Free Math Worksheets

- 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: \(C=\pi d=\pi(4.50\,\text{m})\approx14.14\,\text{m}\). 2. Convert the height: \(12\,\text{cm}=0.12\,\text{m}\). 3. When laid flat, the strip is a rectangle. Its area is \(A=Ch\approx14.137\,\text{m}\cdot0.12\,\text{m}\approx1.70\,\text{m}^2\).

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

- 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) About \(9.55\,\text{cm}\)

- Include the curved lateral surface and only the closed circular base. - Use both radii in the cone-frustum lateral-area formula. - Add the area of the smaller circle after finding the lateral area.

1. The cone-frustum lateral area is \(L=\pi(r_1+r_2)s\). 2. Substituting gives \(L=\pi\left(15+9\right)\cdot10=240\pi\,\text{cm}^2\). 3. The smaller circular base has area \(B=\pi\cdot9^2=81\pi\,\text{cm}^2\). 4. The total exterior area is \(A=240\pi+81\pi=321\pi\approx1{,}008.45\,\text{cm}^2\).

The exterior area is approximately \(1{,}008.45\,\text{cm}^2\).

- 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 \(S=200\pi+100\pi=300\pi\,\text{cm}^2\).

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

- 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: \(S=2B+L\). 2. One base has area \(B=\pi r^2=\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The lateral area is \(L=2\pi rh=2\pi\cdot4\cdot10=80\pi\,\text{cm}^2\). 4. Therefore, \(S=2\cdot16\pi+80\pi=112\pi\approx351.8584\,\text{cm}^2\). 5. Rounded to the nearest tenth, \(S\approx351.9\,\text{cm}^2\).

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

- 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. - Multiply the area for one tree by the number of trees.

1. The trunk's circumference is \(C=\pi\cdot16\,\text{cm}\approx50.27\,\text{cm}\). 2. Including the overlap, one piece must be \(50.27\,\text{cm}+15\,\text{cm}\approx65.27\,\text{cm}\) long, or about \(0.6527\,\text{m}\). 3. The area for one tree is \(0.6527\,\text{m}\cdot1.50\,\text{m}\approx0.9791\,\text{m}^2\). 4. For \(8\) trees, the total area is \(8\cdot0.9791\,\text{m}^2\approx7.83\,\text{m}^2\).

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

- 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 \(L=2\pi rh=2\pi\cdot0.30\cdot4.00\,\text{m}^2=2.4\pi\,\text{m}^2\approx7.54\,\text{m}^2\). 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 number of cans is \(30.16\div12\approx2.51\). Since whole cans are required, at least \(3\) cans are needed.

a) About \(30.16\,\text{m}^2\) 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.

1. Since \(1\,\text{mL}=1\,\text{cm}^3\), the volume is \(425\,\text{cm}^3\). 2. From \(V=\pi r^2h\), \(r=\sqrt{\frac{V}{\pi h}}=\sqrt{\frac{425}{10\pi}}\,\text{cm}\approx3.678\,\text{cm}\). 3. The diameter is \(d=2r\approx7.356\,\text{cm}\), which rounds to \(7.4\,\text{cm}\). 4. Using the unrounded diameter, the lateral area is \(L=\pi dh=\pi\left(2\sqrt{\frac{425}{10\pi}}\right)\cdot10\,\text{cm}^2\approx231.10\,\text{cm}^2\).

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

- Which cylinder surfaces are included in the painted area? - Convert diameter to radius first. - Add the lateral area and one circular base area. - Divide the total area by the coverage per liter.

1. The radius is \(r=1.20\,\text{m}\). 2. The lateral area is \(L=2\pi rh=2\pi\cdot1.20\cdot3.50\,\text{m}^2\approx26.389\,\text{m}^2\). 3. The top area is \(G=\pi(1.20)^2\,\text{m}^2\approx4.524\,\text{m}^2\). 4. The total painted area is \(A=L+G\approx30.913\,\text{m}^2\). 5. The coating needed is \(\frac{30.913\,\text{m}^2}{8.5\,\text{m}^2/\text{L}}\approx3.637\,\text{L}\), which rounds to \(3.6\,\text{L}\).

