Free math worksheets

- A rectangular prism has three pairs of congruent faces. - Find one area for each face type before doubling. - Surface area uses square units, not cubic units.

1. Use the surface area formula for a rectangular prism: \(SA = 2(lw + lh + wh)\). 2. Substitute the dimensions: \(SA = 2(2\cdot5 + 2\cdot8 + 5\cdot8)\). 3. The three different face areas sum to \(10+16+40=66\,\text{cm}^2\). 4. Double the sum to include both faces of each type: \(SA=132\,\text{cm}^2\).

b) \(132\,\text{cm}^2\)

- A cube has \(12\) equal edges. - Its surface is made of six congruent squares. - Its volume is the cube of its edge length.

1. A cube has \(12\) equal edges, so each edge is \(48\div12=4\,\text{cm}\). 2. The surface area is \(6s^2=6(4^2)=96\,\text{cm}^2\). 3. The volume is \(s^3=4^3=64\,\text{cm}^3\).

a) The edge length is \(4\,\text{cm}\). b) The surface area is \(96\,\text{cm}^2\). c) The volume is \(64\,\text{cm}^3\).

- Square the edge length first. - Multiply one face area by \(6\). - Use square units in each answer.

1. For \(a=3\,\text{cm}\), \(SA=6\cdot3^2=54\,\text{cm}^2\). 2. For \(a=0.5\,\text{m}\), \(SA=6\cdot0.5^2=1.5\,\text{m}^2\). 3. For \(a=1.2\,\text{dm}\), \(SA=6\cdot1.2^2=8.64\,\text{dm}^2\).

a) \(54\,\text{cm}^2\) b) \(1.5\,\text{m}^2\) c) \(8.64\,\text{dm}^2\)

- Which pair of face dimensions are equal? - A rectangular prism has three pairs of congruent opposite faces. - Add one area from each pair and then double.

1. The two \(a \times b\) faces are \(6\,\text{cm} \times 6\,\text{cm}\), so they are squares. The other four faces are \(6\,\text{cm} \times 15\,\text{cm}\). 2. The three face areas are \(36\,\text{cm}^2\), \(90\,\text{cm}^2\), and \(90\,\text{cm}^2\). 3. Include both faces in each pair: \(SA=2(36+90+90)=432\,\text{cm}^2\).

a) The rectangular prism has \(2\) square faces. b) \(SA=432\,\text{cm}^2\)

- A closed rectangular prism has three pairs of congruent faces. - Find one area for each pair before doubling.

1. The three different face areas are \(112\,\text{in.}^2\), \(40\,\text{in.}^2\), and \(70\,\text{in.}^2\). 2. Double their sum: \(SA=2(112+40+70)=444\,\text{in.}^2\).

The total surface area is \(444\,\text{in.}^2\).

- Calculate each cube's surface area separately. - Compare the results by division rather than assuming the linear scale factor carries over unchanged.

1. The smaller cube has \(SA_1=6\cdot2^2=24\,\text{cm}^2\). 2. The larger cube has \(SA_2=6\cdot4^2=96\,\text{cm}^2\). 3. Since \(96\div24=4\), the larger cube has four times the surface area, so Maya's claim is false.

Maya's claim is false. The surface areas are \(24\,\text{cm}^2\) and \(96\,\text{cm}^2\), so the larger is \(4\) times as great.

- Find the area of one square face. - How many congruent faces does a cube have?

1. One face has area \(8\cdot8=64\,\text{cm}^2\). 2. A cube has \(6\) congruent faces, so \(SA=6\cdot64=384\,\text{cm}^2\).

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

- Multiply all three edge lengths for volume. - A rectangular prism has three pairs of congruent faces. - Add the areas of all six faces.

a) The volume is \(8\cdot5\cdot4=160\,\text{cm}^3\). b) The surface area is \(2(8\cdot5+5\cdot4+8\cdot4)=2(40+20+32)=184\,\text{cm}^2\).

a) The volume is \(160\,\text{cm}^3\). b) The surface area is \(184\,\text{cm}^2\).

- Count the edges and faces of a cube. - Multiply one edge length by the number of edges. - Find one square face's area, then multiply by six.

a) A cube has \(12\) equal edges, so their total length is \(12\cdot1.5=18\,\text{m}\). b) A cube has six square faces, so its surface area is \(6(1.5^2)=13.5\,\text{m}^2\).

a) The total edge length is \(18\,\text{m}\). b) The painted area is \(13.5\,\text{m}^2\).

- Identify the three different rectangular face areas. - Each type of face occurs twice. - Add all six face areas.

1. The three different face areas are \(8\cdot5=40\,\text{cm}^2\), \(8\cdot12=96\,\text{cm}^2\), and \(5\cdot12=60\,\text{cm}^2\). 2. Each face has a congruent opposite face, so the surface area is \(2(40+96+60)=392\,\text{cm}^2\).

The surface area is \(392\,\text{cm}^2\).

- Find each box's surface area separately. - A cube has six congruent square faces. - Include all six faces of the rectangular prism.

1. Box A's surface area is \(6(10^2)=600\,\text{cm}^2\). 2. Box B's surface area is \(2(15\cdot8+15\cdot8+8\cdot8)=608\,\text{cm}^2\). 3. Since \(608>600\), Box B has the greater surface area.

Box B has the greater surface area: \(608\,\text{cm}^2\), compared with \(600\,\text{cm}^2\) for Box A.

- Divide the total wire length among the cube's \(12\) equal edges. - A cube has six congruent square faces.

1. A cube has \(12\) equal edges, so one edge is \(144\div12=12\,\text{cm}\). 2. The cube's surface area is \(6(12^2)=864\,\text{cm}^2\).

The cube's surface area is \(864\,\text{cm}^2\).

- Separate the square base from the four congruent triangular faces in the net. - For each triangle, use the side of the square as its base and the shown slant height as its perpendicular height. - Add the base area and the areas of all four triangles.

1. The square base has side length \(6\,\text{cm}\), so its area is \(6^2=36\,\text{cm}^2\). 2. Each triangular face has base \(6\,\text{cm}\) and slant height \(5\,\text{cm}\), so its area is \(\frac12\cdot6\cdot5=15\,\text{cm}^2\). 3. The four triangular faces have total area \(4\cdot15=60\,\text{cm}^2\). 4. The total surface area is \(36+60=96\,\text{cm}^2\).

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

- Lateral surface area includes the triangular faces but not the square base. - Read the base and slant height of a triangular face from the net. - Add the square base only when finding total surface area.

1. Each triangular face has base \(8\,\text{cm}\) and slant height \(5\,\text{cm}\), so its area is \(\frac12\cdot8\cdot5=20\,\text{cm}^2\). 2. The lateral surface area is \(4\cdot20=80\,\text{cm}^2\). 3. The square base has area \(8^2=64\,\text{cm}^2\). 4. The total surface area is \(80+64=144\,\text{cm}^2\).

a) \(80\,\text{cm}^2\) b) \(144\,\text{cm}^2\)

- Separate the rectangular base from the four triangular faces. - The two pairs of triangles have different bases and different slant heights. - Add the areas of all five faces for total surface area.

