- Count the grid squares along two sides of each lettered rectangle. - Which labeled rectangles have the same width and height? - Check the areas in square centimeters.

1. Faces \(E\) and \(F\) are each \(2\) by \(3\) grid squares, so each has area \(6\,\text{cm}^2\). 2. Faces \(B\) and \(D\) are each \(4\) by \(2\) grid squares, so each has area \(8\,\text{cm}^2\).

\(6\,\text{cm}^2\): \(E\) and \(F\). \(8\,\text{cm}^2\): \(B\) and \(D\).

- Count the unit squares in the long middle strip. - What extra squares stick out to the sides? - Check that every small square is counted exactly once.

1. The long strip contains \(2\times8=16\) unit squares. 2. The two side sections each contain \(2\times2=4\) unit squares. 3. The net covers \(16+4+4=24\,\text{cm}^2\).

\(24\,\text{cm}^2\).

- Follow the edges where each triangle joins the strip. - Count the separate flat regions that will become faces.

1. The upper and lower triangles share edges with the leftmost rectangle in the strip. 2. The strip has three rectangular regions and the net has two triangular regions. All five become faces after folding.

The left rectangle touches both triangles. The solid has \(5\) faces.

- Look at the shape to which the other pieces attach. - Imagine the outside pieces folding upward around it.

1. The central square becomes the base. 2. Each of the four triangles attached to the square folds upward into a side face.

The square is the base, and there are \(4\) triangular side faces.

- 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 \(10\times5=50\,\text{cm}^2\), \(10\times3=30\,\text{cm}^2\), and \(5\times3=15\,\text{cm}^2\). 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 \(SA=2(50+30+15)=190\,\text{cm}^2\).

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

- Read the base side length and box height from the net. - A square base means that two of the three edge lengths are equal. - Identify which faces in the net have equal areas before adding them.

1. The box dimensions shown are \(20\,\text{cm}\), \(20\,\text{cm}\), and \(5\,\text{cm}\). 2. The top and bottom have total area \(2\times(20\times20)=800\,\text{cm}^2\). 3. One pair of side faces has total area \(2\times(20\times5)=200\,\text{cm}^2\). 4. The other pair of side faces also has total area \(2\times(20\times5)=200\,\text{cm}^2\). 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? - Read the side length from the net and find the area of one square. - Use one face area to find the total area of all six faces.

1. A cube has \(6\) congruent square faces. 2. The labeled side length gives one face area of \(6\times6=36\,\text{cm}^2\). 3. The total surface area is \(SA=6\times36=216\,\text{cm}^2\).

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

- Read the two face sizes from the net and group congruent faces. - Find the area of one face of each type. - Add the areas of all six faces.

1. From the labels in the net, the prism has four \(1\times2\) rectangular faces and two \(1\times1\) square faces. 2. Each rectangular face has area \(1\times2=2\) square units, so the four rectangular faces have total area \(4\times2=8\) square units. 3. Each square face has area \(1\times1=1\) square unit. The two square faces add \(2\) square units, so the surface area is \(8+2=10\) square units.

a) Four larger rectangular faces and two 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 labeled edge lengths form the smallest rectangle in the net? - How do you find the area of a rectangle? - Compare the three possible face-area products.

1. The net has three different face types with areas \(6\times3=18\,\text{cm}^2\), \(6\times2=12\,\text{cm}^2\), and \(3\times2=6\,\text{cm}^2\). 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. - Track which face ends up directly across from the face named in each part. - Opposite faces of a rectangular prism are congruent.

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

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

- Separate the net into the two triangular bases and the three rectangular side faces. - Use the marked right angle to identify a perpendicular base-height pair for a triangular face. - The widths of the three rectangles correspond to the three side lengths of the triangular base.

1. The marked right angle shows that the triangular base has perpendicular legs \(3\,\text{cm}\) and \(4\,\text{cm}\), so one triangle has area \(\frac{1}{2}\times3\times4=6\,\text{cm}^2\). The two triangles have total area \(12\,\text{cm}^2\). 2. The three rectangles have common prism length \(6\,\text{cm}\) and widths \(3\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). 3. Their total area is \(6\times(3+4+5)=72\,\text{cm}^2\). 4. The total surface area is \(12+72=84\,\text{cm}^2\).

\(84\,\text{cm}^2\)

- Separate the square base from the four congruent triangular faces. - Use the height drawn on a triangular face when finding that triangle's area. - Add the area of the base to the combined area of all four triangles.