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 \(L_A=2\pi r_A\cdot40\,\text{cm}\approx5013.26\,\text{cm}^2\). 4. For container B, \(r_B=\sqrt{\frac{50{,}000}{60\pi}}\,\text{cm}\approx16.287\,\text{cm}\). 5. Its lateral area is \(L_B=2\pi r_B\cdot60\,\text{cm}\approx6139.96\,\text{cm}^2\). 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 \(S_A=2\pi\cdot5^2+2\pi\cdot5\cdot\left(\frac{40}{\pi}\right)\,\text{cm}^2=50\pi+400\,\text{cm}^2\approx557.08\,\text{cm}^2\). 4. For model B, \(1000=10^2h_B\), so \(h_B=10\,\text{cm}\). 5. Its surface area is \(S_B=2\cdot10^2+4\cdot10\cdot10\,\text{cm}^2=600\,\text{cm}^2\). 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. \(L_1=2\pi\cdot4\cdot10\,\text{cm}^2=80\pi\,\text{cm}^2\). 3. \(L_2=2\pi\cdot8\cdot2.5\,\text{cm}^2=40\pi\,\text{cm}^2\). 4. Therefore, \(L_1>L_2\), and \(L_1:L_2=80\pi:40\pi=2:1\).

Cylinder \(C_1\) has the greater lateral area, and \(L_1:L_2=2:1\).

- 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 \(L=2\pi\cdot0.40\cdot1.0\,\text{m}^2=0.8\pi\,\text{m}^2\approx2.51\,\text{m}^2\). 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) About \(502.7\,\text{L}\) b) About \(2.51\,\text{m}^2\) c) About \(3.02\,\text{m}^2\)

- 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: \(S=2\pi(R^2-r^2)+2\pi Rh+2\pi rh\). 4. Thus, \(S=2\pi\cdot5+2\pi\cdot3\cdot8+2\pi\cdot2\cdot8\,\text{cm}^2=90\pi\,\text{cm}^2\approx282.74\,\text{cm}^2\).

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 \(S=2\pi r^2+2\pi rh=2\pi\cdot5^2+2\pi\cdot5\cdot12=170\pi\,\text{cm}^2\approx534.07\,\text{cm}^2\). 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 surface area is \(S_{\text{new}}=2\pi\cdot2.5^2+2\pi\cdot2.5\cdot24=132.5\pi\,\text{cm}^2\approx416.26\,\text{cm}^2\), so the surface area decreases.

a) \(300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\) b) \(170\pi\,\text{cm}^2\approx534.07\,\text{cm}^2\) 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 \(L=2\pi rh\). 4. For \(Z_1\), \(L_1=2\pi\cdot4\cdot9=72\pi\,\text{cm}^2\approx226.19\,\text{cm}^2\). 5. For \(Z_2\), \(L_2=2\pi\cdot6\cdot4=48\pi\,\text{cm}^2\approx150.80\,\text{cm}^2\). 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 \(S=2\pi r^2+2\pi rh\). 3. Substitute the values: \(S=2\pi\cdot4.5^2+2\pi\cdot4.5\left(\frac{800}{\pi\cdot4.5^2}\right)\approx482.7901\,\text{cm}^2\). 4. Rounded to the nearest hundredth, \(S\approx482.79\,\text{cm}^2\).

The total surface area is approximately \(482.79\,\text{cm}^2\).

- 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 \(S=2\pi r^2+2\pi rh=2\pi r(r+h)\). Thus, \(S=2\pi\cdot0.50\cdot(0.50+1.50)=2\pi\approx6.28\,\text{m}^2\).

a) The maximum capacity is approximately \(1{,}178.10\,\text{L}\). b) The total exterior surface area is approximately \(6.28\,\text{m}^2\).

- 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: \(L=2\pi rh=2\pi\sqrt{\frac{68}{\pi}}(12.5)\approx365.4007\,\text{cm}^2\), so \(L\approx365.4\,\text{cm}^2\).

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

- 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 \(s=\sqrt{r^2+h^2}=\sqrt{3.5^2+9.3544^2}\approx9.9877\,\text{cm}\). 3. The lateral surface area is \(L=\pi rs=\pi\cdot3.5\cdot9.9877\approx109.8210\,\text{cm}^2\).

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

- 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 \(s=\sqrt{r^2+h^2}=\sqrt{\frac{90}{\pi}+15^2}\approx15.9263\,\text{cm}\). 3. The base area is \(B=\pi r^2=90\,\text{cm}^2\). 4. The lateral area is \(L=\pi rs\approx267.8008\,\text{cm}^2\). 5. The total surface area is \(S=B+L\approx357.8008\,\text{cm}^2\).

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

- 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 \(l=\sqrt{24^2+10^2}=\sqrt{676}=26\,\text{cm}\). 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 area \(L=4\cdot650=2{,}600\,\text{cm}^2\). 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 \(s=\sqrt{21^2+20^2}=\sqrt{841}=29\,\text{cm}\). 3. The lateral area is \(L=\pi(r_1+r_2)s=\pi\left(35+15\right)\cdot29=1{,}450\pi\approx4{,}555.31\,\text{cm}^2\).