1. The rectangular base has area \(10\cdot18=180\,\text{cm}^2\). 2. The two triangular faces with base \(10\,\text{cm}\) have slant height \(15\,\text{cm}\), so together they have area \(2\cdot\frac12\cdot10\cdot15=150\,\text{cm}^2\). 3. The two triangular faces with base \(18\,\text{cm}\) have slant height \(13\,\text{cm}\), so together they have area \(2\cdot\frac12\cdot18\cdot13=234\,\text{cm}^2\). 4. Therefore, \(SA=180+150+234=564\,\text{cm}^2\).

\(564\,\text{cm}^2\)

- Only the four triangular faces are covered. - Pair the congruent triangular faces before adding their areas. - Keep the two different slant heights matched to the correct base lengths.

1. The two faces with \(12\,\text{ft}\) bases have total area \(2\cdot\frac12\cdot12\cdot17=204\,\text{ft}^2\). 2. The two faces with \(30\,\text{ft}\) bases have total area \(2\cdot\frac12\cdot30\cdot10=300\,\text{ft}^2\). 3. The required fabric area is \(204+300=504\,\text{ft}^2\).

\(504\,\text{ft}^2\)

- List only the faces that are actually coated. - The two selected triangular faces are congruent. - Add their combined area to the rectangular base area.

1. The rectangular base has area \(14\cdot36=504\,\text{m}^2\). 2. The two coated triangular faces have total area \(2\cdot\frac12\cdot36\cdot25=900\,\text{m}^2\). 3. The total coated area is \(504+900=1404\,\text{m}^2\).

\(1404\,\text{m}^2\)

- Convert all edge lengths to the same unit. - A rectangular prism has three pairs of congruent faces. - How many square centimeters are in one square decimeter?

1. Convert all edge lengths to centimeters: \(a=12\,\text{cm}\), \(b=5\,\text{cm}\), and \(c=8\,\text{cm}\). 2. Use the surface-area formula: \(SA=2(ab+ac+bc)\). 3. Substitute: \(SA=2(12\cdot5+12\cdot8+5\cdot8)=2(60+96+40)=392\,\text{cm}^2\). 4. Since \(100\,\text{cm}^2=1\,\text{dm}^2\), \(392\,\text{cm}^2=3.92\,\text{dm}^2\).

The surface area is \(3.92\,\text{dm}^2\).

- A closed rectangular prism has six faces. - Which face is missing from an open-top box? - List and find the area of each face that remains.

1. The open box has one base, two side faces measuring \(30\,\text{cm}\times15\,\text{cm}\), and two side faces measuring \(20\,\text{cm}\times15\,\text{cm}\). 2. The base area is \(30\cdot20=600\,\text{cm}^2\). 3. The two longer side faces have total area \(2(30\cdot15)=900\,\text{cm}^2\). 4. The two shorter side faces have total area \(2(20\cdot15)=600\,\text{cm}^2\). 5. The total cardboard area is \(600+900+600=2100\,\text{cm}^2\).

\(2100\,\text{cm}^2\) of cardboard is needed.

- Use the volume formula for a rectangular prism. - List the faces inside the open-top bin. - Do not include a top face in the liner area.

1. The bin's volume is \(V=3\cdot2\cdot1.5=9\,\text{m}^3\). 2. The bottom area is \(3\cdot2=6\,\text{m}^2\). 3. The two longer side walls have total area \(2(3\cdot1.5)=9\,\text{m}^2\). 4. The two shorter side walls have total area \(2(2\cdot1.5)=6\,\text{m}^2\). 5. The total liner area is \(6+9+6=21\,\text{m}^2\).

a) The bin can hold \(9\,\text{m}^3\) of salt. b) The liner must cover \(21\,\text{m}^2\).

- Use volume to find Solid A's missing height. - Use the surface-area formulas for a rectangular prism and a cube. - Calculate and compare the surface areas of both solids.

1. For Solid A, \(216=12\cdot6\cdot h\), so \(h=216\div72=3\,\text{cm}\). 2. Solid A's surface area is \(2(12\cdot6+12\cdot3+6\cdot3)=252\,\text{cm}^2\). 3. Solid B's surface area is \(6(6^2)=216\,\text{cm}^2\). 4. Since \(216<252\), Solid B has the smaller surface area.

a) Solid A is \(3\,\text{cm}\) high. b) Solid B has the smaller surface area: \(216\,\text{cm}^2\), compared with \(252\,\text{cm}^2\) for Solid A.

- What three whole-number factors can have a product of \(24\)? - A rectangular prism has three pairs of congruent faces. - The largest face is the face with the greatest area.

1. One possible prism has dimensions \(4\,\text{cm}\), \(3\,\text{cm}\), and \(2\,\text{cm}\), because \(4\cdot3\cdot2=24\). 2. A second possible prism has dimensions \(6\,\text{cm}\), \(4\,\text{cm}\), and \(1\,\text{cm}\), because \(6\cdot4\cdot1=24\). 3. For the first prism, \(SA=2(4\cdot3+3\cdot2+4\cdot2)=52\,\text{cm}^2\). 4. For the second prism, \(SA=2(6\cdot4+4\cdot1+6\cdot1)=68\,\text{cm}^2\). 5. The prisms have the same volume but different surface areas. 6. The first prism has \(4\cdot3=12\) cubes in its bottom layer. The second prism has \(6\cdot4=24\) cubes in its bottom layer.

One possible answer is: a) Prism 1: \(4\,\text{cm}\), \(3\,\text{cm}\), and \(2\,\text{cm}\); Prism 2: \(6\,\text{cm}\), \(4\,\text{cm}\), and \(1\,\text{cm}\). b) Prism 1: \(52\,\text{cm}^2\); Prism 2: \(68\,\text{cm}^2\). The surface areas are different. c) Prism 1: \(12\) cubes; Prism 2: \(24\) cubes.

- Find the areas of all six faces of each box. - A box that uses less material has a smaller surface area. - Compare the two results.

1. Box A's surface area is \(2(6\cdot2+6\cdot2+2\cdot2)=56\,\text{cm}^2\). 2. Box B's surface area is \(2(4\cdot3+4\cdot2+3\cdot2)=52\,\text{cm}^2\). 3. Since \(52<56\), Box B uses less cardboard.

a) Box A has surface area \(56\,\text{cm}^2\), and Box B has surface area \(52\,\text{cm}^2\). b) Box B uses less cardboard because it has the smaller surface area.

- Use the surface-area formula for a rectangular prism. - Compare the material needed, not the equal volumes. - Think about how the shape affects the aquarium's use.

1. Aquarium A's surface area is \(2(60\cdot30+60\cdot40+30\cdot40)=10{,}800\,\text{cm}^2\). 2. Aquarium B's surface area is \(2(120\cdot20+120\cdot30+20\cdot30)=13{,}200\,\text{cm}^2\). 3. Aquarium B requires more glass. 4. One possible reason to choose it is that its greater length gives fish more horizontal swimming space or creates a wider viewing area.

a) Aquarium A has surface area \(10{,}800\,\text{cm}^2\), and Aquarium B has surface area \(13{,}200\,\text{cm}^2\). b) Aquarium B requires more glass. c) Its longer shape may provide more horizontal swimming space or a wider viewing area. Answers will vary.

- Identify which glass surfaces are below the water level. - Do not count the water's top surface. - Use the water depth, not the aquarium's full height, for the side walls.

1. The wet glass includes the bottom and the four side-wall portions below the water level. 2. The bottom area is \(60\cdot30=1800\,\text{cm}^2\). 3. The two longer wet side portions have area \(2(60\cdot20)=2400\,\text{cm}^2\). 4. The two shorter wet side portions have area \(2(30\cdot20)=1200\,\text{cm}^2\). 5. The total wet glass area is \(1800+2400+1200=5400\,\text{cm}^2\).

