- Count the grid-square width and height of the rectangle labeled \(B\). - Each grid square represents \(1\,\text{cm}\) in each direction.

1. Face \(B\) is \(3\) grid squares wide and \(2\) grid squares high, so its dimensions are \(3\,\text{cm}\) by \(2\,\text{cm}\). 2. Its area is \(A=3\cdot2=6\,\text{cm}^2\).

The area of face \(B\) is \(6\,\text{cm}^2\).

- List the surfaces that are physically present on an open-top cylinder. - The curved side contributes the lateral area; decide how many circular bases are included.

1. The curved side has area \(2\pi rh\). 2. Because the container has no top, only one circular base is included, with area \(\pi r^2\). 3. Therefore, the required material area is \(2\pi rh+\pi r^2\).

\(2\pi rh+\pi r^2\). The term \(2\pi rh\) is the curved side, and \(\pi r^2\) is the bottom.

- Recall the surface-area formula for a sphere. - Substitute the radius and keep \(\pi\) exact.

1. Use \(SA=4\pi r^2\): \(SA=4\pi(2)^2=16\pi\,\text{cm}^2\).

\(SA=16\pi\,\text{cm}^2\)

- Count the six faces on each cube before joining them. - Which faces can no longer be touched or painted? - Area counts faces on the outside, not inside joints.

1. Before joining, the two cubes have \(6+6=12\) square faces. 2. The two faces in contact are hidden and do not count as exposed surfaces. 3. There are \(12-2=10\) exposed faces, each of area \(1\,\text{cm}^2\).

\(10\) exposed faces; \(10\,\text{cm}^2\) of surface area.

- Think about the faces on the outside of the box. - Does paper cover an area or fill a space?

1. Covering the outside requires the total area of the box's outer faces: its surface area. 2. Area is measured in square units, so the answer uses \(\text{cm}^2\), not \(\text{cm}^3\).

Surface area, in \(\text{cm}^2\).

- Which surfaces are the same in both containers? - Identify the one surface present in only one container. - Do you need their heights to compare the areas?

1. The curved walls and bottom disks have equal areas in both containers. 2. The closed one has one extra circular top of area \(\pi(3)^2=9\pi\,\text{cm}^2\). 3. The open one therefore has \(9\pi\,\text{cm}^2\) less surface area.

\(9\pi\,\text{cm}^2\) less.

- Covering the outside requires surface area, not volume. - In the surface area formula, what power of the radius appears? - Compare \((2r)^2\) with \(r^2\).

1. A sphere has surface area \(A=4\pi r^2\). 2. Replacing \(r\) by \(2r\) gives \(4\pi(2r)^2=4(4\pi r^2)\). 3. The required covering area is four times as great.

\(4\) times as much covering material.

- What shape does the strip make when laid flat? - Which circle measure gives the strip's length? - Convert all measurements to the same unit before finding area.

1. The strip's length is the pool's circumference: \(C=\pi d=4.50\pi\,\text{m}\approx14.14\,\text{m}\). 2. Convert the height: \(12\,\text{cm}=0.12\,\text{m}\). 3. When laid flat, the strip is a rectangle. Its area is \(A=Ch=(4.50\pi)(0.12)\,\text{m}^2=0.54\pi\,\text{m}^2\approx1.70\,\text{m}^2\).

a) \(14.14\,\text{m}\) b) \(1.70\,\text{m}^2\)

- How are the area, height, and length of a rectangle related? - What does the label's length represent on the can? - How can you find a circle's diameter from its circumference?

1. When unwrapped, the label is a rectangle whose length equals the can's circumference. Thus, \(C=\frac{180\,\text{cm}^2}{6\,\text{cm}}=30\,\text{cm}\). 2. Since \(C=\pi d\), \(d=\frac{C}{\pi}=\frac{30}{\pi}\,\text{cm}\approx9.55\,\text{cm}\).

a) \(30\,\text{cm}\) b) \(9.55\,\text{cm}\)

- Convert the diameter to a radius first. - Use the sphere volume and surface-area formulas. - Check the exponents carefully when entering the calculations.

1. The radius is \(r=12.4\div2=6.2\,\text{cm}\). 2. The volume is \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\cdot6.2^3\approx998.3060\,\text{cm}^3\). 3. The surface area is \(SA=4\pi r^2=4\pi\cdot6.2^2\approx483.0513\,\text{cm}^2\).

\(V\approx998.3\,\text{cm}^3\); \(SA\approx483.1\,\text{cm}^2\)

- The open end has no disk to include in the outside area. - The given frustum formula counts only the curved side. - Combine the curved area with the base area before rounding.

1. The curved lateral area is \(A_{\text{lat}}=\pi(15+9)\cdot10=240\pi\,\text{cm}^2\). 2. The closed circular base has area \(\pi(9)^2=81\pi\,\text{cm}^2\). 3. The total exterior area is \(321\pi\,\text{cm}^2\approx1008.45\,\text{cm}^2\).

Approximately \(1008.45\,\text{cm}^2\).

- A cylinder has two circular bases and one lateral surface. - Use the radius and height in the total surface area formula. - Keep \(\pi\) in the exact result.

1. The two circular bases have total area \(2\pi r^2=2\pi\cdot10^2=200\pi\,\text{cm}^2\). 2. The lateral surface area is \(2\pi rh=2\pi\cdot10\cdot5=100\pi\,\text{cm}^2\). 3. The total surface area is \(SA=200\pi+100\pi=300\pi\,\text{cm}^2\).

\(300\pi\,\text{cm}^2\)

- The roof consists of four congruent triangular faces. - Do not include the square base attached to the tower. - Use the triangle area formula for one face, then account for all four faces.

1. Only the four triangular lateral faces are covered; the square base is attached to the tower. 2. One triangular face has area \(A=\frac{1}{2}\cdot4\cdot9=18\,\text{m}^2\). 3. The four faces have total area \(4\cdot18=72\,\text{m}^2\).

The roof area is \(72\,\text{m}^2\).

- Which surfaces make up the outside of a cylinder? - Include both circular bases and the curved lateral surface. - How are the base circumference and cylinder height related to the lateral area?

1. A cylinder's total surface area is the sum of two circular bases and the lateral surface: \(SA=2B+A_{\text{lat}}\). 2. One base has area \(B=\pi r^2=\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The lateral area is \(A_{\text{lat}}=2\pi rh=2\pi\cdot4\cdot10=80\pi\,\text{cm}^2\). 4. Therefore, \(SA=2\cdot16\pi+80\pi=112\pi\,\text{cm}^2\approx351.8584\,\text{cm}^2\). 5. Rounded to the nearest tenth, \(SA\approx351.9\,\text{cm}^2\).

The total surface area is approximately \(351.9\,\text{cm}^2\).

- Read the base dimensions and both different triangular-face altitudes from the net. - Pair the two congruent triangles attached to equal-length sides of the base. - Use \(\frac{1}{2}(\text{base})(\text{altitude})\) for each triangular face before adding all five faces.

1. The rectangular base has area \(10\cdot18=180\,\text{cm}^2\). 2. The two triangular faces with base \(10\,\text{cm}\) have displayed altitude \(15\,\text{cm}\). Together they have area \(2\left(\frac{1}{2}\cdot10\cdot15\right)=150\,\text{cm}^2\). 3. The two triangular faces with base \(18\,\text{cm}\) have displayed altitude \(13\,\text{cm}\). Together they have area \(2\left(\frac{1}{2}\cdot18\cdot13\right)=234\,\text{cm}^2\). 4. Therefore, \(SA=180+150+234=564\,\text{cm}^2\).

\(SA=180+150+234=564\,\text{cm}^2\)

- Count the cubes and the six faces each would have separately. - Which faces become hidden when cubes touch? - Each contact hides one face on each of two cubes.

