dimensions from the net: \(5\,\text{dm}\), \(3\,\text{dm}\), and \(2\,\text{dm}\). 2. The net contains two \(5\,\text{dm} \times 3\,\text{dm}\) faces, two \(3\,\text{dm} \times 2\,\text{dm}\) faces, and two \(5\,\text{dm} \times 2\,\text{dm}\) faces. 3. Its total area is \(S = 2(5 \times 3 + 3 \times 2 + 5 \times 2) = 2(15 + 6 + 10) = 62\,\text{dm}^2\). 4. The mass is \(62 \times 5 = 310\,\text{g}\).

The finished box has a mass of \(310\,\text{g}\).

- Which rectangles in the net have the same dimensions? - How many pairs of congruent faces does a rectangular prism have? - Find the area of one rectangle of each type. - Add the areas of all six faces.

1. The net has three pairs of congruent rectangles: two \(5\,\text{cm} \times 2\,\text{cm}\) rectangles, two \(5\,\text{cm} \times 4\,\text{cm}\) rectangles, and two \(2\,\text{cm} \times 4\,\text{cm}\) rectangles. 2. The area of one rectangle in each pair is \(5 \times 2 = 10\,\text{cm}^2\), \(5 \times 4 = 20\,\text{cm}^2\), and \(2 \times 4 = 8\,\text{cm}^2\). 3. Add all six face areas: \(S = 2 \times 10 + 2 \times 20 + 2 \times 8 = 76\,\text{cm}^2\).

a) Two \(5\,\text{cm} \times 2\,\text{cm}\) rectangles, two \(5\,\text{cm} \times 4\,\text{cm}\) rectangles, and two \(2\,\text{cm} \times 4\,\text{cm}\) rectangles b) \(76\,\text{cm}^2\)

- Use the area formula for a square. - A cube has \(6\) congruent faces. - Choose face \(C\) as the bottom and track the farthest face as the net folds.

1. The area of one square face is \(3 \times 3=9\,\text{cm}^2\). 2. A cube has \(6\) congruent faces, so its surface area is \(6 \times 9=54\,\text{cm}^2\). 3. Use face \(C\) as the bottom. Face \(F\) folds over to become the top face, so \(F\) is opposite \(C\).

a) \(9\,\text{cm}^2\) b) \(54\,\text{cm}^2\) c) Face \(F\)

- Begin with square A as the bottom face. - Track the marked corner as square E folds upward. - Determine which corners of the front and top faces reach the same cube vertex.

1. Treat square A as the bottom face. Square E folds upward along A's left edge. 2. The marked top-left corner of E becomes one vertex of the folded cube. 3. Square B folds upward from A, and its top-left corner reaches the same vertex. 4. Square C folds over B to form the top face, and its bottom-left corner also reaches that vertex.

The top-left corner of square B and the bottom-left corner of square C meet the blue corner.

- What three different edge lengths are labeled on the net? - Which faces form congruent pairs? - Find the area of each type of face before adding all six areas. - How can you organize the three pairs in one surface-area expression?

1. Read the three dimensions from the net. The greatest is \(6\,\text{cm}\), the middle dimension is \(4\,\text{cm}\), and the least is \(3\,\text{cm}\). Therefore, the length is \(6\,\text{cm}\), the width is \(4\,\text{cm}\), and the height is \(3\,\text{cm}\). 2. The net contains two \(6\,\text{cm} \times 4\,\text{cm}\) faces, two \(4\,\text{cm} \times 3\,\text{cm}\) faces, and two \(6\,\text{cm} \times 3\,\text{cm}\) faces. 3. Add their areas: \(S = 2(6 \times 4 + 4 \times 3 + 6 \times 3) = 2(24 + 12 + 18) = 108\,\text{cm}^2\).

a) Length: \(6\,\text{cm}\); width: \(4\,\text{cm}\); height: \(3\,\text{cm}\) b) \(108\,\text{cm}^2\)

- How do you find the area of each rectangle in the net? - How many faces does a rectangular prism have? - How many square centimeters are in one square decimeter? - Once the area is in square decimeters, how can you use the mass rate?