The water touches \(5400\,\text{cm}^2\) of glass.

- Write the areas of all six faces. - Use a variable for the unknown width. - Set the expression equal to the given surface area.

1. Let the unknown width be \(w\). The surface-area equation is \(280=2(6w+6\cdot10+10w)\). 2. Simplify: \(280=2(16w+60)=32w+120\). 3. Then \(160=32w\), so \(w=5\,\text{cm}\).

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

- Lateral area includes only the four side faces. - Compare the original and changed products. - When all lengths are doubled, each face has two doubled dimensions.

1. The lateral area is \(2(3+4)(5)=70\,\text{cm}^2\). 2. Doubling only the height gives \(2(3+4)(10)=140\,\text{cm}^2\), so the lateral area doubles. 3. The original total surface area is \(2(3\cdot4+3\cdot5+4\cdot5)=94\,\text{cm}^2\). 4. Doubling all dimensions gives surface area \(2(6\cdot8+6\cdot10+8\cdot10)=376\,\text{cm}^2\). 5. Since \(376\div94=4\), the total surface area is multiplied by \(4\).

a) The lateral area is \(70\,\text{cm}^2\). Doubling only the height doubles it to \(140\,\text{cm}^2\). b) The total surface area is \(94\,\text{cm}^2\). Doubling all dimensions multiplies it by \(4\), giving \(376\,\text{cm}^2\).

- Use the rectangular-prism formulas for volume and surface area. - In this model, which geometric measure represents packaging material? - Compare the two surface areas directly.

1. The volumes are \(V_A=10^3\,\text{cm}^3=1000\,\text{cm}^3\) and \(V_B=25\cdot8\cdot5\,\text{cm}^3=1000\,\text{cm}^3\). 2. Design A has \(SA_A=6(10\,\text{cm})^2=600\,\text{cm}^2\). 3. Design B has \(SA_B=2(25\cdot8+25\cdot5+8\cdot5)\,\text{cm}^2=730\,\text{cm}^2\). 4. Since \(600<730\), Design A has \(130\,\text{cm}^2\) less surface area.

a) \(V_A=1000\,\text{cm}^3\) and \(V_B=1000\,\text{cm}^3\) b) \(SA_A=600\,\text{cm}^2\) and \(SA_B=730\,\text{cm}^2\) c) Design A uses less material because its surface area is \(130\,\text{cm}^2\) smaller.

- Convert all dimensions to centimeters first. - Find the three different rectangular face areas. - Each face type occurs twice.

1. Convert \(40\,\text{mm}=4\,\text{cm}\). 2. The three different face areas are \(12.5\cdot8=100\), \(12.5\cdot4=50\), and \(8\cdot4=32\), in square centimeters. 3. Double their sum: \(SA=2(100+50+32)=364\,\text{cm}^2\).

The surface area of the package is \(364\,\text{cm}^2\).

- Convert all lengths to the same unit before calculating. - Use the rectangular-prism surface-area formula for Solid A. - A cube has six congruent square faces.

1. Convert Solid A's dimensions: \(30\,\text{mm}=3\,\text{cm}\) and \(0.5\,\text{dm}=5\,\text{cm}\). 2. \(SA_A=2(8\cdot3+3\cdot5+8\cdot5)=158\,\text{cm}^2\). 3. \(SA_B=6(6\cdot6)=216\,\text{cm}^2\). 4. Since \(216>158\), Solid B has the greater surface area.

Solid B has the greater surface area: \(216\,\text{cm}^2\), compared with \(158\,\text{cm}^2\) for Solid A.

- Imagine unfolding each box into a net. - Find one area for each distinct face type. - Compare the two total surface areas.

1. Box 1: \(SA_1=6(15\cdot15)=1350\,\text{cm}^2\). 2. Box 2: \(SA_2=2(20\cdot10+10\cdot12+20\cdot12)=1120\,\text{cm}^2\). 3. Since \(1350>1120\), Box 1 requires more paper.

Box 1 requires more paper because its surface area is \(1350\,\text{cm}^2\), while Box 2 has surface area \(1120\,\text{cm}^2\).

- A cube has six congruent square faces. - Compare surface areas by division. - How does squaring a linear scale factor affect the result?

1. Use \(SA=6a^2\). Then \(SA_1=54\,\text{cm}^2\), \(SA_2=216\,\text{cm}^2\), and \(SA_3=486\,\text{cm}^2\). 2. \(216\div54=4\), so doubling the edge length multiplies surface area by \(4\). 3. \(486\div54=9\), so tripling the edge length multiplies surface area by \(9\).

a) \(SA_1=54\,\text{cm}^2\); \(SA_2=216\,\text{cm}^2\); \(SA_3=486\,\text{cm}^2\) b) \(4\) c) \(9\)

- Divide the total surface area by \(6\) to get one face area. - A face is a square, so recover its side length from its area. - Then triple the edge length and recalculate surface area.

1. One face has area \(150\div6=25\,\text{cm}^2\). 2. Since \(a^2=25\), the edge length is \(a=5\,\text{cm}\). 3. The second cube has edge length \(15\,\text{cm}\), so \(SA=6\cdot15^2=1350\,\text{cm}^2\).

a) \(25\,\text{cm}^2\) b) \(5\,\text{cm}\) c) \(1350\,\text{cm}^2\)

- Find each cube's surface area separately. - Compare the smaller surface area with the original by division.

1. The original surface area is \(SA=6\cdot10^2=600\,\text{cm}^2\). 2. The new edge length is \(5\,\text{cm}\), so \(SA=6\cdot5^2=150\,\text{cm}^2\). 3. Since \(150\div600=\frac14\), halving the edge length reduces surface area to one-fourth.

a) \(600\,\text{cm}^2\) b) \(150\,\text{cm}^2\) c) The surface area becomes one-fourth of the original.

- How many edges does a cube have? - Use the total edge length to find one edge. - Then use the six square faces.

1. A cube has \(12\) congruent edges, so one edge is \(144\div12=12\,\text{cm}\). 2. One face has area \(12^2=144\,\text{cm}^2\). 3. Therefore, \(SA=6\cdot144=864\,\text{cm}^2\).

The surface area is \(864\,\text{cm}^2\).

- Use \(SA=2(ab+ac+bc)\). - In part b, only one dimension changes. - Compare the new result with twice the original.

1. \(SA=2(5\cdot3+5\cdot2+3\cdot2)=62\,\text{cm}^2\). 2. The new surface area is \(SA_{\text{new}}=2(10\cdot3+10\cdot2+3\cdot2)=112\,\text{cm}^2\). 3. Twice the original would be \(124\,\text{cm}^2\), so the surface area did not double.

a) \(62\,\text{cm}^2\) b) \(112\,\text{cm}^2\) c) No; \(112\) is not twice \(62\).

- First find the area of one face of the cube. - What edge length produces that square area? - How many edges does a cube have? - Multiply one edge length by the total number of edges.

1. Divide the surface area by \(6\) to find the area of one face: \(600 \div 6 = 100\,\text{cm}^2\). 2. Each face is a square. Since \(a^2 = 100\), the edge length is \(a = 10\,\text{cm}\). 3. A cube has \(12\) edges, so their total length is \(12 \cdot 10 = 120\,\text{cm}\).

The total length of all the edges is \(120\,\text{cm}\).