1. There are four cubes, so they have \(4\cdot6=24\) unit-square faces before joining. 2. The three cubes in the bottom row have two shared faces. The upper cube shares one more face with the middle bottom cube. 3. Each of the three contacts hides two faces, leaving \(24-2\cdot3=18\) exposed unit-square faces. 4. Each has area \(1\,\text{cm}^2\), giving total surface area \(18\,\text{cm}^2\).

\(18\,\text{cm}^2\)

- Match pairs of congruent rectangles in the net. - How many times does each distinct face size appear? - The net has no overlap when unfolded.

1. The six faces form three equal opposite pairs. 2. Their total area is \(2(3\cdot2)+2(3\cdot4)+2(2\cdot4)=12+24+16=52\,\text{cm}^2\). 3. Folding changes where the faces meet but does not change their total area.

\(52\,\text{cm}^2\). Folding preserves the sum of the six face areas.

- Compare the outer areas, not the volumes. - Use the radius shown for each sphere. - Subtract the smaller surface area from the larger.

1. The smaller sphere has radius \(3\,\text{cm}\) and area \(4\pi(3)^2=36\pi\,\text{cm}^2\). 2. The larger has radius \(4\,\text{cm}\) and area \(4\pi(4)^2=64\pi\,\text{cm}^2\). 3. The difference is \(64\pi-36\pi=28\pi\,\text{cm}^2\).

\(28\pi\,\text{cm}^2\) more.

- You can calculate the area of each separate solid first. - A contact face belongs to both solids but is not on the outside. - How many copies of the contact area must be removed?

1. The larger block has area \(2(4\cdot3+4\cdot2+3\cdot2)=52\,\text{cm}^2\). 2. The cube has area \(6\cdot1^2=6\,\text{cm}^2\). 3. The two joined faces, each \(1\,\text{cm}^2\), are no longer exposed. 4. The combined surface area is \(52+6-2=56\,\text{cm}^2\).

\(56\,\text{cm}^2\)

- First find the mesh length needed to go exactly once around the trunk. - How does the overlap change the required length? - Use consistent units when finding area. - Keep the unrounded circumference available until the final area is found.

1. The trunk's circumference is \(C=16\pi\,\text{cm}\). 2. Including the overlap, one piece must be \((16\pi+15)\,\text{cm}\approx65.27\,\text{cm}\) long. 3. In meters, the exact setup for one piece is \(\frac{16\pi+15}{100}\,\text{m}\). The area for eight trees is \(8\cdot1.50\cdot\frac{16\pi+15}{100}\,\text{m}^2\approx7.83\,\text{m}^2\).

a) \(65.27\,\text{cm}\) b) \(7.83\,\text{m}^2\)

- Write formulas for prism volume, lateral area, and total surface area. - Which surfaces depend on the prism's height? - The two bases do not change when the height changes. - Compare the new surface-area expression with three times the original expression.

1. The original volume is \(V=Bh\). With height \(3h\), the new volume is \(V''=B(3h)=3V\). 2. The original lateral area is \(A_{\text{lat}}=ph\). The new lateral area is \(A''_{\text{lat}}=p(3h)=3A_{\text{lat}}\). 3. The original total surface area is \(SA=2B+A_{\text{lat}}\), while the new total surface area is \(SA''=2B+3A_{\text{lat}}\). 4. Three times the original surface area would be \(3SA=6B+3A_{\text{lat}}\). Since \(B>0\), \(2B+3A_{\text{lat}}<6B+3A_{\text{lat}}\). 5. Therefore, the total surface area increases but does not triple because the two bases do not change.

a) The volume triples: \(V''=3V\). b) The lateral area triples: \(A''_{\text{lat}}=3A_{\text{lat}}\). c) The claim is false. The lateral area triples, but the two unchanged bases prevent the total surface area from tripling.

- What is the formula for a cylinder's lateral area? - Convert the diameter to a radius in meters. - Multiply by the number of columns. - Why must the final number of paint cans be rounded up?

1. The radius is \(0.30\,\text{m}\). 2. One column has lateral area \(A_{\text{lat}}=2\pi rh=2\pi\cdot0.30\cdot4.00\,\text{m}^2=2.4\pi\,\text{m}^2\). 3. Four columns have total lateral area \(4\cdot2.4\pi\,\text{m}^2=9.6\pi\,\text{m}^2\approx30.16\,\text{m}^2\). 4. The exact number of cans required by area is \(9.6\pi\div12=0.8\pi\approx2.51\). Since whole cans are required, at least \(3\) cans are needed.

a) \(30.16\,\text{m}^2\) b) \(3\) cans

- How are milliliters and cubic centimeters related? - Rearrange the volume formula to solve for the radius. - The label's dimensions are the can's circumference and height. - Keep unrounded values until the final rounding step in each part.

1. Since \(1\,\text{mL}=1\,\text{cm}^3\), the volume is \(425\,\text{cm}^3\). 2. From \(V=\pi r^2h\), \(r=\sqrt{\frac{425}{10\pi}}\,\text{cm}\approx3.678\,\text{cm}\). Thus, \(d=2r\approx7.356\,\text{cm}\), which rounds to \(7.4\,\text{cm}\). 3. Using the unrounded radius, the lateral area is \(A_{\text{lat}}=2\pi rh\approx231.0997\,\text{cm}^2\), which rounds to \(231.1\,\text{cm}^2\).

a) About \(7.4\,\text{cm}\) b) About \(231.1\,\text{cm}^2\)

- Which parts of the cylinder will receive coating? - Convert the diameter into a radius before calculating area. - The bottom does not count; the top does. - Divide square meters by square meters covered per liter.

1. The radius is half the diameter: \(r=1.20\,\text{m}\). 2. The lateral area is \(2\pi rh=2\pi\cdot1.20\cdot3.50=8.40\pi\,\text{m}^2\). 3. The top area is \(\pi r^2=1.44\pi\,\text{m}^2\). 4. Total coated area is \(9.84\pi\,\text{m}^2\). 5. Divide by the coverage rate: \(\frac{9.84\pi}{8.5}\,\text{L}\approx3.6\,\text{L}\) to the nearest tenth.

About \(3.6\,\text{L}\).

- Use volume and height to find each radius. - What is the formula for lateral surface area? - Compare the two calculated areas.

1. Convert the volume: \(50\,\text{L}=50{,}000\,\text{cm}^3\). 2. For container A, \(r_A=\sqrt{\frac{50{,}000}{40\pi}}\,\text{cm}\approx19.947\,\text{cm}\). 3. Its lateral area is \(A_{\text{lat},A}=2\pi r_A\cdot40\,\text{cm}\approx5013.26\,\text{cm}^2\). 4. For container B, \(r_B=\sqrt{\frac{50{,}000}{60\pi}}\,\text{cm}\approx16.287\,\text{cm}\). 5. Its lateral area is \(A_{\text{lat},B}=2\pi r_B\cdot60\,\text{cm}\approx6139.96\,\text{cm}^2\). 6. Since \(5013.26<6139.96\), container A has the smaller lateral area.

Container A has the smaller lateral area: about \(5013.26\,\text{cm}^2\), compared with about \(6139.96\,\text{cm}^2\) for container B.

- First solve each volume formula for the missing height. - Total surface area includes the lateral area and both bases. - Compare the final surface areas.