1. The square base has area \(6\times6=36\,\text{cm}^2\). 2. Each triangular face has base \(6\,\text{cm}\) and perpendicular face height \(5\,\text{cm}\), so its area is \(\frac{1}{2}\times6\times5=15\,\text{cm}^2\). 3. The four triangular faces have total area \(4\times15=60\,\text{cm}^2\). 4. The total surface area is \(36+60=96\,\text{cm}^2\).

\(96\,\text{cm}^2\)

- Keep track of where face \(A\) points as you fold the row of squares. - Which square would face away from \(A\) when the four sides close?

1. The four squares in the row wrap around four sides of the cube. 2. Faces \(A\) and \(C\) are two positions apart in the row, so they end up opposite each other.

Face \(C\). Squares \(A\) and \(C\) face opposite directions when the strip wraps around the cube.

- Trace the edge where each triangle joins the strip. - Find the widths of the rectangular faces and their shared length. - Check the right angles in the two triangular ends before finding their areas.

1. The upper triangle is attached to the \(6\,\text{cm}\)-wide rectangle; the lower triangle is attached to the \(8\,\text{cm}\)-wide rectangle. 2. The three rectangles all have length \(3\,\text{cm}\) and widths \(6\), \(8\), and \(10\,\text{cm}\). Their total area is \((6+8+10)\times3=72\,\text{cm}^2\). 3. Each right-triangle end has perpendicular legs \(6\) and \(8\,\text{cm}\), so each area is \(\frac{1}{2}\times6\times8=24\,\text{cm}^2\). 4. The surface area is \(72+2\times24=120\,\text{cm}^2\).

Upper triangle: the rectangle \(6\,\text{cm}\) wide. Lower triangle: the rectangle \(8\,\text{cm}\) wide. Rectangular faces: \(72\,\text{cm}^2\). Two triangular faces: \(48\,\text{cm}^2\). Total: \(120\,\text{cm}^2\).

- Imagine the four rectangles wrapping around to close the sides of the box. - Think about which edges the upper flap meets after folding, not just before. - Which face pairs have equal areas once the net is folded?

1. Yes. The four side rectangles wrap around the box, so the upper flap \(E\) meets the top edges of \(A\), \(B\), \(C\), and \(D\), even though it is joined only to \(A\) in the flat net. 2. In the folded box, \(A\) and \(C\) are opposite, as are \(B\) and \(D\). Their combined area is \(2\times15+2\times20=70\,\text{cm}^2\). 3. The two remaining faces together have area \(94-70=24\,\text{cm}^2\). The opposite flaps have equal areas, so each is \(12\,\text{cm}^2\).

a) Yes. \(E\) caps the four side faces when the box folds, so it also meets \(C\), although they do not touch in the flat net. b) \(E=12\,\text{cm}^2\) and \(F=12\,\text{cm}^2\).

- 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.

- Identify the three different rectangle types directly from the net. - Each rectangle type occurs twice in a rectangular-prism net. - For the scaled net, apply the scale factor to lengths before forming the new rectangle dimensions.

1. The net shows prism edge lengths \(10\,\text{cm}\), \(6\,\text{cm}\), and \(4\,\text{cm}\). The three face types are \(10\times6\), \(10\times4\), and \(6\times4\) rectangles. 2. The surface area is \(2(10\times6+10\times4+6\times4)=2(60+40+24)=248\,\text{cm}^2\). 3. A scale of \(1{:}2\) halves each length, giving edge lengths \(5\,\text{cm}\), \(3\,\text{cm}\), and \(2\,\text{cm}\). 4. The scaled rectangle types are \(5\,\text{cm}\times3\,\text{cm}\), \(5\,\text{cm}\times2\,\text{cm}\), and \(3\,\text{cm}\times2\,\text{cm}\).

a) \(248\,\text{cm}^2\) b) \(5\,\text{cm}\times3\,\text{cm}\), \(5\,\text{cm}\times2\,\text{cm}\), 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 \(6\,\text{cm}\times4\,\text{cm}\) rectangles, two \(6\,\text{cm}\times2\,\text{cm}\) rectangles, and two \(4\,\text{cm}\times2\,\text{cm}\) rectangles. 2. Nabila still needs one \(6\,\text{cm}\times2\,\text{cm}\) rectangle and one \(4\,\text{cm}\times2\,\text{cm}\) rectangle. 3. The total area is \(2\times(6\times4)+2\times(6\times2)+2\times(4\times2)=48+24+16=88\,\text{cm}^2\).

a) One \(6\,\text{cm}\times2\,\text{cm}\) rectangle and one \(4\,\text{cm}\times2\,\text{cm}\) 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) \(76\,\text{cm}^2\)

- Find the area of one square face. - Count how many congruent faces appear in the net. - The area of the complete net equals the surface area of the folded cube.