- Which faces are present when a rectangular-prism container has no top? - Find the area of each face before adding them. - How many groups of \(100\,\text{cm}^2\) are in the total area?

1. Because the tray has no top, its surface consists of one base, two long side faces, and two short side faces. 2. Find the areas: the base is \(40 \cdot 30 = 1200\,\text{cm}^2\), the two long sides total \(2(40 \cdot 5) = 400\,\text{cm}^2\), and the two short sides total \(2(30 \cdot 5) = 300\,\text{cm}^2\). 3. The total sheet-metal area is \(1200 + 400 + 300 = 1900\,\text{cm}^2\). 4. This area contains \(1900 \div 100 = 19\) groups of \(100\,\text{cm}^2\), so the mass is \(19 \cdot 5 = 95\,\text{g}\).

The empty tray has a mass of \(95\,\text{g}\).

- How many square faces remain when the bottom face is missing? - First find the total area in square centimeters. - Use the given conversion to express that area as part of a square meter.

1. A cube-shaped shell without a bottom has \(5\) congruent square faces. 2. The area of one face is \(20 \cdot 20 = 400\,\text{cm}^2\). 3. The total material area is \(5 \cdot 400 = 2000\,\text{cm}^2\). 4. Convert the area to square meters: \(2000 \div 10{,}000 = 0.2\,\text{m}^2\). 5. Find the mass: \(0.2 \cdot 500 = 100\,\text{g}\).

The pedestal has a mass of \(100\,\text{g}\).

- Find one area for each pair of congruent faces. - Convert square inches to square feet only after finding the total surface area. - Multiply the converted area by the cost per square foot.

1. The three face areas are \(864\), \(648\), and \(432\) square inches. 2. \(SA=2(864+648+432)=3888\,\text{in.}^2\). 3. Convert: \(3888\div144=27\,\text{ft}^2\). 4. The cost is \(27\cdot\$0.35=\$9.45\).

a) \(3888\,\text{in.}^2\) b) \(27\,\text{ft}^2\); \(\$9.45\)

- Which rectangles represent the walls, and which rectangle represents the roof? - Which area must be excluded from the painted surface? - When the quotient is not a whole number of cans, can you purchase only part of a can?

1. The two longer walls have total area \(2(10 \cdot 7) = 140\,\text{ft}^2\), and the two shorter walls have total area \(2(8 \cdot 7) = 112\,\text{ft}^2\). Before subtracting the door, the wall area is \(140 + 112 = 252\,\text{ft}^2\). 2. Exclude the door: \(252 - 21 = 231\,\text{ft}^2\). 3. The flat roof area is \(10 \cdot 8 = 80\,\text{ft}^2\). 4. The total area to paint is \(231 + 80 = 311\,\text{ft}^2\). 5. The number of cans needed is \(311 \div 100 = 3.11\). Because paint is purchased in whole cans, \(4\) cans are required.

a) \(231\,\text{ft}^2\) b) The roof area is \(80\,\text{ft}^2\), and the total area is \(311\,\text{ft}^2\). c) \(4\) cans

- Find the floor area of each pool first. - Then find the areas of the four side walls and add them to the floor area. - Do not include the open top. - Compare the two total areas.

1. For Pool A, the floor area is \(20 \cdot 12 = 240\,\text{ft}^2\). The side-wall area is \(2(20 \cdot 5) + 2(12 \cdot 5) = 200 + 120 = 320\,\text{ft}^2\). The total tiled area is \(240 + 320 = 560\,\text{ft}^2\). 2. For Pool B, the floor area is \(16 \cdot 16 = 256\,\text{ft}^2\). The side-wall area is \(4(16 \cdot 5) = 320\,\text{ft}^2\). The total tiled area is \(256 + 320 = 576\,\text{ft}^2\). 3. Since \(576\,\text{ft}^2 > 560\,\text{ft}^2\), Pool B requires more tile.

Pool B requires more tile. Pool A has \(560\,\text{ft}^2\) of interior surface to tile, while Pool B has \(576\,\text{ft}^2\).

- Convert all lengths to the same unit before calculating. - How many faces are square, and how many are long rectangles? - Add the areas of all six faces.

1. Convert the length to inches: \(8\,\text{ft} = 96\,\text{in.}\). 2. The beam is a rectangular prism with dimensions \(96\,\text{in.}\), \(6\,\text{in.}\), and \(6\,\text{in.}\). 3. The two square ends have total area \(2(6 \cdot 6) = 72\,\text{in.}^2\). The four long faces have total area \(4(96 \cdot 6) = 2304\,\text{in.}^2\). 4. The total surface area is \(72 + 2304 = 2376\,\text{in.}^2\).

The total surface area is \(2376\,\text{in.}^2\).

- Find each surface area separately. - Use the cube shortcut for one solid and the rectangular-prism formula for the other. - Compare before subtracting.

1. The cube has \(SA_{\text{cube}}=6\cdot10^2=600\,\text{cm}^2\). 2. The prism has \(SA_{\text{prism}}=2(20\cdot8+20\cdot5+8\cdot5)=600\,\text{cm}^2\). 3. The surface areas are equal, so the difference is \(0\,\text{cm}^2\).

Both solids have surface area \(600\,\text{cm}^2\). The difference is \(0\,\text{cm}^2\).

- Use the triangle area formula for the base. - A prism's volume is its base area times its length. - The lateral surface area is the base perimeter times the prism length. - Add both bases to the lateral area for total surface area.

1. \(B=\frac12\cdot12\cdot8=48\,\text{cm}^2\). 2. \(V=Bh=48\cdot15=720\,\text{cm}^3\). 3. The base perimeter is \(12+10+10=32\,\text{cm}\), so \(L=32\cdot15=480\,\text{cm}^2\). 4. \(SA=2B+L=2\cdot48+480=576\,\text{cm}^2\).

a) \(48\,\text{cm}^2\) b) \(720\,\text{cm}^3\) c) \(480\,\text{cm}^2\) d) \(576\,\text{cm}^2\)

- Find the area and perimeter of the trapezoidal base. - Multiply base area by prism length for volume. - Multiply base perimeter by prism length for lateral area. - Add the two bases for total surface area.

1. The base area is \(B=\frac{12+6}{2}\cdot4=36\,\text{cm}^2\). 2. The volume is \(V=Bh=36\cdot15=540\,\text{cm}^3\). 3. The base perimeter is \(12+5+6+5=28\,\text{cm}\), so the lateral surface area is \(L=28\cdot15=420\,\text{cm}^2\). 4. The total surface area is \(SA=2B+L=2\cdot36+420=492\,\text{cm}^2\).

The volume is \(540\,\text{cm}^3\), and the total surface area is \(492\,\text{cm}^2\).

- Add the two prism volumes. - Add their separate surface areas. - Subtract the shared contact area twice.

1. The volumes are \(6\cdot4\cdot2=48\,\text{cm}^3\) and \(3\cdot4\cdot3=36\,\text{cm}^3\), so the total volume is \(84\,\text{cm}^3\). 2. The lower prism's surface area is \(2(6\cdot4+6\cdot2+4\cdot2)=88\,\text{cm}^2\). 3. The upper prism's surface area is \(2(3\cdot4+3\cdot3+4\cdot3)=66\,\text{cm}^2\). 4. Their contact area is \(3\cdot4=12\,\text{cm}^2\), hidden on both prisms. 5. The exterior surface area is \(88+66-2(12)=130\,\text{cm}^2\).