1. Convert the volume: \(1\,\text{L}=1000\,\text{cm}^3\). 2. For model A, \(1000=\pi(5)^2h_A\), so \(h_A=\frac{40}{\pi}\,\text{cm}\approx12.73\,\text{cm}\). 3. Its surface area is \(SA_A=2\pi\cdot5^2+2\pi\cdot5\cdot\left(\frac{40}{\pi}\right)\,\text{cm}^2=50\pi+400\,\text{cm}^2\approx557.08\,\text{cm}^2\). 4. For model B, \(1000=10^2h_B\), so \(h_B=10\,\text{cm}\). 5. Its surface area is \(SA_B=2\cdot10^2+4\cdot10\cdot10\,\text{cm}^2=600\,\text{cm}^2\). 6. Since \(557.08<600\), model A uses less material.

Model A has the smaller surface area: about \(557.08\,\text{cm}^2\), compared with \(600\,\text{cm}^2\) for model B.

- Convert all measurements to the same unit before comparing areas. - Use the cylinder lateral-area formula. - You do not need a decimal approximation for \(\pi\) to simplify the ratio.

1. Convert all measurements to centimeters: \(r_1=4\,\text{cm}\), \(h_1=10\,\text{cm}\), \(r_2=8\,\text{cm}\), and \(h_2=2.5\,\text{cm}\). 2. \(A_{\text{lat},1}=2\pi\cdot4\cdot10\,\text{cm}^2=80\pi\,\text{cm}^2\). 3. \(A_{\text{lat},2}=2\pi\cdot8\cdot2.5\,\text{cm}^2=40\pi\,\text{cm}^2\). 4. Therefore, \(A_{\text{lat},1}>A_{\text{lat},2}\), and \(A_{\text{lat},1}:A_{\text{lat},2}=80\pi:40\pi=2:1\).

Cylinder \(C_1\) has the greater lateral area, and \(A_{\text{lat},1}:A_{\text{lat},2}=2:1\).

- Convert measurements to compatible units before calculating. - Use cubic decimeters when the volume is requested in liters. - Which cylinder surfaces are included when the top is excluded?

1. For volume, use \(r=4\,\text{dm}\) and \(h=10\,\text{dm}\): \(V=\pi\cdot4^2\cdot10\,\text{dm}^3=160\pi\,\text{dm}^3\approx502.7\,\text{L}\). 2. In meters, \(r=0.40\,\text{m}\) and \(h=1.0\,\text{m}\). The lateral area is \(A_{\text{lat}}=2\pi\cdot0.40\cdot1.0\,\text{m}^2=0.8\pi\,\text{m}^2\approx2.51\,\text{m}^2\). 3. The painted area includes the lateral surface and one circular base: \(A=0.8\pi+\pi(0.40)^2\,\text{m}^2=0.96\pi\,\text{m}^2\approx3.02\,\text{m}^2\).

a) \(502.7\,\text{L}\) b) \(2.51\,\text{m}^2\) c) \(3.02\,\text{m}^2\)

- Convert diameter to radius. - Add the wall thickness to the inside radius to get the outside radius. - List the two annular ends and both curved surfaces. - Convert all measurements to centimeters first.

1. Convert to centimeters. The inside radius is \(r=2\,\text{cm}\), the outside radius is \(R=3\,\text{cm}\), and the length is \(h=8\,\text{cm}\). 2. The steel volume is \(V=\pi(R^2-r^2)h=\pi(3^2-2^2)\cdot8\,\text{cm}^3=40\pi\,\text{cm}^3\approx125.66\,\text{cm}^3\). 3. The total surface area includes two annular ends, the outside lateral surface, and the inside lateral surface: \(SA=2\pi(R^2-r^2)+2\pi Rh+2\pi rh\). 4. Thus, \(SA=2\pi\cdot5+2\pi\cdot3\cdot8+2\pi\cdot2\cdot8\,\text{cm}^2=90\pi\,\text{cm}^2\approx282.74\,\text{cm}^2\).

The volume is \(40\pi\,\text{cm}^3\approx125.66\,\text{cm}^3\), and the total surface area is \(90\pi\,\text{cm}^2\approx282.74\,\text{cm}^2\).

- Use the cylinder formulas for volume and total surface area. - For part c, substitute the new radius and height into both formulas. - Remember that the radius is squared in the volume formula. - Compare each new result with its original value.

1. The volume is \(V=\pi r^2h=\pi\cdot5^2\cdot12=300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\). 2. The total surface area is \(SA=2\pi r^2+2\pi rh=2\pi\cdot5^2+2\pi\cdot5\cdot12=170\pi\,\text{cm}^2\approx534.07\,\text{cm}^2\). 3. After the changes, the radius is \(2.5\,\text{cm}\) and the height is \(24\,\text{cm}\). 4. The new volume is \(V_{\text{new}}=\pi\cdot2.5^2\cdot24=150\pi\,\text{cm}^3\approx471.24\,\text{cm}^3\), which is half the original volume. 5. The new total surface area is \(SA_{\text{new}}=2\pi\cdot2.5^2+2\pi\cdot2.5\cdot24=132.5\pi\,\text{cm}^2\approx416.26\,\text{cm}^2\), so the surface area decreases.

a) \(300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\) b) \(SA=170\pi\,\text{cm}^2\approx534.07\,\text{cm}^2\) c) The volume is halved to \(150\pi\,\text{cm}^3\approx471.24\,\text{cm}^3\), and the total surface area decreases to \(132.5\pi\,\text{cm}^2\approx416.26\,\text{cm}^2\).

- Find the volume of the first cylinder. - Set the two volume expressions equal to find the missing radius. - Use the lateral surface area formula for each cylinder.

1. The volume of \(Z_1\) is \(V_1=\pi\cdot4^2\cdot9=144\pi\,\text{cm}^3\). 2. Equal volumes give \(144\pi=\pi r_2^2\cdot4\). Therefore, \(r_2^2=36\), so \(r_2=6\,\text{cm}\). 3. The lateral surface area of a cylinder is \(A_{\text{lat}}=2\pi rh\). 4. For \(Z_1\), \(A_{\text{lat},1}=2\pi\cdot4\cdot9=72\pi\,\text{cm}^2\approx226.19\,\text{cm}^2\). 5. For \(Z_2\), \(A_{\text{lat},2}=2\pi\cdot6\cdot4=48\pi\,\text{cm}^2\approx150.80\,\text{cm}^2\). 6. Since \(72\pi>48\pi\), cylinder \(Z_1\) has the greater lateral surface area.

a) The radius of \(Z_2\) is \(6\,\text{cm}\). b) Cylinder \(Z_1\) has the greater lateral surface area: \(72\pi\,\text{cm}^2\approx226.19\,\text{cm}^2\), compared with \(48\pi\,\text{cm}^2\approx150.80\,\text{cm}^2\) for \(Z_2\).

- Which measurement must you find before you can calculate the total surface area? - Rearrange the cylinder volume formula to find that measurement. - Which surfaces make up a closed cylinder? - Avoid rounding intermediate values too early.

1. Use \(V=\pi r^2h\) to find the height: \(h=\frac{800}{\pi\cdot4.5^2}\approx12.5752\,\text{cm}\). 2. A closed cylinder has total surface area \(SA=2\pi r^2+2\pi rh\). 3. Substitute the values: \(SA=2\pi\cdot4.5^2+2\pi\cdot4.5\left(\frac{800}{\pi\cdot4.5^2}\right)\approx482.7901\,\text{cm}^2\). 4. Rounded to the nearest hundredth, \(SA\approx482.79\,\text{cm}^2\).

The total surface area is \(SA\approx482.79\,\text{cm}^2\).

- First determine the cylinder radius from the relationship between the diameter and height. - Recall the conversion between cubic meters and liters. - Which formulas give the volume and total surface area of a cylinder? - Decide which surfaces are included in the total exterior area.