1. The area of one square face is \(3\times3=9\,\text{cm}^2\). 2. The net contains \(6\) congruent square faces, so its total area, and therefore the cube's surface area, is \(6\times9=54\,\text{cm}^2\).

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

- 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. \(416\,\text{g}\)

- Mentally fold along the shared edges. - Check whether two squares are forced onto the same cube face. - A row that wraps all the way around the four side positions cannot continue to a fifth distinct side face.

1. Net 1 is valid. Its squares can fold to make one face surrounded by four side faces and a final closing face without overlap. 2. Net 2 contains a \(2\times2\) block positioned so that faces overlap when folded instead of occupying six distinct cube faces. 3. Net 3 contains five squares in one straight row. Wrapping that row around a cube forces the fifth square to overlap the first.

a) Net 1 b) Net 2 causes overlapping faces when folded. Net 3 has five squares in one row, forcing an overlap when wrapped around the cube.

- First determine which labeled face is opposite bottom face \(B\). - Classify the three film-covered faces as top, bottom, or side faces after folding. - Only then apply the full-, half-, or zero-painted fraction to their face areas.

1. With \(B\) as the bottom, face \(D\) folds to the opposite top face. Faces \(A\), \(C\), \(E\), and \(F\) become side faces. 2. Of the film-covered faces \(A\), \(C\), and \(D\), faces \(A\) and \(C\) are side faces and are half painted. Face \(D\) is the top and stays dry. 3. One face has area \(4\times4=16\,\text{cm}^2\), so half of one face is \(8\,\text{cm}^2\). 4. Two film-covered side faces are half painted, so the painted film area is \(2\times8=16\,\text{cm}^2\).

Faces \(A\) and \(C\) get blue paint. The total painted area of protective film is \(16\,\text{cm}^2\).

- Use the net to identify the bottom face and the two congruent pairs of side faces. - The paint reaches halfway up each side face, so only part of each side rectangle is counted. - Add the painted bottom area to the painted portions of all four side faces.

1. From the net, the bottom face is an \(8\,\text{cm}\times5\,\text{cm}\) rectangle, so its area is \(8\times5=40\,\text{cm}^2\). 2. The prism height is \(4\,\text{cm}\), so half the height is \(4\div2=2\,\text{cm}\). 3. The two long side faces contribute painted strips with total area \(2(8\times2)=32\,\text{cm}^2\). 4. The two short side faces contribute painted strips with total area \(2(5\times2)=20\,\text{cm}^2\). 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 \(148\,\text{cm}^2\).

- 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 \(60\,\text{cm}^3\).

- Check the folding condition separately from the area condition. - Picture the row of four squares wrapping around the cube. - Then find the area of one square and compare the six-face total with the required surface area.

1. The four squares in the horizontal row can wrap around to make the four side faces of a cube. 2. The squares above and below the same interior square close the two remaining faces without overlap, so the arrangement is a valid cube net. 3. Each square face has area \(3\times3=9\,\text{cm}^2\). 4. The six faces have total area \(6\times9=54\,\text{cm}^2\). 5. The proposal works because it satisfies both the folding requirement and the required surface area.

Yes. The arrangement is a valid cube net, and its six \(3\,\text{cm}\times3\,\text{cm}\) faces have total surface area \(54\,\text{cm}^2\).

- 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 \(384\,\text{cm}^2\).

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

1. The labeled edge length gives one square face area of \(5\times5=25\,\text{cm}^2\). 2. The net has \(6\) congruent squares, so its total area is \(6\times25=150\,\text{cm}^2\). 3. The net contains \(150\div10=15\) groups of \(10\,\text{cm}^2\). 4. Its mass is \(15\times2\,\text{g}=30\,\text{g}\).

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 \(75\,\text{cm}^2\).

- 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 \(444\,\text{cm}^2\).