The total volume is \(84\,\text{cm}^3\), and the exterior surface area is \(130\,\text{cm}^2\).

- Subtract the opening from the outer rectangle to get the base area. - Trace every edge of the U-shaped base to get its perimeter. - Include both bases and all lateral faces.

1. The U-shaped base area is \(B=8\cdot6-4\cdot4=32\,\text{cm}^2\). 2. The U-shaped base perimeter is \(8+6+2+4+4+4+2+6=36\,\text{cm}\). 3. The lateral surface area is \(L=36\cdot12=432\,\text{cm}^2\). 4. The total surface area is \(SA=2B+L=2\cdot32+432=496\,\text{cm}^2\).

\(496\,\text{cm}^2\)

- Count the unit squares in the cross section. - Count the unit-length segments around the cross-shaped boundary. - Use the cross-sectional area for volume and its perimeter for lateral surface area. - Include both end faces.

a) The cross-sectional area is \(B=5\cdot1^2=5\,\text{dm}^2\), so the volume is \(V=5\cdot20=100\,\text{dm}^3\). b) The cross-shaped base has twelve boundary segments of length \(1\,\text{dm}\), so its perimeter is \(12\,\text{dm}\). The lateral surface area is \(12\cdot20=240\,\text{dm}^2\). Including both ends gives \(2\cdot5+240=250\,\text{dm}^2\).

a) \(100\,\text{dm}^3\) b) \(250\,\text{dm}^2\)

- Find the square base area first. - Subtract the top and bottom from the total surface area. - Use the four congruent side faces to solve for the height.

a) The base area is \(4^2=16\,\text{cm}^2\). b) The top and bottom have total area \(2\cdot16=32\,\text{cm}^2\), so the four side faces have total area \(160-32=128\,\text{cm}^2\). If the height is \(h\), the four side faces have area \(4(4h)=16h\). Therefore, \(16h=128\), so \(h=8\,\text{cm}\). c) The volume is \(16\cdot8=128\,\text{cm}^3\).

a) The base area is \(16\,\text{cm}^2\). b) The height is \(8\,\text{cm}\). c) The volume is \(128\,\text{cm}^3\).

- Decompose the L-shaped base into rectangles. - Trace the entire boundary to find the base perimeter. - Use the base area for volume and the base perimeter for lateral surface area. - Include both end faces.

1. Split the L-shaped base into a \(2\,\text{cm}\times4\,\text{cm}\) rectangle and a \(3\,\text{cm}\times2\,\text{cm}\) rectangle. 2. The base area is \(B=2\cdot4+3\cdot2=14\,\text{cm}^2\), so the volume is \(V=14\cdot8=112\,\text{cm}^3\). 3. The base perimeter is \(5+2+3+2+2+4=18\,\text{cm}\), so the lateral surface area is \(18\cdot8=144\,\text{cm}^2\). 4. The total surface area is \(2\cdot14+144=172\,\text{cm}^2\).

The volume is \(112\,\text{cm}^3\), and the total surface area is \(172\,\text{cm}^2\).

- Split the house-shaped base into a rectangle and a triangle. - Find the roof triangle's height from the two given vertical heights. - Add the outside edge lengths to find the base perimeter. - Include both house-shaped ends.

1. The rectangular part of the base has area \(8\cdot3=24\,\text{cm}^2\). 2. The roof triangle has height \(6-3=3\,\text{cm}\), so its area is \(\frac{1}{2}\cdot8\cdot3=12\,\text{cm}^2\). 3. The total base area is \(36\,\text{cm}^2\), and the volume is \(36\cdot12=432\,\text{cm}^3\). 4. The base perimeter is \(8+3+5+5+3=24\,\text{cm}\), so the lateral surface area is \(24\cdot12=288\,\text{cm}^2\). 5. The total surface area is \(2\cdot36+288=360\,\text{cm}^2\).

The volume is \(432\,\text{cm}^3\), and the total surface area is \(360\,\text{cm}^2\).

- Convert all dimensions to centimeters first. - Find the surface area of each rectangular prism. - Subtract the shared contact area twice because it is not exposed.

1. Convert the dimensions to centimeters. The lower prism measures \(6\,\text{cm}\times3\,\text{cm}\times2\,\text{cm}\), and the upper prism measures \(2\,\text{cm}\times3\,\text{cm}\times4\,\text{cm}\). 2. The lower prism has surface area \(2(6\cdot3+6\cdot2+3\cdot2)=72\,\text{cm}^2\). 3. The upper prism has surface area \(2(2\cdot3+2\cdot4+3\cdot4)=52\,\text{cm}^2\). 4. Their contact area is \(2\cdot3=6\,\text{cm}^2\). This area is counted once on each prism, so subtract it twice. 5. The total surface area is \(72+52-2(6)=112\,\text{cm}^2\).

The total surface area is \(112\,\text{cm}^2\).

- Find the area and perimeter of the triangular base. - Multiply the base perimeter by the prism length. - Add both triangular end faces.

1. The right triangular base has area \(B=\frac{1}{2}\cdot3\cdot4=6\,\text{cm}^2\). 2. The base perimeter is \(3+4+5=12\,\text{cm}\). 3. The lateral surface area is \(12\cdot10=120\,\text{cm}^2\). 4. The total surface area is \(2\cdot6+120=132\,\text{cm}^2\).

\(132\,\text{cm}^2\)

- Find the C-shaped base area by subtracting the rectangular cutout. - Trace the entire indented perimeter of the base. - Use the prism length shown in the diagram for the lateral area.

1. The base is a \(6\,\text{cm} \times 6\,\text{cm}\) rectangle with a \(4\,\text{cm} \times 2\,\text{cm}\) cutout, so \(B=6\cdot6-4\cdot2=28\,\text{cm}^2\). 2. The C-shaped base perimeter is \(6+6+6+2+4+2+4+2=32\,\text{cm}\). 3. The diagram gives prism length \(10\,\text{cm}\), so the lateral area is \(32\cdot10=320\,\text{cm}^2\). 4. The total surface area is \(SA=2B+320=376\,\text{cm}^2\).

The total surface area is \(376\,\text{cm}^2\).

- Find the surface area of each rectangular prism. - Subtract the shared contact face twice. - Multiply the exterior area by the cost per square centimeter.

1. The lower \(4\,\text{cm}\times4\,\text{cm}\times2\,\text{cm}\) prism has surface area \(2(4\cdot4+4\cdot2+4\cdot2)=64\,\text{cm}^2\). 2. The upper \(1\,\text{cm}\times4\,\text{cm}\times2\,\text{cm}\) prism has surface area \(2(1\cdot4+1\cdot2+4\cdot2)=28\,\text{cm}^2\). 3. Their contact area is \(1\cdot4=4\,\text{cm}^2\), so subtract it twice. The exterior area is \(64+28-2(4)=84\,\text{cm}^2\). 4. The coating cost is \(84\cdot\$0.05=\$4.20\).

The total coating cost is \(\$4.20\).

- Count four edges of each dimension. - A rectangular prism has three pairs of congruent faces. - Use the square-unit conversion between square centimeters and square meters.

a) A rectangular prism has four edges of each dimension, so the total edge length is \(4(60+40+35)=540\,\text{cm}\). b) The surface area is \(2(60\cdot40+40\cdot35+60\cdot35)=11{,}800\,\text{cm}^2\). Since \(10{,}000\,\text{cm}^2=1\,\text{m}^2\), this is \(1.18\,\text{m}^2\).

a) The total edge length is \(540\,\text{cm}\). b) The surface area is \(1.18\,\text{m}^2\).