1. The net contains three pairs of congruent rectangles. Its total area is \(S = 2(10 \times 15 + 10 \times 20 + 15 \times 20) = 2(150 + 200 + 300) = 1300\,\text{cm}^2\). 2. Convert the area to square decimeters: \(1300\,\text{cm}^2 = 13\,\text{dm}^2\). 3. Find the mass: \(13 \times 32 = 416\,\text{g}\).

1. \(1300\,\text{cm}^2\) 2. \(416\,\text{g}\)

- Picture each square folding along its shared edges. - Check whether any faces would overlap. - After finding the valid net, add the areas of its six square faces.

1. Net 1 is valid. Its squares can fold to make one bottom face, four side faces, and one top face. 2. Net 2 contains a \(2 \times 2\) block. When folded, some faces overlap instead of forming distinct faces of a cube. 3. Net 3 contains five squares in one straight row. When wrapped around the cube, the fifth square overlaps the first, so the net cannot close correctly. 4. Each square has area \(2 \times 2=4\,\text{cm}^2\). The valid net has \(6\) squares, so the cube's surface area is \(6 \times 4=24\,\text{cm}^2\).

a) Net 1 b) Net 2 contains a \(2 \times 2\) block that causes overlapping faces. Net 3 has five squares in a row, so one face overlaps another when folded. c) \(24\,\text{cm}^2\)

- First determine which faces will be opposite after folding. - Each opposite pair must add to \(7\). - Use the area of one square face to find the total area of all six faces.

1. Mentally fold the net. The opposite pairs are \(1\) and \(y\), \(2\) and \(x\), and \(3\) and \(z\). 2. Use the rule that opposite faces add to \(7\): \(1+y=7\), so \(y=6\); \(2+x=7\), so \(x=5\); and \(3+z=7\), so \(z=4\). 3. The area of one face is \(2 \times 2=4\,\text{cm}^2\). 4. The cube has \(6\) faces, so its surface area is \(6 \times 4=24\,\text{cm}^2\).

a) \(x=5\), \(y=6\), and \(z=4\) b) \(4\,\text{cm}^2\) c) \(24\,\text{cm}^2\)

- Identify the face opposite each labeled red symbol after folding. - In the four-square row, faces two positions apart become opposite.

1. The square at the far right of the horizontal row is opposite RS, so it must contain BS. 2. The square directly below RS is opposite RC, so it must contain BC. 3. The square directly to the right of RS is opposite RT, so it must contain BT.

Place BS in the far-right square of the horizontal row, BC directly below RS, and BT directly to the right of RS.

- Find the face opposite \(B\) after folding. - The other four faces become the side faces. - Add the area of the fully painted face to half the area of each side face.

1. Face \(D\) folds to the position opposite face \(B\), so \(D\) is the top face. 2. Faces \(A\), \(C\), \(E\), and \(F\) fold up around \(B\) to form the four side faces, so those four faces are half blue. 3. One face has area \(4 \times 4=16\,\text{cm}^2\). The fully painted bottom contributes \(16\,\text{cm}^2\). Each half-painted side contributes \(8\,\text{cm}^2\), so the four sides contribute \(4 \times 8=32\,\text{cm}^2\). 4. The total painted area is \(16+32=48\,\text{cm}^2\).

a) Face \(D\) b) Faces \(A\), \(C\), \(E\), and \(F\) c) \(48\,\text{cm}^2\)

- Which parts of the surface are painted? Do not forget the bottom face. - How high does the paint reach on each side face? - Find the bottom area and the four rectangular side strips separately.

1. The bottom face has area \(8 \times 5 = 40\,\text{cm}^2\). 2. Half the prism's height is \(4 \div 2 = 2\,\text{cm}\), so each painted side strip is \(2\,\text{cm}\) high. 3. The two long side strips have total area \(2(8 \times 2) = 32\,\text{cm}^2\). 4. The two short side strips have total area \(2(5 \times 2) = 20\,\text{cm}^2\). 5. The total painted area is \(40 + 32 + 20 = 92\,\text{cm}^2\).