1. The height is \(h=1.50\,\text{m}\). The diameter is \(d=\frac{2}{3}\cdot1.50=1.00\,\text{m}\), so the radius is \(r=0.50\,\text{m}\). 2. The volume is \(V=\pi r^2h=\pi\cdot0.50^2\cdot1.50\approx1.17810\,\text{m}^3\). 3. Since \(1\,\text{m}^3=1{,}000\,\text{L}\), the capacity is approximately \(1{,}178.10\,\text{L}\). 4. The total surface area is \(SA=2\pi r^2+2\pi rh=2\pi r(r+h)\). Thus, \(SA=2\pi\cdot0.50\cdot(0.50+1.50)=2\pi\approx6.28\,\text{m}^2\).

a) The maximum capacity is approximately \(1{,}178.10\,\text{L}\). b) The total exterior surface area is \(SA\approx6.28\,\text{m}^2\).

- How are cylinder volume, base area, and height related? - Which formula relates a circle's area to its radius? - How can circumference and height be used to find lateral surface area? - Use unrounded values in later calculations.

1. Use \(V=Bh\) to find the base area: \(B=V\div h=850\div12.5=68\,\text{cm}^2\). 2. Since \(B=\pi r^2\), \(r=\sqrt{\frac{B}{\pi}}=\sqrt{\frac{68}{\pi}}\approx4.6524\,\text{cm}\), so \(r\approx4.7\,\text{cm}\). 3. Use the unrounded radius for the lateral area: \(A_{\text{lat}}=2\pi rh=2\pi\sqrt{\frac{68}{\pi}}(12.5)\approx365.4007\,\text{cm}^2\), so \(A_{\text{lat}}\approx365.4\,\text{cm}^2\).

\(r\approx4.7\,\text{cm}\); \(B=68.0\,\text{cm}^2\); \(A_{\text{lat}}\approx365.4\,\text{cm}^2\)

- Rearrange the cone volume formula to isolate the height. - The lateral area requires the slant height. - Use the Pythagorean theorem with the radius and vertical height. - Round only after completing the calculations.

1. From \(V=\frac{1}{3}\pi r^2h\), solve for height: \(h=\frac{3V}{\pi r^2}=\frac{360}{\pi\cdot3.5^2}\approx9.3544\,\text{cm}\). 2. The slant height is \(\ell=\sqrt{r^2+h^2}\approx9.9877\,\text{cm}\). 3. The lateral surface area is \(A_{\text{lat}}=\pi r\ell\approx109.8210\,\text{cm}^2\).

a) \(h\approx9.35\,\text{cm}\) b) \(A_{\text{lat}}\approx109.82\,\text{cm}^2\)

- Rearrange the cone volume formula to solve for \(r^2\), then take a square root. - The total surface area includes the circular base and the curved lateral surface. - Find the slant height before calculating the lateral area. - Keep unrounded values through the final step.

1. From \(V=\frac{1}{3}\pi r^2h\), \(r^2=\frac{3V}{\pi h}=\frac{90}{\pi}\), so \(r=\sqrt{\frac{90}{\pi}}\approx5.3524\,\text{cm}\). 2. The slant height is \(\ell=\sqrt{r^2+h^2}=\sqrt{\frac{90}{\pi}+15^2}\approx15.9263\,\text{cm}\). 3. The base area is \(B=\pi r^2=90\,\text{cm}^2\). 4. The lateral area is \(A_{\text{lat}}=\pi r\ell\approx267.8008\,\text{cm}^2\). 5. The total surface area is \(SA=B+A_{\text{lat}}\approx357.8008\,\text{cm}^2\).

a) \(r\approx5.35\,\text{cm}\) b) \(SA\approx357.80\,\text{cm}^2\)

- A parallel cross-section creates a smaller similar pyramid. - Find the original slant height with the Pythagorean theorem. - Surface areas scale by the square of the linear scale factor. - Include both parallel bases in the frustum’s total surface area.

1. The slant height of the original pyramid is \(\sqrt{3^2+4^2}=5\,\text{cm}\). 2. The small pyramid has linear scale factor \(\frac{1}{2}\), so its base edge is \(4\,\text{cm}\) and its slant height is \(2.5\,\text{cm}\). 3. Its lateral area is \(4\left(\frac{1}{2}\cdot4\cdot2.5\right)=20\,\text{cm}^2\). 4. The original pyramid’s lateral area is \(4\left(\frac{1}{2}\cdot8\cdot5\right)=80\,\text{cm}^2\), so the frustum’s lateral area is \(80-20=60\,\text{cm}^2\). 5. Add the two square bases: \(60+8^2+4^2=60+64+16=140\,\text{cm}^2\).

Small pyramid lateral area: \(20\,\text{cm}^2\) Frustum total surface area: \(140\,\text{cm}^2\)

- Each lateral face is a trapezoid. - First find the slant height of a lateral face. - A right triangle formed by the vertical height and the horizontal offset can be used to find the slant height. - Convert the final area to square decimeters.

1. The horizontal offset on each side is \(\frac{35-15}{2}=10\,\text{cm}\). 2. The slant height of each trapezoidal face is \(\ell=\sqrt{24^2+10^2}=\sqrt{676}=26\,\text{cm}\). 3. One lateral face is a trapezoid with area \(A=\frac{35+15}{2}\cdot26=650\,\text{cm}^2\). 4. The four faces have total lateral area \(A_{\text{lat}}=4\cdot650=2{,}600\,\text{cm}^2\). 5. Since \(100\,\text{cm}^2=1\,\text{dm}^2\), \(2{,}600\,\text{cm}^2=26\,\text{dm}^2\).

The lampshade requires \(26\,\text{dm}^2\) of fabric.

- Only the lateral surface is included because both ends are open. - Use the vertical height and the difference of the radii to find the slant height. - Apply the Pythagorean theorem in the frustum's cross section.

1. The horizontal difference between the radii is \(35-15=20\,\text{cm}\). 2. The slant height is \(\ell=\sqrt{21^2+20^2}=\sqrt{841}=29\,\text{cm}\). 3. The lateral area is \(A_{\text{lat}}=\pi(r_1+r_2)\ell=\pi(35+15)\cdot29=1{,}450\pi\approx4{,}555.31\,\text{cm}^2\).

The lampshade requires approximately \(4{,}555.31\,\text{cm}^2\) of material.

- A regular tetrahedron has four congruent equilateral triangular faces. - Relate the area of one equilateral triangle to its side length. - Use the Pythagorean theorem or the standard equilateral-triangle altitude formula.

1. A regular tetrahedron has four equilateral triangular faces, so its total surface area is \(SA=4\left(\frac{\sqrt{3}}{4}a^2\right)=a^2\sqrt{3}\). 2. Set \(a^2\sqrt{3}=144\sqrt{3}\). Then \(a^2=144\), so \(a=12\,\text{cm}\). 3. The altitude of an equilateral triangle is \(h_a=\frac{\sqrt{3}}{2}a\). 4. Therefore, \(h_a=\frac{\sqrt{3}}{2}(12)=6\sqrt{3}\,\text{cm}\approx10.39\,\text{cm}\).

a) \(a=12\,\text{cm}\) b) \(h_a=6\sqrt{3}\,\text{cm}\approx10.39\,\text{cm}\)

- Begin with the area of one equilateral triangular face. - Account for the number of congruent faces. - Square the decimal edge length before multiplying by \(\sqrt{3}\).