- Use the net to separate the known pair of \(8\)-by-\(3\) faces from the four faces that involve the height. - After removing the known face areas, think about how much surface area one unit of height contributes across the remaining face pairs. - The faces involving the height come in two congruent pairs.

1. The top and bottom are two \(8\,\text{cm}\times3\,\text{cm}\) faces, with combined area \(2\times24=48\,\text{cm}^2\). 2. The other four faces therefore have combined area \(158-48=110\,\text{cm}^2\). 3. For each \(1\,\text{cm}\) of height, the two \(8\)-by-height faces contribute \(16\,\text{cm}^2\), and the two \(3\)-by-height faces contribute \(6\,\text{cm}^2\), for \(22\,\text{cm}^2\) altogether. 4. The height is \(110\div22=5\,\text{cm}\).

The height is \(5\,\text{cm}\).

- A rectangular prism has three pairs of congruent opposite faces. Can you identify the two pairs whose areas are already shown? - Compare the area accounted for by those four faces with the total surface area. - What must be true about the two faces that remain?

1. In a rectangular prism, opposite faces are congruent, so the two faces labeled \(28\) form one pair and the two faces labeled \(21\) form another pair. 2. Their combined area is \(2\times28+2\times21=56+42=98\,\text{cm}^2\). 3. The two remaining faces form the third congruent pair, so together they have area \(122-98=24\,\text{cm}^2\). 4. Each remaining face has area \(24\div2=12\,\text{cm}^2\).

Each face marked \(?\) has area \(12\,\text{cm}^2\).

- Find the area of one right-triangular base first. - The three rectangles have widths equal to the three side lengths of the triangular base. - Separate the surface area into the two triangles and the three rectangles.

1. One triangular base has area \(\frac{1}{2}\times5\times12=30\,\text{cm}^2\). 2. The two triangular bases have total area \(60\,\text{cm}^2\). 3. The three rectangular faces have widths \(5\,\text{cm}\), \(12\,\text{cm}\), and \(13\,\text{cm}\), and common length \(x\,\text{cm}\). Their total area is \((5+12+13)x=30x\). 4. The surface-area equation is \(60+30x=300\). 5. Therefore \(30x=240\), so \(x=8\).

\(8\,\text{cm}\)

- Find the area of one right-triangular base. - Each side of the triangular base corresponds to the width of one rectangular face. - After finding the total surface area, use the cost per square centimeter.

1. One triangular base has area \(\frac{1}{2}\times6\times8=24\,\text{cm}^2\). 2. The two triangular bases have total area \(48\,\text{cm}^2\). 3. The three rectangular faces have total area \(12\times(6+8+10)=288\,\text{cm}^2\). 4. The total cardboard area is \(48+288=336\,\text{cm}^2\). 5. The cost is \(336\times\$0.02=\$6.72\).

Area: \(336\,\text{cm}^2\) Cost: \(\$6.72\)

- Compare the shapes and side lengths of the faces, not their positions in the net. - Each side of a triangular face is the width of one rectangular face. - Does changing the order of the same five faces change the sum of their areas?

1. Each net has the same two triangular faces. One triangle has area \(\frac{1}{2}\times8\times15=60\,\text{cm}^2\), so the two triangles have total area \(120\,\text{cm}^2\). 2. The widths of the three rectangular faces are the three side lengths of the triangular face: \(8\,\text{cm}\), \(15\,\text{cm}\), and \(17\,\text{cm}\). 3. The rectangles therefore have areas \(5\times8\), \(5\times15\), and \(5\times17\) square centimeters, regardless of the order in which they appear in the net. 4. Their total area is \(5\times(8+15+17)=200\,\text{cm}^2\). 5. Each net has total area \(120+200=320\,\text{cm}^2\).

The nets have the same total area because they contain the same two triangular faces and the same three rectangular faces; only the order of the faces changes. Total surface area: \(320\,\text{cm}^2\)

- Compare the pyramid's vertical height in the text with the dashed perpendicular height shown on one triangular face of the net. - Use the side length of the square as the base of each triangular face. - Add the square base area and the areas of all four triangular faces.

1. The \(3\,\text{cm}\) measurement is the pyramid's vertical height, not the perpendicular height of a triangular face. 2. The net shows that each triangular face has base \(8\,\text{cm}\) and perpendicular height \(5\,\text{cm}\). 3. The square base has area \(8\times8=64\,\text{cm}^2\). 4. One triangular face has area \(\frac{1}{2}\times8\times5=20\,\text{cm}^2\), so the four triangular faces have total area \(80\,\text{cm}^2\). 5. The correct surface area is \(64+80=144\,\text{cm}^2\).