The painted surface area is \(92\,\text{cm}^2\).

- Find the surface area of each solid separately. - All six faces of the cube are congruent squares. - The rectangular prism has three pairs of congruent rectangles.

1. For the cube in a), the edge length is \(5\,\text{cm}\). Its surface area is \(S_{\text{cube}} = 6 \times 5^2 = 150\,\text{cm}^2\). 2. For the rectangular prism in b), the dimensions are \(6\,\text{cm}\), \(5\,\text{cm}\), and \(4\,\text{cm}\). Its surface area is \(S_{\text{prism}} = 2(6 \times 5 + 6 \times 4 + 5 \times 4) = 2(30 + 24 + 20) = 148\,\text{cm}^2\). 3. Since \(150\,\text{cm}^2 > 148\,\text{cm}^2\), the cube has the greater surface area.

The cube in a) has the greater surface area. Its surface area is \(150\,\text{cm}^2\), while the rectangular prism in b) has surface area \(148\,\text{cm}^2\).

- How many edges does a cube have? - Once you know one edge length, find the area of one square face. - How many square faces are in the net?

1. A cube has \(12\) congruent edges, so one edge is \(96 \div 12 = 8\,\text{cm}\) long. 2. The net consists of \(6\) congruent squares. 3. Its total area is \(6 \times 8^2 = 6 \times 64 = 384\,\text{cm}^2\).

The area of the net is \(384\,\text{cm}^2\).

- Identify the three edge lengths from the net. - Add the areas of all six rectangles for surface area. - Multiply the three dimensions for volume.

1. The prism's dimensions are \(5\,\text{cm}\), \(4\,\text{cm}\), and \(3\,\text{cm}\). 2. Its surface area is \(2(5\times4+4\times3+5\times3)=94\,\text{cm}^2\). 3. Its volume is \(5\times4\times3=60\,\text{cm}^3\).

a) The surface area is \(94\,\text{cm}^2\). b) The volume is \(60\,\text{cm}^3\).

- A cube net contains \(6\) squares. - Picture the row of four squares wrapping around the cube. - Check whether the two remaining squares close different faces. - Use the area of one square to find the area of all six faces.

1. The diagram has three squares in a horizontal row and one square above the middle square. 2. Adding one square to the left end of the row makes a row of four squares. Adding one square below the middle square gives two additional faces on opposite sides of the row. 3. The four squares in the row fold around to form the side faces. The square above and the square below close the top and bottom without overlapping, so the arrangement is a valid cube net. 4. Each face has area \(3 \times 3=9\,\text{cm}^2\). A cube has \(6\) faces, so its surface area is \(6 \times 9=54\,\text{cm}^2\).

a) Yes. b) The four squares in the row form the side faces, and the squares above and below close the top and bottom without overlap. c) \(54\,\text{cm}^2\)

- Count the equal-length segments along the outside boundary of the net. - Once you know one square's side length, find its area. - How many squares make up the cube's surface?

1. The outside boundary of this cube net contains \(14\) segments, each equal to one square's side length. 2. The side length is \(112 \div 14 = 8\,\text{cm}\). 3. The area of one square is \(8 \times 8 = 64\,\text{cm}^2\). 4. The net has \(6\) squares, so the cube's surface area is \(6 \times 64 = 384\,\text{cm}^2\).

The cube's surface area is \(384\,\text{cm}^2\).

- Find the area of one square face. - Multiply by the number of faces in a cube net. - Use the given mass for each \(10\,\text{cm}^2\) of cardstock.

1. One square face has area \(5\,\text{cm} \times 5\,\text{cm}=25\,\text{cm}^2\). 2. The net has \(6\) congruent squares, so its total area is \(6 \times 25\,\text{cm}^2=150\,\text{cm}^2\). 3. The net contains \(150 \div 10=15\) groups of \(10\,\text{cm}^2\). 4. Its mass is \(15 \times 2\,\text{g}=30\,\text{g}\).