1. An equilateral triangle with side length \(a\) has altitude \(\frac{\sqrt{3}}{2}a\). 2. Its area is \(A=\frac{1}{2}a\left(\frac{\sqrt{3}}{2}a\right)=\frac{\sqrt{3}}{4}a^2\). 3. A regular tetrahedron has four such faces, so \(SA=4\left(\frac{\sqrt{3}}{4}a^2\right)=a^2\sqrt{3}\). 4. For \(a=8.5\,\text{cm}\), \(SA=(8.5)^2\sqrt{3}=72.25\sqrt{3}\approx125.1407\,\text{cm}^2\). 5. Rounded to the nearest tenth, \(SA\approx125.1\,\text{cm}^2\).

a) \(SA=4\left(\frac{\sqrt{3}}{4}a^2\right)=a^2\sqrt{3}\) b) \(SA\approx125.1\,\text{cm}^2\)

- Convert the diameters to radii. - Convert cubic centimeters to liters after finding the volume. - Because the bucket is open, include the lateral surface and bottom only. - Use the height and radius difference to find the slant height.

1. The radii are \(14\,\text{cm}\) and \(10\,\text{cm}\). 2. The volume is \(V=\frac{1}{3}\pi\cdot24\left(14^2+14\cdot10+10^2\right)=3{,}488\pi\approx10{,}957.88\,\text{cm}^3\). 3. Therefore, the capacity is approximately \(10.9579\,\text{L}\), which rounds to \(11.0\,\text{L}\). 4. The slant height is \(\ell=\sqrt{24^2+(14-10)^2}=\sqrt{592}\approx24.3311\,\text{cm}\). 5. The lateral area is \(A_{\text{lat}}=\pi(14+10)\ell\approx1{,}834.52\,\text{cm}^2\), and the bottom area is \(B=\pi\cdot10^2\approx314.16\,\text{cm}^2\). 6. The exterior area is \(A_{\text{lat}}+B\approx2{,}148.68\,\text{cm}^2\).

a) \(V\approx11.0\,\text{L}\) b) \(A\approx2{,}148.68\,\text{cm}^2\)

- Identify the square base and the four triangular faces. - A pyramid has one-third the volume of a prism with the same base and height. - Use a right triangle formed by the vertical height and half a base edge to find the slant height. - Add the base area and all four triangular face areas.

1. The square base area is \(B=6^2=36\,\text{cm}^2\), so \(V=\frac{1}{3}Bh=\frac{1}{3}\cdot36\cdot4=48\,\text{cm}^3\). 2. The slant height is \(\ell=\sqrt{4^2+3^2}=5\,\text{cm}\). 3. The four triangular faces have total lateral area \(A_{\text{lat}}=4\left(\frac{1}{2}\cdot6\cdot5\right)=60\,\text{cm}^2\). 4. Therefore, \(SA=B+A_{\text{lat}}=36+60=96\,\text{cm}^2\).

a) \(V=48\,\text{cm}^3\) b) \(\ell=5\,\text{cm}\); \(SA=96\,\text{cm}^2\)

- Use the cylinder volume formula for each can. - A closed can's total surface area includes two circular bases and the lateral surface. - Compare the calculated areas, not only the dimensions.

1. For can A, \(V_A=\pi\cdot4^2\cdot10=160\pi\,\text{cm}^3\approx502.7\,\text{cm}^3\). 2. For can B, \(V_B=\pi\cdot5^2\cdot6.4=160\pi\,\text{cm}^3\approx502.7\,\text{cm}^3\). The cans have equal volumes. 3. Can A has total surface area \(SA_A=2\pi\cdot4^2+2\pi\cdot4\cdot10=112\pi\,\text{cm}^2\approx351.9\,\text{cm}^2\). 4. Can B has total surface area \(SA_B=2\pi\cdot5^2+2\pi\cdot5\cdot6.4=114\pi\,\text{cm}^2\approx358.1\,\text{cm}^2\). 5. Because \(SA_A<SA_B\), can A requires less material.

a) Both volumes are \(160\pi\,\text{cm}^3\approx502.7\,\text{cm}^3\). b) Can A requires less material. Its total surface area is about \(351.9\,\text{cm}^2\), compared with about \(358.1\,\text{cm}^2\) for can B.

- The label covers only the cylinder's lateral surface. - Convert the diameter to a radius. - Use the cylinder volume formula. - Convert cubic centimeters to liters.

1. The radius is \(r=10\div2=5\,\text{cm}\). 2. The label area is the lateral surface area: \(A_{\text{lat}}=2\pi rh=2\pi\cdot5\cdot12=120\pi\,\text{cm}^2\approx377.0\,\text{cm}^2\). 3. The volume is \(V=\pi r^2h=\pi\cdot5^2\cdot12=300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{L}\), the volume is about \(0.94\,\text{L}\).

a) \(377.0\,\text{cm}^2\) b) \(0.94\,\text{L}\)

- Use the rectangular base area in the pyramid volume formula. - The pyramid has two pairs of lateral faces with different slant heights. - For each slant height, use the vertical height and half of the other base dimension. - Include the base and all four triangular faces in the total surface area.

1. The base area is \(B=18\cdot10=180\,\text{cm}^2\), so \(V=\frac{1}{3}\cdot180\cdot12=720\,\text{cm}^3\). 2. For the faces with base \(18\,\text{cm}\), \(\ell_a=\sqrt{12^2+5^2}=13\,\text{cm}\). 3. For the faces with base \(10\,\text{cm}\), \(\ell_b=\sqrt{12^2+9^2}=15\,\text{cm}\). 4. The lateral area is \(A_{\text{lat}}=2\left(\frac{1}{2}\cdot18\cdot13\right)+2\left(\frac{1}{2}\cdot10\cdot15\right)=234+150=384\,\text{cm}^2\). Therefore, \(SA=180+384=564\,\text{cm}^2\).

a) \(V=720\,\text{cm}^3\) b) \(\ell_a=13\,\text{cm}\); \(\ell_b=15\,\text{cm}\) c) \(SA=564\,\text{cm}^2\)

- Use the cylinder formulas for volume and lateral surface area. - One cubic decimeter equals one liter. - The barrel is open, so include only one circular base in part c. - Round only the final results.

1. The maximum volume is \(V=\pi r^2h=\pi\cdot3^2\cdot8=72\pi\,\text{dm}^3\approx226.2\,\text{dm}^3\). Since \(1\,\text{dm}^3=1\,\text{L}\), the capacity is about \(226.2\,\text{L}\). 2. The lateral surface area is \(A_{\text{lat}}=2\pi rh=2\pi\cdot3\cdot8=48\pi\,\text{dm}^2\approx150.8\,\text{dm}^2\). 3. Because the barrel has no top, the material area is one circular base plus the lateral surface: \(A=\pi\cdot3^2+48\pi=57\pi\,\text{dm}^2\approx179.1\,\text{dm}^2\).

a) \(226.2\,\text{L}\) b) \(150.8\,\text{dm}^2\) c) \(179.1\,\text{dm}^2\)

- The lateral surface consists of the faces that are not the two triangular bases. - You need all three side lengths of the triangular base. - Use the Pythagorean theorem to find the missing side.

1. Find the hypotenuse of the triangular base: \(c=\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{cm}\). 2. The perimeter of the base is \(P=6+8+10=24\,\text{cm}\). 3. Let the prism length be \(L=12\,\text{cm}\). The lateral surface area of a right prism is the base perimeter times the prism length: \(A_{\text{lat}}=PL=24\cdot12=288\,\text{cm}^2\).

The lateral surface area is \(288\,\text{cm}^2\).

- In the net, which sector length becomes the cone's slant height? - The sector's curved edge must equal the circumference of the circular base. - Compare that arc length with the circumference of a full circle whose radius is the sector radius. - Use the same fraction of the full circle's area to get the sector area.