Haruka confused the pyramid's vertical height with the perpendicular height of a triangular face. Correct surface area: \(144\,\text{cm}^2\)

- Imagine each square folding along the edge it shares with a neighbor. - Keep track of which direction each face points after folding. - Would any two pieces cover the same side of the cube?

1. In a), the four-square strip wraps around four sides. The two squares to its left and right fold onto different faces, so none overlap. 2. In b), both extra squares attach to the same side of the strip, at different positions. When folded, these two squares try to cover the same face, so this is not a cube net.

a) folds into a cube. In b), the two squares sticking out to the left of the four-square strip fold onto the same face and overlap.

- A pyramid's square base and its four triangular sides use different area formulas. - Use the triangle heights marked on the unfolded net, not the vertical height inside the pyramid. - A cube has six identical square faces.

1. The pyramid's base area is \(8\times8=64\,\text{cm}^2\). 2. Each triangular face has base \(8\,\text{cm}\) and face height \(5\,\text{cm}\), so its area is \(\frac{1}{2}\times8\times5=20\,\text{cm}^2\). Four triangles cover \(80\,\text{cm}^2\); the whole pyramid has surface area \(64+80=144\,\text{cm}^2\). 3. The cube has six square faces of area \(5\times5=25\,\text{cm}^2\) each, giving \(6\times25=150\,\text{cm}^2\). 4. The cube's surface area is \(150-144=6\,\text{cm}^2\) greater.

The cube has \(6\,\text{cm}^2\) more surface area (\(150\,\text{cm}^2\) versus \(144\,\text{cm}^2\)). The pyramid has \(64\,\text{cm}^2\) of base area and \(80\,\text{cm}^2\) of triangular faces.

- Imagine folding the row one square at a time around the cube. - Track what happens if the row continues after it has wrapped around the side faces. - After finding an upper bound, check whether a valid cube net can actually attain it.

1. Only four faces can wrap around the four side positions of a cube. A fifth square in the same straight row would fold onto the first square, causing overlap. Adding a sixth square cannot remove that overlap. 2. A valid cube net can have four squares in one straight row, with the remaining two squares attached appropriately to the row. 3. Therefore, the greatest possible number of squares in one straight row is \(4\).

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 straight row.

- 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

- First decide which arrangements fold without overlap. - Ignore the side lengths of arrangements that cannot make cubes. - For each valid arrangement, use six congruent square faces, then combine the resulting surface areas.

1. Arrangements (1) and (3) fold into closed cubes without overlap. 2. In arrangement (2), faces overlap when folded, so it is not a valid cube net. Arrangement (4) has six squares in one straight row and cannot fold into a cube. 3. For arrangement (1), one face has area \(5\times5=25\,\text{cm}^2\), so the cube surface area is \(6\times25=150\,\text{cm}^2\). 4. For arrangement (3), one face has area \(3\times3=9\,\text{cm}^2\), so the cube surface area is \(6\times9=54\,\text{cm}^2\). 5. The combined surface area of the cubes that can actually be built is \(150+54=204\,\text{cm}^2\).

Arrangements (1) and (3) are valid. Their cube surface areas are \(150\,\text{cm}^2\) and \(54\,\text{cm}^2\), for a combined surface area of \(204\,\text{cm}^2\).

- 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.

- For each net, organize the faces into congruent pairs before adding their areas. - Equal surface area does not imply equal edge lengths or equal volume. - After establishing the surface-area comparison, use the three edge lengths of each box to compare capacity.

1. Box a has dimensions \(1\,\text{cm}\), \(3\,\text{cm}\), and \(6\,\text{cm}\). Its surface area is \(2(1\times3+1\times6+3\times6)=2(3+6+18)=54\,\text{cm}^2\). 2. Box b is a cube with side length \(3\,\text{cm}\). Its surface area is \(6\times3^2=54\,\text{cm}^2\). 3. The boxes therefore use the same total area of cardboard. 4. Box a has volume \(1\times3\times6=18\,\text{cm}^3\). 5. Box b has volume \(3\times3\times3=27\,\text{cm}^3\), which is \(27-18=9\,\text{cm}^3\) greater.

Both boxes have surface area \(54\,\text{cm}^2\). Box b has the greater volume by \(9\,\text{cm}^3\).