The net has area \(150\,\text{cm}^2\), and the finished model has mass \(30\,\text{g}\).

- How many complete face areas are painted altogether? - What fraction of each side face is painted when the cube is submerged halfway? - Find the area of one square face first. - Combine all the painted pieces.

1. The painted region includes one complete square face and the lower half of each of four side faces. 2. One complete face has area \(5 \times 5 = 25\,\text{cm}^2\). 3. The four half-faces have the same total area as \(4 \times \frac{1}{2} = 2\) complete faces. Altogether, the painted region equals \(3\) complete faces. 4. The painted area is \(3 \times 25 = 75\,\text{cm}^2\).

The painted surface area is \(75\,\text{cm}^2\).

- Hold U fixed as the bottom and fold its four neighboring faces upward. - The face attached beyond H folds over the top. - The back face is opposite the chosen front face.

1. With U on the bottom, faces L, R, V, and H fold upward to form the four side faces. 2. Face O is attached to H and folds over to form the top, so O is opposite U. 3. If V is the front face, H is the opposite back face.

Face O is the top. If V is the front, H is the back.

- Determine the three edge lengths one at a time from the relationships in the problem. - Add the areas of the six rectangles in the net, or organize them as three congruent pairs. - Check the placement of the decimal point in each product.

1. The length is \(10\,\text{cm}\). The width is \(10 \div 2 = 5\,\text{cm}\), and the height is \(5 \div 2 = 2.5\,\text{cm}\). 2. The three different face areas are \(10 \times 5 = 50\,\text{cm}^2\), \(10 \times 2.5 = 25\,\text{cm}^2\), and \(5 \times 2.5 = 12.5\,\text{cm}^2\). 3. Each face type appears twice, so \(S = 2(50 + 25 + 12.5) = 175\,\text{cm}^2\).

The surface area is \(175\,\text{cm}^2\).

- How many rectangles make up the surface of a rectangular prism? - How many pairs of congruent faces are there? - The problem includes both the interior and the exterior. - First find the area of one side of the material, then account for both sides.

1. The three different face areas are \(12 \times 5 = 60\,\text{cm}^2\), \(12 \times 3 = 36\,\text{cm}^2\), and \(5 \times 3 = 15\,\text{cm}^2\). 2. The exterior surface area is \(S = 2(60 + 36 + 15) = 222\,\text{cm}^2\). 3. The interior has the same area as the exterior, so the total painted area is \(2 \times 222 = 444\,\text{cm}^2\).

The total area to be painted is \(444\,\text{cm}^2\).

- Imagine wrapping the row around the four side faces of a cube. - Track which faces do not share an edge after folding. - In the flat cross, compare how many squares separate each candidate opposite pair.

1. Only four faces can wrap around the sides of a cube. A fifth square in the same row folds onto the first square, causing an overlap. Adding a sixth square cannot remove that overlap. 2. Therefore, at most \(4\) squares can lie in one straight row of a cube net. 3. In the described cross-shaped net, each pair of opposite faces has exactly one square between the two faces in the flat net.

a) The fifth square would overlap the first when the row is folded around the cube. b) At most \(4\) squares may lie in one row. c) Exactly \(1\) square lies between them.

- Track each marked corner as the net folds. - Exactly three faces meet at every cube vertex.

1. Fold the second square in the horizontal row as the front face. The square above becomes the top, the far-left square becomes the left face, the far-right square becomes the back, and the square below becomes the bottom. 2. The marked top-left corner of the top face, the marked top-left corner of the far-left face, and the marked top-right corner of the far-right face meet at one cube vertex. 3. The marked bottom-right corner of the lower square reaches a different vertex.

\(3\) marked corners meet at one vertex.

- Choose the central square as a reference face and fold its neighbors upward. - Track each marked corner separately. - Determine which corner of A shares the blue mark's final vertex.