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 \(S=4\left(\frac{\sqrt{3}}{4}a^2\right)=a^2\sqrt{3}\). 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 \(S=4\left(\frac{\sqrt{3}}{4}a^2\right)=a^2\sqrt{3}\). 4. For \(a=8.5\,\text{cm}\), \(S=(8.5)^2\sqrt{3}=72.25\sqrt{3}\approx125.1407\,\text{cm}^2\). 5. Rounded to the nearest tenth, \(S\approx125.1\,\text{cm}^2\).

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

- 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 \(s=\sqrt{h^2+(r_1-r_2)^2}=\sqrt{5^2+3^2}=\sqrt{34}\approx5.83\,\text{cm}\). 3. The total surface area is the two circular bases plus the lateral area: \(S=\pi r_1^2+\pi r_2^2+\pi(r_1+r_2)s\). 4. Thus, \(S=\pi\left(25+4+7\sqrt{34}\right)\approx219.34\,\text{cm}^2\).

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

- 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 \(s=\sqrt{24^2+(14-10)^2}=\sqrt{592}\approx24.3311\,\text{cm}\). 5. The lateral area is \(L=\pi\left(14+10\right)s\approx1{,}834.52\,\text{cm}^2\), and the bottom area is \(B=\pi\cdot10^2\approx314.16\,\text{cm}^2\). 6. The exterior area is \(L+B\approx2{,}148.68\,\text{cm}^2\).

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 \(B=6^2=36\,\text{cm}^2\). 2. The volume is \(V=\frac{1}{3}Bh=\frac{1}{3}\cdot36\cdot4=48\,\text{cm}^3\). 3. The slant height is \(l=\sqrt{4^2+3^2}=5\,\text{cm}\). 4. The four triangular faces have total area \(L=4\left(\frac{1}{2}\cdot6\cdot5\right)=60\,\text{cm}^2\). 5. The total surface area is \(S=B+L=36+60=96\,\text{cm}^2\).

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

- 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 \(S_A=2\pi\cdot4^2+2\pi\cdot4\cdot10=112\pi\,\text{cm}^2\approx351.9\,\text{cm}^2\). 4. Can B has total surface area \(S_B=2\pi\cdot5^2+2\pi\cdot5\cdot6.4=114\pi\,\text{cm}^2\approx358.1\,\text{cm}^2\). 5. Because \(S_A<S_B\), 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: \(L=2\pi rh=2\pi\cdot5\cdot12=120\pi\,\text{cm}^2\approx377.0\,\text{cm}^2\). 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 rectangular base area is \(B=18\cdot10=180\,\text{cm}^2\). 2. The volume is \(V=\frac{1}{3}\cdot180\cdot12=720\,\text{cm}^3\). 3. For the faces with base \(18\,\text{cm}\), \(l_a=\sqrt{12^2+5^2}=13\,\text{cm}\). 4. For the faces with base \(10\,\text{cm}\), \(l_b=\sqrt{12^2+9^2}=15\,\text{cm}\). 5. The lateral area is \(L=2\left(\frac{1}{2}\cdot18\cdot13\right)+2\left(\frac{1}{2}\cdot10\cdot15\right)=234+150=384\,\text{cm}^2\). 6. The total surface area is \(S=180+384=564\,\text{cm}^2\).

a) \(V=720\,\text{cm}^3\) b) \(l_a=13\,\text{cm}\); \(l_b=15\,\text{cm}\) c) \(S=564\,\text{cm}^2\)

- 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 \(L=2\pi rh=2\pi\cdot3\cdot8=48\pi\,\text{dm}^2\approx150.8\,\text{dm}^2\). 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. The lateral surface area of a right prism is the base perimeter times the prism length: \(L=P\cdot l=24\cdot12=288\,\text{cm}^2\).

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

- 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 \(S_i=2\pi r_i^2+2\pi r_i\cdot1\). 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) \((440\pi+240)\,\text{cm}^2\approx1622.30\,\text{cm}^2\)

- 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 \(L=ph\). The new lateral area is \(L'=p(3h)=3L\). 3. The original total surface area is \(S=2B+L\), while the new total surface area is \(S'=2B+3L\). 4. Three times the original surface area would be \(3S=6B+3L\). Since \(B>0\), \(2B+3L<6B+3L\). 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: \(L'=3L\). c) The claim is false. The lateral area triples, but the two unchanged bases prevent the total surface area from tripling.