- The three edges meeting at one vertex are the prism's length, width, and height. - Find the missing edge length first. - Then use the surface area formula for a rectangular prism.

1. Let the third edge length be \(c\). Then \(8+7+c=20\), so \(c=5\,\text{cm}\). 2. The surface area is \(2(8\cdot7+8\cdot5+7\cdot5)=2(56+40+35)=262\,\text{cm}^2\).

The surface area is \(262\,\text{cm}^2\).

- Count four edges of each dimension. - Find the areas of the bottom and four side walls. - Do not include a top face.

1. A rectangular prism has four edges of each dimension, so the total edge length is \(4(60+30+40)=520\,\text{cm}\). 2. The bottom area is \(60\cdot30=1800\,\text{cm}^2\). 3. The two long side walls have total area \(2(60\cdot40)=4800\,\text{cm}^2\). 4. The two short side walls have total area \(2(30\cdot40)=2400\,\text{cm}^2\). 5. The total glass area is \(1800+4800+2400=9000\,\text{cm}^2\).

a) The total edge length is \(520\,\text{cm}\). b) The total glass area is \(9000\,\text{cm}^2\).

- Separate the covered base from the covered triangular faces. - Use the net to identify the base and slant height of one triangular face. - Count only the triangular faces that are actually covered.

1. The square base has area \(10^2=100\,\text{cm}^2\). 2. Each triangular face has base \(10\,\text{cm}\) and slant height \(13\,\text{cm}\), so its area is \(\frac12\cdot10\cdot13=65\,\text{cm}^2\). 3. Three triangular faces are covered, so their covered area is \(3\cdot65=195\,\text{cm}^2\). 4. The total covered area is \(100+195=295\,\text{cm}^2\).

\(295\,\text{cm}^2\)

- Remove the base area and the two known triangular-face areas from the total surface area. - Express the combined area of the two remaining congruent triangles in terms of \(s\). - Solve the resulting one-step equation only after the known face areas are accounted for.

1. The base area is \(14\cdot20=280\,\text{cm}^2\). 2. The two known triangular faces have total area \(2\cdot\frac12\cdot14\cdot26=364\,\text{cm}^2\). 3. The two remaining faces therefore have total area \(1144-280-364=500\,\text{cm}^2\). 4. Their combined area is \(2\cdot\frac12\cdot20\cdot s=20s\), so \(20s=500\) and \(s=25\,\text{cm}\).

\(s=25\,\text{cm}\)

- Pair the triangular faces by the side of the rectangular base they use. - Check whether the two pairs have the same slant height before combining them. - Find the area of each pair separately.

1. A rectangular base has two different side lengths, so the two pairs of triangular faces can have different slant heights. Mateo cannot use one slant height for every face. 2. The pair with \(30\,\text{cm}\) bases has total area \(2\cdot\frac12\cdot30\cdot29=870\,\text{cm}^2\). 3. The pair with \(42\,\text{cm}\) bases has total area \(2\cdot\frac12\cdot42\cdot25=1050\,\text{cm}^2\). 4. The correct lateral area is \(870+1050=1920\,\text{cm}^2\).

Mateo incorrectly used one slant height for both pairs of triangular faces. The correct lateral area is \(1920\,\text{cm}^2\).

- “Closed” means the rectangular base must be included. - Add the two pairs of triangular-face areas before comparing with the sheet area. - Compare the required surface area with the available material only at the end.

1. The base area is \(18\cdot10=180\,\text{cm}^2\). 2. The four triangular faces have total area \(2\cdot117+2\cdot75=384\,\text{cm}^2\). 3. The total surface area is \(180+384=564\,\text{cm}^2\). 4. Since \(600-564=36\,\text{cm}^2\), one sheet is enough and \(36\,\text{cm}^2\) remains.

Yes. One sheet is enough, with \(36\,\text{cm}^2\) remaining.

- Start with the area of the rectangular base. - Combine each congruent pair of triangular faces before adding all parts. - Keep the two slant heights attached to the correct base lengths.

1. The rectangular base has area \(ab\). 2. The two triangular faces with base \(a\) have combined area \(2\cdot\frac12 a s_a=a s_a\). 3. The two triangular faces with base \(b\) have combined area \(2\cdot\frac12 b s_b=b s_b\). 4. Therefore, \(SA=ab+a s_a+b s_b\).

\(SA=ab+a s_a+b s_b\)

- First find the surface area of the solid whose dimensions are all known. - The two solids have equal surface areas. - Write an equation that includes the unknown height.

1. Convert the cube's edge length: \(40\,\text{mm}=4\,\text{cm}\). 2. The cube's surface area is \(SA=6s^2=6(4^2)=96\,\text{cm}^2\). 3. Set the rectangular prism's surface area equal to \(96\): \(96=2(6\cdot2+6h+2h)\). 4. Simplify and solve: \(96=2(12+8h)\), so \(48=12+8h\), \(36=8h\), and \(h=4.5\,\text{cm}\).

The rectangular prism's height is \(4.5\,\text{cm}\).

- Count the small cubes in the \(3\times3\times3\) arrangement. - Use \(SA=6s^2\) for a cube. - Think about which faces are hidden before the cube is taken apart.

1. There are \(3^3=27\) small cubes, so their total volume is \(27\cdot1=27\,\text{cm}^3\). The large cube also has volume \(3^3=27\,\text{cm}^3\). 2. The large cube has edge length \(3\,\text{cm}\), so its surface area is \(6(3^2)=54\,\text{cm}^2\). 3. Each small cube has edge length \(1\,\text{cm}\) and surface area \(6(1^2)=6\,\text{cm}^2\). All \(27\) small cubes have total surface area \(27\cdot6=162\,\text{cm}^2\). 4. The separated cubes have more surface area because faces that were hidden inside the large cube become exposed.

a) Both volumes are \(27\,\text{cm}^3\). b) The large cube has surface area \(54\,\text{cm}^2\), while the separated cubes have total surface area \(162\,\text{cm}^2\). The difference comes from interior faces becoming exposed.

- Count how many edges have each dimension. - Express every edge length in terms of the square base side. - After finding the dimensions, use the surface-area formula for a rectangular prism.

1. Let \(a\) be the side length of the square base. Then the height is \(2a\). 2. The eight horizontal edges total \(8a\), and the four vertical edges total \(4(2a)\). Therefore, \(8a+8a=120\). 3. Solving \(16a=120\) gives \(a=7.5\,\text{cm}\), so the dimensions are \(7.5\,\text{cm}\times7.5\,\text{cm}\times15\,\text{cm}\). 4. The surface area is \(2(7.5\cdot7.5+7.5\cdot15+7.5\cdot15)=562.5\,\text{cm}^2\).

a) The dimensions are \(7.5\,\text{cm}\times7.5\,\text{cm}\times15\,\text{cm}\). b) The surface area is \(562.5\,\text{cm}^2\).

- Relate liters to cubic decimeters. - Find whole-number factor triples with product \(40\). - Use all six faces when finding the surface area.