1. Use the central square as one face and fold the four adjacent squares upward. 2. The blue mark reaches one cube vertex, while the yellow mark reaches the vertex diagonally opposite it. Therefore, the marks do not meet. 3. The blue mark meets the top-left corner of square A.

a) No. The blue and yellow marks reach opposite vertices of the cube. b) The blue mark meets the top-left corner of square A.

- Identify the two square faces and four rectangular faces in the net. - Convert square centimeters to square decimeters. - Multiply the area by the mass per square decimeter.

1. The net has two \(15\,\text{cm}\times15\,\text{cm}\) faces and four \(60\,\text{cm}\times15\,\text{cm}\) faces. 2. The surface area is \(2(15\times15)+4(60\times15)=4050\,\text{cm}^2\). 3. Convert the area: \(4050\,\text{cm}^2=40.5\,\text{dm}^2\). 4. The mass is \(40.5\times150=6075\,\text{g}=6.075\,\text{kg}\).

The surface area is \(40.5\,\text{dm}^2\). The mass is \(6075\,\text{g}\), or \(6.075\,\text{kg}\).

- A rectangular prism has \(6\) faces in \(3\) congruent pairs. - Check whether touching edges have matching lengths after folding. - Check for overlapping faces or gaps. - Add the areas of all six faces in each valid net.

1. Figure a) has three pairs of congruent rectangular faces with dimensions \(2 \times 3\), \(1 \times 3\), and \(2 \times 1\). Its matching edges align when folded, so it is a valid net. 2. In figure b), two square faces fold into the same position, leaving another side open. It is not a valid net. 3. In figure c), the attached \(1 \times 1\) squares do not match the full \(2\)-unit edges of the larger rectangles. The solid would have gaps, so it is not a valid net. 4. Figure d) is a valid cube net. Because a cube is a rectangular prism, it is also a valid rectangular-prism net. 5. Figure a) represents a \(2 \times 1 \times 3\) rectangular prism. Its surface area is \(2 \times 3+2 \times 3+1 \times 3+1 \times 3+2 \times 1+2 \times 1=22\) square units. 6. Figure d) represents a cube with side length \(2\) units. Its surface area is \(6 \times (2 \times 2)=24\) square units.

a) Figures a) and d) are valid. Figure b) causes overlapping faces and leaves a side open. Figure c) has mismatched edge lengths that create gaps. b) \(22\) square units c) \(24\) square units

- Mentally fold each arrangement and look for overlaps or missing faces. - A cube has three pairs of opposite faces. - Its surface consists of six congruent squares.

1. Arrangements (1) and (3) fold into closed cubes without overlap. In (2), two faces overlap and one face remains open. In (4), the six squares form one straight row and cannot fold into a cube. 2. In net (1), the opposite-face pairs are A and C, B and D, and E and F. 3. One face has area \(5\times5=25\,\text{cm}^2\). The cube's surface area is \(6\times25=150\,\text{cm}^2\).

a) Arrangements (1) and (3) are valid cube nets. b) The opposite pairs are A–C, B–D, and E–F. c) The surface area is \(150\,\text{cm}^2\).

- Check that the six faces form three pairs of congruent rectangles. - Mentally fold the strip and watch for overlaps. - Edges that meet after folding must have equal lengths.

1. Net a) is valid. It contains three pairs of congruent rectangles—\(2\times3\), \(2\times1\), and \(1\times3\)—arranged so that all matching edges meet without overlap. 2. Net b) is invalid. The two \(1\times3\) side faces are attached on the same side of the strip and overlap when folded, leaving the opposite side open. 3. Net c) is invalid. It does not contain three congruent pairs of rectangles: a \(2\times2\) face is paired with a \(2\times1\) face. 4. Net d) is invalid. Although the face sizes form pairs, the order of the faces makes unequal edge lengths meet when folded.

a) Valid. b) Invalid. The side faces overlap. c) Invalid. The faces do not form three congruent pairs. d) Invalid. Unequal edge lengths would have to meet.