1. The net shows base radius \(r=5\,\text{cm}\) and sector radius \(\ell=13\,\text{cm}\). The sector radius becomes the cone's slant height. 2. The sector's arc becomes the circumference of the cone's base, so its arc length is \(2\pi(5)=10\pi\,\text{cm}\). 3. A full circle of radius \(13\,\text{cm}\) has circumference \(2\pi(13)=26\pi\,\text{cm}\). Therefore, the sector is \(\frac{10\pi}{26\pi}=\frac{5}{13}\) of a full circle. 4. The sector area is \(\frac{5}{13}\pi(13)^2=65\pi\,\text{cm}^2\). 5. The base area is \(\pi(5)^2=25\pi\,\text{cm}^2\), so \(SA=65\pi+25\pi=90\pi\,\text{cm}^2\).

The sector is \(\frac{5}{13}\) of a full circle, and the cone's exact total surface area is \(90\pi\,\text{cm}^2\).

- Pay attention to the direction of the fold lines in the strip. - Track how the orientation of each successive square changes as you imagine folding. - Check whether six distinct cube-face positions can be reached without overlap.

1. Every shared fold edge in the straight strip is parallel to the others. 2. After folding through right angles, every square still contains a direction parallel to those hinge edges. The squares can wrap around the four side-face positions of a cube, but none can turn into either end face. 3. Once four distinct side positions have been used, another square must repeat an already occupied face position instead of creating a new cube face. 4. Therefore, the six-square strip is not a net of a cube.

No. Because all fold edges are parallel, the strip can only wrap around side-face positions; it cannot form the two end faces. After four distinct side positions, another square must repeat an occupied position.

- First reject any design that cannot hold enough volume. - Use the areas of all six faces to compare cardboard use. - A smaller surface area is not enough if the box fails the capacity test.

1. The volumes are \(V_A=8\cdot10\cdot12=960\,\text{cm}^3\), \(V_B=9\cdot9\cdot13=1053\,\text{cm}^3\), and \(V_C=7\cdot12\cdot12=1008\,\text{cm}^3\). Only B and C meet the capacity requirement. 2. For a closed rectangular prism, \(A=2(lw+lh+wh)\). 3. Design A has surface area \(2(80+96+120)=592\,\text{cm}^2\), which is the smallest area but does not meet the capacity requirement. 4. Design B has area \(2(81+117+117)=630\,\text{cm}^2\). 5. Design C has area \(2(84+84+144)=624\,\text{cm}^2\). 6. C uses the least cardboard among the qualifying designs.

Designs B and C meet the capacity requirement. Design C is best, with surface area \(624\,\text{cm}^2\); A has less area but is too small.

- Which pieces of the net have different shapes? - The vertical height is not the height of a sloping triangular face. How could you work out the latter? - How many copies of each type of face need to be counted?

1. The square base is \(6\cdot6=36\,\text{cm}^2\). 2. The slant height of each triangular face is \(\sqrt{4^2+3^2}=5\,\text{cm}\). 3. Each triangular face has area \(\frac{1}{2}\cdot6\cdot5=15\,\text{cm}^2\), and four faces contribute \(60\,\text{cm}^2\). 4. The total surface area is \(36+60=96\,\text{cm}^2\).

One triangular face: \(15\,\text{cm}^2\). Total net area: \(96\,\text{cm}^2\), from four triangles and one square.

- Think about which part of the folded cone each piece of paper covers. - The cone's vertical height does not run along its sloping side. How could you find the length that does? - Which surface would still need covering after the curved side was covered?

1. The radius is \(3\,\text{cm}\), the perpendicular height is \(4\,\text{cm}\), and the slant height is \(\sqrt{3^2+4^2}=5\,\text{cm}\). 2. The curved surface unfolds to a sector with area \(\pi r\ell=\pi\cdot3\cdot5=15\pi\,\text{cm}^2\). 3. The circular base adds \(\pi r^2=9\pi\,\text{cm}^2\). 4. Total: \(24\pi\,\text{cm}^2\).

Curved sector: \(15\pi\,\text{cm}^2\). Circular base: \(9\pi\,\text{cm}^2\). Total: \(24\pi\,\text{cm}^2\).

- Which parts of the two halves were already on the surface of the full sphere? - A cut through the center creates a flat circle on each half. - Count the new surfaces without counting the curved outside twice.

1. The curved surfaces of the two halves together make the original sphere's area: \(4\pi(5)^2=100\pi\,\text{cm}^2\). 2. Cutting exposes one circular face on each half, each with area \(\pi(5)^2=25\pi\,\text{cm}^2\). 3. The combined area is \(100\pi+2(25\pi)=150\pi\,\text{cm}^2\). 4. The increase is \(150\pi-100\pi=50\pi\,\text{cm}^2\).

Total area: \(150\pi\,\text{cm}^2\). Increase: \(50\pi\,\text{cm}^2\).

- Let one radius be unknown. How is the other radius related to it? - Write the difference of the two sphere areas before simplifying. - Check that the resulting radii are positive and differ by the given amount.

1. Let the smaller radius be \(r\,\text{cm}\). The larger radius is \(r+3\,\text{cm}\). 2. The area difference is \(4\pi((r+3)^2-r^2)=96\pi\). 3. Divide by \(4\pi\): \(6r+9=24\), so \(r=2.5\,\text{cm}\) and \(r+3=5.5\,\text{cm}\). 4. Check: \(4\pi(5.5)^2-4\pi(2.5)^2=4\pi(30.25-6.25)=96\pi\,\text{cm}^2\).

Smaller radius: \(2.5\,\text{cm}\). Larger radius: \(5.5\,\text{cm}\).

- From directly above, what total horizontal area is visible? - Which surfaces become hidden when the disks are stacked? - List the radii before adding the lateral areas.

1. Each disk has surface area \(SA_i=2\pi r_i^2+2\pi r_i\cdot1\). 2. The sum for the separate disks is \(2\pi[(5^2+5)+(6^2+6)+(7^2+7)+(8^2+8)+(9^2+9)]\,\text{cm}^2=580\pi\,\text{cm}^2\approx1822.12\,\text{cm}^2\). 3. In the stack, the visible horizontal surfaces on top combine to the area of the largest disk, \(81\pi\,\text{cm}^2\). The visible bottom also has area \(81\pi\,\text{cm}^2\). 4. All five lateral surfaces remain visible and have total area \(2\pi\cdot1\cdot(5+6+7+8+9)\,\text{cm}^2=70\pi\,\text{cm}^2\). 5. The visible surface area is \(81\pi+81\pi+70\pi=232\pi\,\text{cm}^2\approx728.85\,\text{cm}^2\).

a) \(580\pi\,\text{cm}^2\approx1822.12\,\text{cm}^2\) b) \(232\pi\,\text{cm}^2\approx728.85\,\text{cm}^2\)

- What fraction of the cylinder remains after a \(60^\circ\) wedge is removed? - Which surfaces were already exposed, and which new surfaces are created by the cuts? - What shapes are the two cut faces?

1. The radius is \(r=12\,\text{cm}\). Removing \(60^\circ\) leaves \(300^\circ\), or \(\frac{5}{6}\), of the cylinder. 2. The remaining volume is \(V=\frac{5}{6}\pi\cdot12^2\cdot10\,\text{cm}^3=1200\pi\,\text{cm}^3\approx3769.91\,\text{cm}^3\). 3. The combined top and bottom area is \(2\cdot\frac{5}{6}\pi\cdot12^2\,\text{cm}^2=240\pi\,\text{cm}^2\). 4. The remaining curved lateral area is \(\frac{5}{6}\cdot2\pi\cdot12\cdot10\,\text{cm}^2=200\pi\,\text{cm}^2\). 5. The two rectangular cut faces have combined area \(2\cdot12\cdot10\,\text{cm}^2=240\,\text{cm}^2\). 6. The total surface area is \(240\pi+200\pi+240\,\text{cm}^2=440\pi+240\,\text{cm}^2\approx1622.30\,\text{cm}^2\).

a) \(1200\pi\,\text{cm}^3\approx3769.91\,\text{cm}^3\) b) \((440\pi+240)\,\text{cm}^2\approx1622.30\,\text{cm}^2\)

- Solve the cylinder volume formula for height. - Total surface area includes two circles and the lateral rectangle. - Compare the unrounded surface areas before choosing the design for part c. - Multiply by the number of cans, include the waste percentage, and then convert square centimeters to square meters.

