Free math worksheets

- Check whether each choice describes the entire solid or only selected parts. - Pay attention to the difference between “every” and “several.” - Decide whether the condition must hold at all heights.

1. Cavalieri's principle requires equal heights and equal areas for all corresponding cross sections parallel to the base plane. 2. Only choice B states both requirements completely.

B

- Area has square units; ask what multiplying by the layer thickness does to the units. - After finding the volume of one layer, use the fact that the five layers do not overlap.

1. One layer has volume equal to its area times its thickness: \(12\,\text{cm}^2\cdot1\,\text{cm}=12\,\text{cm}^3\). 2. There are \(5\) nonoverlapping identical layers, so the total volume is \(5\cdot12=60\,\text{cm}^3\).

Multiplying the layer area by its \(1\,\text{cm}\) thickness gives \(12\,\text{cm}^3\) per layer. Five such layers have volume \(60\,\text{cm}^3\).

- Congruent solids have equal volumes. - If three equal-volume pieces exactly fill a whole, what fraction of the whole belongs to one piece?

1. Because the three pyramids are congruent, they have equal volumes. 2. Because they fill the prism with no gaps or overlaps, their three equal volumes add to the prism's volume. Therefore, one pyramid has \(\frac{1}{3}\) of the prism's volume. 3. Thus, one pyramid has volume \(\frac{1}{3}\cdot81=27\,\text{cm}^3\).

One pyramid occupies \(\frac{1}{3}\) of the prism's volume because three congruent pyramids partition the prism. Its volume is \(27\,\text{cm}^3\).

- Substitute \(B_1\) for every occurrence of \(B_2\). - Use the fact that a base area cannot be negative when simplifying the square root. - Compare the simplified expression with familiar solid-volume formulas.

1. Substitute \(B_2=B_1\): \(V=\frac{h}{3}\left(B_1+\sqrt{B_1^2}+B_1\right)\). 2. Because an area is nonnegative, \(\sqrt{B_1^2}=B_1\). Thus, \(V=\frac{h}{3}(3B_1)=B_1h\). 3. The formula \(V=B_1h\) is the prism volume formula. Here \(B_2=B_1\) is interpreted as the zero-taper limit: as the upper base approaches the same size as the lower base, the frustum formula approaches the volume of a prism with base area \(B_1\) and height \(h\).

1. \(V=B_1h\) 2. This is the prism volume formula. The equality \(B_2=B_1\) represents the zero-taper limiting case of the frustum formula.

- What is true about every cross section of a prism that is parallel to its bases? - Check both requirements of Cavalieri's principle before calculating a volume. - After justifying equal volumes, use the right prism as the comparison solid.

1. Every cross section parallel to the bases of either prism is congruent to its base, so corresponding cross sections have equal area \(36\,\text{cm}^2\). 2. The prisms also have the same perpendicular height, \(14\,\text{cm}\). Therefore, Cavalieri's principle gives equal volumes. 3. The right prism has volume \(V=Bh=36\cdot14=504\,\text{cm}^3\), so the oblique prism has the same volume. 4. The \(17\,\text{cm}\) lateral-edge length is not needed because the comparison uses the perpendicular height and corresponding cross-sectional areas.

Corresponding cross sections parallel to the bases all have area \(36\,\text{cm}^2\), and the prisms have the same perpendicular height, so Cavalieri's principle applies. The oblique prism's volume is \(504\,\text{cm}^3\). The \(17\,\text{cm}\) lateral-edge length is not needed.

- What shape and area does a cross section parallel to either base have? - Which two conditions must match before Cavalieri's principle can compare the cylinders? - After the volume comparison is justified, convert the diameter to a radius and use the right cylinder's volume. - Convert cubic centimeters to cubic decimeters only at the end.

1. Every cross section of the oblique cylinder parallel to its bases is a circle congruent to the base, so its area is \(\pi(6)^2=36\pi\,\text{cm}^2\). The corresponding cross section of the comparison right cylinder has the same area. 2. The two cylinders also have the same perpendicular height, \(15\,\text{cm}\). Therefore, Cavalieri's principle gives equal volumes. 3. The comparison cylinder has volume \(V=\pi r^2h=\pi\cdot6^2\cdot15=540\pi\,\text{cm}^3\). 4. Since \(1{,}000\,\text{cm}^3=1\,\text{dm}^3\), \(V=0.54\pi\,\text{dm}^3\approx1.70\,\text{dm}^3\). 5. The \(18\,\text{cm}\) generator length is not needed because Cavalieri's comparison depends on the base cross-sectional areas and the perpendicular height.