1. Since \(1\,\text{L}=1\,\text{dm}^3\), the box's volume is \(40\,\text{dm}^3\). 2. Three possible dimension sets are \(10\,\text{dm}\times2\,\text{dm}\times2\,\text{dm}\), \(8\,\text{dm}\times5\,\text{dm}\times1\,\text{dm}\), and \(5\,\text{dm}\times4\,\text{dm}\times2\,\text{dm}\). 3. For \(5\,\text{dm}\times4\,\text{dm}\times2\,\text{dm}\), the surface area is \(2(5\cdot4+5\cdot2+4\cdot2)=76\,\text{dm}^2\). 4. Each small cube has volume \(1\,\text{dm}^3\), so exactly \(40\) cubes fit.

a) Sample dimensions are \(10\,\text{dm}\times2\,\text{dm}\times2\,\text{dm}\), \(8\,\text{dm}\times5\,\text{dm}\times1\,\text{dm}\), and \(5\,\text{dm}\times4\,\text{dm}\times2\,\text{dm}\). b) For \(5\,\text{dm}\times4\,\text{dm}\times2\,\text{dm}\), the surface area is \(76\,\text{dm}^2\). Answers will vary. c) \(40\) unit cubes fit exactly.

- Divide the total surface area by six to find one face's area. - Volumes add when the cubes are stacked. - At each contact, two square faces become hidden.

1. One face has area \(150\div6=25\,\text{cm}^2\), so the cube's edge length is \(5\,\text{cm}\). 2. One cube's volume is \(5^3=125\,\text{cm}^3\). 3. The tower's volume is \(4\cdot125=500\,\text{cm}^3\). 4. Four separate cubes have \(24\) faces. The three contact points hide \(2\) faces each, leaving \(24-6=18\) exposed faces. 5. The tower's surface area is \(18\cdot25=450\,\text{cm}^2\).

a) The edge length is \(5\,\text{cm}\), and one cube's volume is \(125\,\text{cm}^3\). b) The tower's volume is \(500\,\text{cm}^3\), and its surface area is \(450\,\text{cm}^2\).

- Find the original surface area first. - Determine the dimensions of each half. - Account for the two new faces created by the cut.

1. The original surface area is \(2(10\cdot8+10\cdot6+8\cdot6)=376\,\text{cm}^2\). 2. Each new prism measures \(10\,\text{cm}\times8\,\text{cm}\times3\,\text{cm}\). 3. One new prism's surface area is \(2(10\cdot8+10\cdot3+8\cdot3)=268\,\text{cm}^2\). 4. The sum is \(2\cdot268=536\,\text{cm}^2\). 5. Equivalently, the cut creates two new \(10\,\text{cm}\times8\,\text{cm}\) faces, increasing the original area by \(160\,\text{cm}^2\).

a) The original surface area is \(376\,\text{cm}^2\). b) The sum of the new surface areas is \(536\,\text{cm}^2\).

- Find the surface area of each solid separately. - The glued region is not exposed on either solid. - Subtract the contact area twice.

1. The lower prism's surface area is \(2(10\cdot10+10\cdot4+10\cdot4)=360\,\text{cm}^2\). 2. The cube's surface area is \(6(5^2)=150\,\text{cm}^2\). 3. The glued contact area is \(5\cdot5=25\,\text{cm}^2\). This area is hidden on both solids. 4. The exterior surface area is \(360+150-2(25)=460\,\text{cm}^2\).

The pedestal's exterior surface area is \(460\,\text{cm}^2\).

- A rectangular prism has \(12\) edges. With a square base, \(8\) of those edges have the base side length. - The remaining \(4\) edges are the vertical edges. - The surface area is the sum of the areas of all six faces.

1. The square base and square top have a total of \(8\) edges of length \(12\,\text{cm}\), using \(8 \cdot 12 = 96\,\text{cm}\) of wire. 2. The remaining wire for the \(4\) vertical edges is \(140 - 96 = 44\,\text{cm}\). 3. The height is \(44 \div 4 = 11\,\text{cm}\). 4. The two square faces have total area \(2(12 \cdot 12) = 288\,\text{cm}^2\). The four rectangular side faces have total area \(4(12 \cdot 11) = 528\,\text{cm}^2\). 5. The total surface area is \(288 + 528 = 816\,\text{cm}^2\).

a) The height is \(11\,\text{cm}\). b) The minimum area of paper needed is \(816\,\text{cm}^2\).

- Use the wire length to determine all missing edge lengths first. - A rectangular prism has four edges of each dimension. - Then compare the two surface areas.

1. Solid A has edge length \(60\div12=5\,\text{cm}\), so \(SA_A=6\cdot5^2=150\,\text{cm}^2\). 2. For Solid B, \(4(l+w+h)=60\), so \(l+w+h=15\). Thus \(h=3\,\text{cm}\). 3. \(SA_B=2(7\cdot5+7\cdot3+5\cdot3)=142\,\text{cm}^2\). 4. Since \(150>142\), Solid A has the greater surface area.

Solid A has the greater surface area: \(150\,\text{cm}^2\), compared with \(142\,\text{cm}^2\) for Solid B.

- Find the original surface area first. - Recalculate after each dimension change. - Compare each increase, not just the new totals.

1. The original surface area is \(SA=2(4\cdot3+4\cdot1+3\cdot1)=38\,\text{cm}^2\). 2. Change 1 gives dimensions \(8,6,2\) cm, so \(SA_1=152\,\text{cm}^2\), an increase of \(114\,\text{cm}^2\). 3. Change 2 gives dimensions \(4,3,4\) cm, so \(SA_2=80\,\text{cm}^2\), an increase of \(42\,\text{cm}^2\). 4. Change 1 produces the greater increase.

Change 1. It increases surface area by \(114\,\text{cm}^2\), compared with \(42\,\text{cm}^2\) for Change 2.

- Which pair of faces can you calculate from the base dimensions? - After subtracting the base and top areas, how much surface area remains for the four side faces? - How are the four side faces related to the base perimeter and the height?

1. Find the combined area of the base and top: \(2(12 \cdot 8) = 192\,\text{in.}^2\). 2. The four side faces have total area \(392 - 192 = 200\,\text{in.}^2\). 3. The perimeter of the base is \(2(12 + 8) = 40\,\text{in.}\). 4. The total area of the side faces equals the base perimeter times the height, so \(40h = 200\). Therefore, \(h = 5\,\text{in.}\).

The box is \(5\,\text{in.}\) high.

- Find the surface area shared by the two solids first. - Subtract the area of the pair of faces whose dimensions are already known. - Write the remaining four-face area in terms of the unknown dimension.

1. The cube has \(SA=6\cdot10^2=600\,\text{cm}^2\), so the prism also has surface area \(600\,\text{cm}^2\). 2. The two \(5\,\text{cm}\times20\,\text{cm}\) faces total \(200\,\text{cm}^2\), leaving \(400\,\text{cm}^2\) for the other four faces. 3. If the third dimension is \(c\), those faces total \(2(5c)+2(20c)=50c\). 4. Solve \(50c=400\) to get \(c=8\,\text{cm}\).

The third dimension is \(8\,\text{cm}\).

- Start with the surface area of each cube before gluing. - The contact region is hidden on both cubes. - Combine the remaining like terms.

1. Before gluing, the cubes have total surface area \(6a^2+6b^2\). 2. The glued face of the smaller cube contributes hidden area \(b^2\), and an equal area on the larger cube is covered. 3. Subtract both hidden areas: \(SA=6a^2+6b^2-2b^2=6a^2+4b^2\).