1. Since \(425\,\text{mL}=425\,\text{cm}^3\), \(h_A=\frac{425}{\pi(3.5)^2}\,\text{cm}\approx11.04\,\text{cm}\), and \(h_B=\frac{425}{\pi(4.2)^2}\,\text{cm}\approx7.67\,\text{cm}\). 2. Using \(SA=2\pi r^2+2\pi rh\), can A has \(SA_A\approx319.83\,\text{cm}^2\), and can B has \(SA_B\approx313.22\,\text{cm}^2\). Can B uses less metal. 3. Using the unrounded surface area for can B, the metal for \(1000\) cans including \(15\%\) waste is \(1000\cdot SA_B\cdot1.15\approx360{,}198.79\,\text{cm}^2\). Since \(10{,}000\,\text{cm}^2=1\,\text{m}^2\), this is approximately \(36.02\,\text{m}^2\).

a) \(h_A\approx11.04\,\text{cm}\); \(h_B\approx7.67\,\text{cm}\) b) \(SA_A\approx319.83\,\text{cm}^2\); \(SA_B\approx313.22\,\text{cm}^2\) c) About \(36.02\,\text{m}^2\) of metal for can B

- Identify which original surfaces change when the hole is drilled. - Find the surface area of the solid cylinder first. - The wall of the hole is a new lateral surface. - Use \(\frac{\text{new}-\text{original}}{\text{original}}\cdot100\%\) for percent increase.

1. The original cylinder has surface area \(SA_{\text{old}}=2\pi r^2+2\pi rh=2\pi\cdot6^2+2\pi\cdot6\cdot15=252\pi\,\text{cm}^2\). 2. After drilling, the two circular faces become annuli with total area \(2\pi(6^2-2^2)=64\pi\,\text{cm}^2\). 3. The outside lateral area remains \(2\pi\cdot6\cdot15=180\pi\,\text{cm}^2\), and the hole adds an inside lateral area of \(2\pi\cdot2\cdot15=60\pi\,\text{cm}^2\). 4. The new total surface area is \(SA_{\text{new}}=64\pi+180\pi+60\pi=304\pi\,\text{cm}^2\). 5. The percent increase is \(\frac{304\pi-252\pi}{252\pi}\cdot100\%\approx20.6\%\).

The total surface area increases by \(20.6\%\).

- Subtract the cylindrical hole from the prism when finding volume. - For surface area, include both drilled square faces, the four outside rectangular faces, and the inside wall of the hole. - Convert the hole's diameter to its radius.

1. The square cross-sectional area is \(10^2=100\,\text{cm}^2\). The hole has radius \(3\,\text{cm}\) and cross-sectional area \(9\pi\,\text{cm}^2\). 2. The remaining volume is \(V=(100-9\pi)\cdot20=2000-180\pi\,\text{cm}^3\approx1434.51\,\text{cm}^3\). 3. The two drilled square faces have total area \(2\cdot(100-9\pi)\,\text{cm}^2\). 4. The four outside rectangular faces have total area \(4\cdot10\cdot20=800\,\text{cm}^2\), and the inside wall of the hole has area \(2\pi\cdot3\cdot20=120\pi\,\text{cm}^2\). 5. Therefore, \(SA=2\cdot(100-9\pi)+800+120\pi=1000+102\pi\,\text{cm}^2\approx1320.44\,\text{cm}^2\).

a) \(2000-180\pi\,\text{cm}^3\approx1434.51\,\text{cm}^3\) b) \(SA=1000+102\pi\,\text{cm}^2\approx1320.44\,\text{cm}^2\)

- Which parts make up the total surface area of a closed cylinder? - Which of those areas can you calculate directly from the radius? - Once you know the lateral area, how can you find the height? - Which measurements are needed to calculate the volume?

1. The area of one circular base is \(B=\pi r^2=\pi\cdot4.2^2\approx55.4177\,\text{cm}^2\). 2. The lateral area is \(A_{\text{lat}}=SA-2B=540-2\pi\cdot4.2^2\approx429.1646\,\text{cm}^2\). 3. Since \(A_{\text{lat}}=2\pi rh\), \(h=\frac{A_{\text{lat}}}{2\pi r}=\frac{540-2\pi\cdot4.2^2}{2\pi\cdot4.2}\approx16.2628\,\text{cm}\). Thus, \(h\approx16.3\,\text{cm}\). 4. The volume is \(V=\pi r^2h\approx\pi\cdot4.2^2\cdot16.2628\approx901.2457\,\text{cm}^3\). Thus, \(V\approx901.2\,\text{cm}^3\).

\(h\approx16.3\,\text{cm}\); \(V\approx901.2\,\text{cm}^3\)

- Use the ratio of the cross-sectional area to the base area to find the linear scale factor. - Distinguish the distance from the apex from the distance from the base. - Use a right triangle containing the vertical height and half a base edge to find the slant height. - The total surface area includes the square base and four triangular faces.

1. The original base area is \(B=12^2=144\,\text{cm}^2\). 2. Let \(k\) be the linear scale factor from the original pyramid to the smaller pyramid above the cut. Since areas scale by \(k^2\), \(k^2=36\div144=\frac{1}{4}\), so \(k=\frac{1}{2}\). 3. The smaller pyramid's height is \(\frac{1}{2}\cdot8=4\,\text{cm}\). Therefore, the cross section is \(8-4=4\,\text{cm}\) above the original base. 4. The original pyramid's slant height is \(\ell=\sqrt{8^2+6^2}=10\,\text{cm}\). 5. Its lateral area is \(A_{\text{lat}}=4\left(\frac{1}{2}\cdot12\cdot10\right)=240\,\text{cm}^2\). 6. The total surface area is \(SA=144+240=384\,\text{cm}^2\).

The cross section is \(4\,\text{cm}\) from the base. The original pyramid's total surface area is \(384\,\text{cm}^2\).

- Convert the volume into cubic centimeters before finding the height. - How can the two square base areas and the unknown perpendicular height give the volume? - Each sloping face is a trapezoid. What length is its perpendicular height within that face? - The two base faces are not painted.

1. Convert the volume: \(28\,\text{dm}^3=28{,}000\,\text{cm}^3\). 2. Let \(a=40\,\text{cm}\) and \(b=20\,\text{cm}\) be the lower and upper square edges. For a square-pyramid frustum, \(V=\frac{h}{3}(a^2+ab+b^2)\). Thus, \(28{,}000=\frac{h}{3}(40^2+40\cdot20+20^2)=\frac{2800h}{3}\). 3. Solve for the perpendicular height: \(h=30\,\text{cm}\). 4. The horizontal offset on each side face is \(\frac{40-20}{2}=10\,\text{cm}\). Its slant height is \(\ell=\sqrt{30^2+10^2}=10\sqrt{10}\,\text{cm}\). 5. Each sloping face is a trapezoid of area \(\frac{40+20}{2}\ell=30\ell\). All four total \(120\ell=1200\sqrt{10}\,\text{cm}^2\approx3794.7\,\text{cm}^2\).

a) \(h=30\,\text{cm}\) b) \(3794.7\,\text{cm}^2\) (approximately)

- Use similarity to find the dimensions of the smaller pyramid. - Areas scale by the square of the linear scale factor. - The frustum's lateral area can be found by subtracting the small pyramid's lateral area from the original pyramid's lateral area. - Include the cut face in the surface area of both resulting solids.