- Solve the cylinder volume formula for height. - Total surface area includes two circles and the lateral rectangle. - Multiply by the number of cans, then include the waste percentage. - 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 \(S=2\pi r^2+2\pi rh\), can A has \(S_A\approx319.83\,\text{cm}^2\). 3. Can B has \(S_B\approx313.22\,\text{cm}^2\), so can B uses less metal. 4. Using the unrounded surface area for can B, the metal for \(1000\) cans including \(15\%\) waste is \(1000\cdot S_B\cdot1.15\approx360{,}198.79\,\text{cm}^2\). 5. Since \(10{,}000\,\text{cm}^2=1\,\text{m}^2\), this is about \(36.02\,\text{m}^2\).

a) \(h_A\approx11.04\,\text{cm}\); \(h_B\approx7.67\,\text{cm}\) b) \(S_A\approx319.83\,\text{cm}^2\); \(S_B\approx313.22\,\text{cm}^2\) 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 \(S_{\text{old}}=2\pi r^2+2\pi rh=2\pi\cdot6^2+2\pi\cdot6\cdot15=252\pi\,\text{cm}^2\). 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 \(S_{\text{new}}=64\pi+180\pi+60\pi=304\pi\,\text{cm}^2\). 5. The percent increase is \(\frac{304\pi-252\pi}{252\pi}\cdot100\%\approx20.6\%\).

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

- 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, the total surface area is \(S=2\cdot(100-9\pi)+800+120\pi=1000+102\pi\,\text{cm}^2\approx1320.44\,\text{cm}^2\).

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

- 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 \(L=S-2B=540-2\pi\cdot4.2^2\approx429.1646\,\text{cm}^2\). 3. Since \(L=2\pi rh\), \(h=\frac{L}{2\pi r}=\frac{540-2\pi\cdot4.2^2}{2\pi\cdot4.2}\approx16.2628\,\text{cm}\). 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 \(l=\sqrt{8^2+6^2}=10\,\text{cm}\). 5. Its lateral area is \(L=4\left(\frac{1}{2}\cdot12\cdot10\right)=240\,\text{cm}^2\). 6. The total surface area is \(S=144+240=384\,\text{cm}^2\).

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

- Use the volume formula for a pyramid frustum and solve for the unknown height. - Convert all measurements to compatible units before calculating. - For part b, use the Pythagorean theorem to find the slant height of a lateral face. - Only the four lateral faces are painted.

1. Convert the volume: \(28\,\text{dm}^3=28{,}000\,\text{cm}^3\). 2. For a square-pyramid frustum, \(V=\frac{h}{3}\left(a^2+ab+b^2\right)\). Thus, \(28{,}000=\frac{h}{3}\left(40^2+40\cdot20+20^2\right)=\frac{2{,}800h}{3}\). 3. Solving gives \(h=30\,\text{cm}\). 4. The horizontal offset for a lateral face is \(\frac{40-20}{2}=10\,\text{cm}\), so its slant height is \(l=\sqrt{30^2+10^2}=10\sqrt{10}\approx31.6228\,\text{cm}\). 5. One lateral face has area \(\frac{40+20}{2}l=30l\). The four faces have total area \(L=120l=1{,}200\sqrt{10}\approx3{,}794.7\,\text{cm}^2\).

a) \(h=30\,\text{cm}\) b) The area to be painted is approximately \(3{,}794.7\,\text{cm}^2\).

- 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 \(l=\sqrt{8^2+6^2}=10\,\text{cm}\). 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 \(l=\sqrt{12^2+9^2}=15\,\text{cm}\). 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 \(840\,\text{cm}^2\).

- 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 \(L=2\pi rh\), and one base has area \(B=\pi r^2\). 2. Since \(L=3B\), \(2\pi rh=3\pi r^2\). Because \(r>0\), divide by \(\pi r\) to get \(2h=3r\). 3. Substitute \(h=15\): \(2\cdot15=3r\), so \(r=10\,\text{cm}\). 4. The volume is \(V=\pi\cdot10^2\cdot15=1500\pi\,\text{cm}^3\approx4712.4\,\text{cm}^3\). 5. The total surface area is \(S=2\pi\cdot10^2+2\pi\cdot10\cdot15=500\pi\,\text{cm}^2\approx1570.8\,\text{cm}^2\).

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 \(250\pi=\pi r^2(2r)=2\pi r^3\). 2. Divide by \(2\pi\): \(r^3=125\), so \(r=5\,\text{cm}\). 3. The height is \(h=2\cdot5=10\,\text{cm}\). 4. The total surface area is \(S=2\pi r^2+2\pi rh=2\pi\cdot5^2+2\pi\cdot5\cdot10=150\pi\,\text{cm}^2\). 5. Numerically, \(S\approx471.24\,\text{cm}^2\).

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