\(SA=6a^2+4b^2\)

- Split the L-shaped face into two rectangles. - Multiply the face area by the prism length for volume. - Use the face perimeter times the prism length for the lateral area.

a) Split the L-shaped face into rectangles: \(3\cdot4+3\cdot2=18\,\text{cm}^2\). The volume is \(18\cdot5=90\,\text{cm}^3\). b) The perimeter of the L-shaped face is \(6+2+3+2+3+4=20\,\text{cm}\). The lateral area is \(20\cdot5=100\,\text{cm}^2\). The total surface area is \(2(18)+100=136\,\text{cm}^2\).

a) The volume is \(90\,\text{cm}^3\). b) The surface area is \(136\,\text{cm}^2\).

- Add the volumes of the three prisms. - For surface area, count only faces visible from outside. - View the staircase from the front, side, top, and bottom.

1. The total volume is \(30\cdot10\cdot10+20\cdot10\cdot10+10\cdot10\cdot10=6000\,\text{cm}^3\). 2. The front and back stair-shaped faces each have area \(30\cdot10+20\cdot10+10\cdot10=600\,\text{cm}^2\), for \(1200\,\text{cm}^2\) total. 3. The vertical end and riser faces total \(30\cdot10+3(10\cdot10)=600\,\text{cm}^2\). 4. The bottom area is \(30\cdot10=300\,\text{cm}^2\), and the horizontal tread areas also total \(300\,\text{cm}^2\). 5. The exterior surface area is \(1200+600+300+300=2400\,\text{cm}^2\).

The total volume is \(6000\,\text{cm}^3\), and the exterior surface area is \(2400\,\text{cm}^2\).

- Add the volumes of the three prisms. - Add their individual surface areas. - Subtract both sides of each contact area.

a) The two posts have total volume \(2(2\cdot3\cdot4)=48\,\text{cm}^3\). The top beam has volume \(8\cdot3\cdot2=48\,\text{cm}^3\). The total is \(96\,\text{cm}^3\). b) One post has surface area \(2(2\cdot3+3\cdot4+2\cdot4)=52\,\text{cm}^2\), and the top beam has surface area \(2(8\cdot3+3\cdot2+8\cdot2)=92\,\text{cm}^2\). There are two contact areas of \(2\cdot3=6\,\text{cm}^2\), each hidden on both touching solids. Thus the exterior surface area is \(2(52)+92-4(6)=172\,\text{cm}^2\).

a) The total volume is \(96\,\text{cm}^3\). b) The exterior surface area is \(172\,\text{cm}^2\).

- Add the three prism volumes. - Include all vertical side faces. - For horizontal visible areas, account for covered contact regions and omit the ground face.

a) The total volume is \(6\cdot4\cdot2+4\cdot2\cdot2+2\cdot2\cdot2=48+16+8=72\,\text{m}^3\). b) The lateral areas are \(2(6+4)(2)=40\,\text{m}^2\), \(2(4+2)(2)=24\,\text{m}^2\), and \(2(2+2)(2)=16\,\text{m}^2\). The visible horizontal top-facing areas together equal the bottom prism's top area, \(6\cdot4=24\,\text{m}^2\). Excluding the ground face, the visible area is \(40+24+16+24=104\,\text{m}^2\).

a) The total volume is \(72\,\text{m}^3\). b) The area to be cleaned is \(104\,\text{m}^2\).

- Start with the surface area of the \(3\times3\times1\) base. - Compare the face covered by the top cube with the new exterior faces it creates. - Include the bottom faces as directed.

1. A \(3\times3\times1\) rectangular prism has surface area \(2(3\cdot3+3\cdot1+3\cdot1)=30\,\text{cm}^2\). 2. Placing one cube on the center covers \(1\,\text{cm}^2\) of the prism's top and exposes the other five faces of that cube. The surface area increases by \(5-1=4\,\text{cm}^2\). 3. The total surface area is \(30+4=34\,\text{cm}^2\).

The surface area is \(34\,\text{cm}^2\).

- Include both the outside and inside wall surfaces. - Subtract twice the wall thickness to find the side length of the opening. - Each end face is a large square with a smaller square removed.

1. The four outer side faces have total area \(4(30\cdot20)=2400\,\text{cm}^2\). 2. The opening has side length \(30-2(5)=20\,\text{cm}\). 3. The four inner side faces have total area \(4(20\cdot20)=1600\,\text{cm}^2\). 4. Each end face has area \(30^2-20^2=500\,\text{cm}^2\), so the front and back contribute \(2(500)=1000\,\text{cm}^2\). 5. The total surface area is \(2400+1600+1000=5000\,\text{cm}^2\).

The total surface area is \(5000\,\text{cm}^2\).

- Add the surface areas of the three separate boxes first. - The contact height between adjacent boxes is the height of the shorter box. - Subtract each shared contact face twice.

1. The three separate surface areas total \(210+290+250=750\,\text{dm}^2\). 2. The \(8\)-dm and \(12\)-dm boxes share a \(5\,\text{dm} \times 8\,\text{dm}\) face. The \(12\)-dm and \(10\)-dm boxes share a \(5\,\text{dm} \times 10\,\text{dm}\) face. 3. Each contact face was counted twice, so subtract both copies: \(750-2(40)-2(50)=570\,\text{dm}^2\).

The combined surface area is \(570\,\text{dm}^2\).

- Remove the square base area from the total surface area first. - The remaining lateral area is shared equally by four congruent triangular faces. - In one triangular face, the square’s side is the triangle’s base and the unknown slant height is its altitude.

1. The base area is \(12^2=144\,\text{cm}^2\). 2. The total lateral area is \(384-144=240\,\text{cm}^2\). 3. Each of the four congruent triangular faces has area \(240\div4=60\,\text{cm}^2\). 4. If \(s\) is the slant height, then \(\frac12\cdot12\cdot s=60\). Thus \(6s=60\), so \(s=10\,\text{cm}\).

The slant height is \(10\,\text{cm}\).

- Find all whole-number factor triples of \(12\). - An open-top box has five faces. - For each factor triple, test each distinct choice of height.

1. The unordered whole-number factor triples with product \(12\) are \((12, 1, 1)\), \((6, 2, 1)\), \((4, 3, 1)\), and \((3, 2, 2)\), in decimeters. 2. For a base of \(4\,\text{dm}\times3\,\text{dm}\) and height \(1\,\text{dm}\), the wood area is \(4\cdot3+2(4\cdot1)+2(3\cdot1)=26\,\text{dm}^2\). 3. For an open box with base dimensions \(l\) and \(w\) and height \(h\), the wood area is \(A=lw+2lh+2wh\). 4. Testing each distinct orientation gives a minimum of \(26\,\text{dm}^2\). This occurs for a \(4\,\text{dm}\times3\,\text{dm}\) base with height \(1\,\text{dm}\), and for a \(3\,\text{dm}\times2\,\text{dm}\) base with height \(2\,\text{dm}\).

a) The combinations are \((12, 1, 1)\), \((6, 2, 1)\), \((4, 3, 1)\), and \((3, 2, 2)\), in decimeters. b) The wood area is \(26\,\text{dm}^2\). c) The minimum is \(26\,\text{dm}^2\), for dimensions \(4\,\text{dm}\times3\,\text{dm}\times1\,\text{dm}\) with height \(1\,\text{dm}\), or \(3\,\text{dm}\times2\,\text{dm}\times2\,\text{dm}\) with height \(2\,\text{dm}\).