1. The small pyramid has linear scale factor \(k=6\div8=\frac{3}{4}\). 2. Its base edge is \(\frac{3}{4}\cdot12=9\,\text{cm}\), so the cross-sectional area is \(9^2=81\,\text{cm}^2\). 3. The original pyramid's slant height is \(\ell=\sqrt{8^2+6^2}=10\,\text{cm}\). 4. The small pyramid's slant height is \(\frac{3}{4}\cdot10=7.5\,\text{cm}\). 5. The small pyramid's total surface area is \(81+4\left(\frac{1}{2}\cdot9\cdot7.5\right)=81+135=216\,\text{cm}^2\). 6. The original lateral area is \(4\left(\frac{1}{2}\cdot12\cdot10\right)=240\,\text{cm}^2\), and the small pyramid's lateral area is \(135\,\text{cm}^2\). Thus, the frustum's lateral area is \(105\,\text{cm}^2\). 7. The frustum's total surface area is \(12^2+9^2+105=144+81+105=330\,\text{cm}^2\).

Cross-sectional area: \(81\,\text{cm}^2\) Small pyramid surface area: \(216\,\text{cm}^2\) Frustum surface area: \(330\,\text{cm}^2\)

- Find the side length of the square cross section from its area. - Use the ratio of corresponding edge lengths to find the smaller pyramid's scale factor. - Distinguish the distance from the apex from the height above the base. - Include both square bases and the four trapezoidal lateral faces in the frustum's surface area.

1. The cross-section edge length is \(\sqrt{36}=6\,\text{cm}\). 2. The small pyramid above the cut has linear scale factor \(k=6\div18=\frac{1}{3}\). 3. Its height is \(\frac{1}{3}\cdot12=4\,\text{cm}\), so the cut is \(12-4=8\,\text{cm}\) above the original base. 4. The original pyramid's slant height is \(\ell=\sqrt{12^2+9^2}=15\,\text{cm}\). 5. The small pyramid's slant height is \(\frac{1}{3}\cdot15=5\,\text{cm}\). 6. The frustum's lateral area is \(4\left(\frac{1}{2}\cdot18\cdot15\right)-4\left(\frac{1}{2}\cdot6\cdot5\right)=540-60=480\,\text{cm}^2\). 7. The total surface area is \(18^2+6^2+480=840\,\text{cm}^2\).

The cut is \(8\,\text{cm}\) above the base. The frustum's total surface area is \(840\,\text{cm}^2\).

- Solve the cone-frustum volume formula for the height. - In a cross section, use the height and the difference of the radii to find the slant height. - Include both circular bases and the curved lateral surface in the total surface area.

1. Use \(V=\frac{1}{3}\pi h(r_1^2+r_1r_2+r_2^2)\): \(65\pi=\frac{1}{3}\pi h(25+10+4)=13\pi h\), so \(h=5\,\text{cm}\). 2. The slant height is \(\ell=\sqrt{h^2+(r_1-r_2)^2}=\sqrt{5^2+3^2}=\sqrt{34}\approx5.83\,\text{cm}\). 3. The total surface area is the two circular bases plus the lateral area: \(SA=\pi r_1^2+\pi r_2^2+\pi(r_1+r_2)\ell\). 4. Thus, \(SA=\pi\left(25+4+7\sqrt{34}\right)\approx219.34\,\text{cm}^2\).

a) \(h=5\,\text{cm}\) b) \(\ell\approx5.83\,\text{cm}\) c) \(SA\approx219.34\,\text{cm}^2\)

- Write formulas for lateral surface area and the area of one base. - Translate the stated relationship into an equation. - Simplify the equation before substituting the height. - After finding the radius, use the volume and total surface area formulas.

1. The lateral surface area is \(A_{\text{lat}}=2\pi rh\), and one base has area \(B=\pi r^2\). Since \(A_{\text{lat}}=3B\), \(2\pi rh=3\pi r^2\). 2. Because \(r>0\), divide by \(\pi r\) to get \(2h=3r\). With \(h=15\), \(r=10\,\text{cm}\). 3. The volume is \(V=\pi\cdot10^2\cdot15=1500\pi\,\text{cm}^3\approx4712.4\,\text{cm}^3\). 4. The total surface area is \(SA=2\pi\cdot10^2+2\pi\cdot10\cdot15=500\pi\,\text{cm}^2\approx1570.8\,\text{cm}^2\).

a) \(10\,\text{cm}\) b) \(4712.4\,\text{cm}^3\) c) \(1570.8\,\text{cm}^2\)

- Replace the height in the volume formula with an expression involving the radius. - Isolate \(r^3\), then take a cube root. - Total surface area includes two bases and the lateral surface. - Round only the decimal approximation.

1. Since \(h=2r\), the volume equation becomes \(250\pi=\pi r^2(2r)=2\pi r^3\). Divide by \(2\pi\): \(r^3=125\), so \(r=5\,\text{cm}\). 2. The height is \(h=2\cdot5=10\,\text{cm}\). The total surface area is \(SA=2\pi r^2+2\pi rh=2\pi\cdot5^2+2\pi\cdot5\cdot10=150\pi\,\text{cm}^2\approx471.24\,\text{cm}^2\).

1) \(r=5\,\text{cm}\) 2) \(SA=150\pi\,\text{cm}^2\approx471.24\,\text{cm}^2\)

- Express total surface area as the two square bases plus four rectangular side faces. - After isolating \(h\), use positive dimensions to determine the physically meaningful domain. - Substitute the height expression into \(V=a^2h\) before evaluating either requested base length.

1. The surface area equation is \(1000=2a^2+4ah\). Solving for height gives \(h(a)=\frac{250}{a}-0.5a\). 2. A physical prism requires \(a>0\) and \(h(a)>0\). Thus \(\frac{250}{a}-0.5a>0\), which gives \(a^2<500\). Therefore \(0<a<10\sqrt{5}\). 3. Since \(V=a^2h\), substitute the height expression: \(V(a)=a^2\left(\frac{250}{a}-0.5a\right)=250a-0.5a^3\). 4. Evaluate: \(V(5)=1187.5\,\text{cm}^3\) and \(V(15)=2062.5\,\text{cm}^3\). The prism with \(a=15\,\text{cm}\) has the greater volume.

a) \(h(a)=\frac{250}{a}-0.5a\), for \(0<a<10\sqrt{5}\) b) \(V(a)=250a-0.5a^3\) c) \(V(5)=1187.5\,\text{cm}^3\); \(V(15)=2062.5\,\text{cm}^3\). The prism with \(a=15\,\text{cm}\) has the greater volume.

- For one cut through the center, how many new flat faces appear? - Each cutting plane meets the sphere in a circle. What happens where two of the planes meet? - Use the given area of the newly exposed faces before adding the original outer area.

1. Each plane passes through the center and cuts out a disk of radius \(r\). The two sides of that cut give area \(2\pi r^2\). 2. Three perpendicular planes produce \(3\cdot2\pi r^2=6\pi r^2\) of new faces. Their intersections are lines, which have zero area, so no surface area is double-counted. 3. Thus \(6\pi r^2=216\pi\), giving \(r^2=36\) and the positive radius \(r=6\,\text{cm}\). 4. The original outside area is \(4\pi(6)^2=144\pi\,\text{cm}^2\). 5. The total painted area is \(144\pi+216\pi=360\pi\,\text{cm}^2\).

Radius: \(6\,\text{cm}\). Total painted area: \(360\pi\,\text{cm}^2\). Each cut adds two disk faces; their intersections have no area.