Corresponding cross sections parallel to the bases have the same area, and the cylinders have the same perpendicular height, so Cavalieri's principle gives equal volumes. The oblique cylinder's volume is \(0.54\pi\,\text{dm}^3\approx1.70\,\text{dm}^3\). The \(18\,\text{cm}\) generator length is not needed.

- Compare one horizontal slice before and after the cards are shifted. - Which measurements of that slice change, if any? - Use the unshifted stack only after establishing the Cavalieri comparison.

1. At every height through either stack, a horizontal cross section is an \(8\,\text{cm}\times8\,\text{cm}\) square with area \(64\,\text{cm}^2\). 2. The shifted and unshifted stacks have the same total height, \(22\,\text{cm}\), and equal corresponding cross-sectional areas. 3. Therefore, Cavalieri's principle gives equal volumes. 4. The unshifted stack is a right rectangular prism, so \(V=64\cdot22=1{,}408\,\text{cm}^3\). The shifted stack has the same volume.

The horizontal cross sections have equal area at every height and the stacks have the same height, so Cavalieri's principle shows that shifting the cards does not change the volume. The volume is \(1{,}408\,\text{cm}^3\).

- Compare the column with a simpler solid having the same cross-sectional areas. - The area of each cross section matters more than its exact shape. - Use the common height.

1. Compare the column with a right prism that is \(10\,\text{cm}\) tall and has constant cross-sectional area \(18\,\text{cm}^2\). 2. The two solids have equal cross-sectional areas at every height, so Cavalieri's principle gives equal volumes. 3. Thus, \(V=18\cdot10=180\,\text{cm}^3\).

The column volume is \(180\,\text{cm}^3\).

- Compare all corresponding cross sections, not just one. - A constant cross-sectional area factor produces the same volume factor when heights are equal. - Apply the factor to the known volume.

1. Because the same area scale factor applies to all corresponding cross sections and the heights are equal, the volume has the same scale factor. 2. Therefore, \(V_B=1.25V_A=1.25\cdot320=400\,\text{cm}^3\).

\(V_B=400\,\text{cm}^3\)

- Transfer the ratio of corresponding cross-sectional areas to the volumes. - Identify which part of the ratio corresponds to the known volume. - The unknown volume should be larger than \(270\,\text{cm}^3\).

1. By Cavalieri's principle, a constant ratio of \(3\) to \(2\) for corresponding cross-sectional areas gives the same ratio of volumes. 2. The given \(270\,\text{cm}^3\) corresponds to the smaller, \(2\)-part volume. 3. Therefore, \(V=270\left(\frac{3}{2}\right)=405\,\text{cm}^3\).

The other solid has volume \(405\,\text{cm}^3\).

- Determine the cylinder's cross-sectional area. - Because the heights are equal, compare the cross-sectional areas directly. - The unknown is an area, not a volume.

1. Both solids have constant cross-sectional area throughout their height. 2. To make corresponding cross sections equal, the prism base area must equal the cylinder's circular base area. 3. Therefore, \(B=\pi r^2=\pi\cdot3^2=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\).

\(B=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\)

- Find the known prism volume first. - Replace the unusual column mentally with a prism having the same cross-sectional area. - Work backward from volume to height.

1. The prism volume is \(V=50\cdot12=600\,\text{cm}^3\). 2. By Cavalieri's principle, the column has the same volume as a prism with base area \(40\,\text{cm}^2\) and the same height as the column. 3. Let the column height be \(h\). Then \(40h=600\), so \(h=15\,\text{cm}\).

The column is \(15\,\text{cm}\) tall.

- Decide whether information about only the bottom of a solid determines the entire solid. - Think of solids that narrow at different rates. - State a condition that applies throughout the full height.

1. The claim is false for arbitrary solids because equal base areas and equal heights do not determine the areas of the other cross sections. 2. The missing condition is that, at every corresponding height, cross sections parallel to the base planes must have equal areas.

The claim is false. The solids must have equal heights and equal areas for corresponding cross sections at every height.

- Compare the two area expressions at an arbitrary height. - Distinguish cross-sectional shape from cross-sectional area. - Confirm that both solids use the same height interval.

1. For every \(z\) from \(0\) to \(8\), the two cross-sectional areas are equal. 2. The solids also have equal heights. 3. By Cavalieri's principle, \(V_K=V_L\). The different cross-sectional shapes do not matter because their areas are equal.

Yes. The volumes are equal: \(V_K=V_L\).

- For a cross section made of separate regions, add their areas. - Compare the total cross-sectional areas of the two solids. - Use a simple comparison solid with the same height and constant cross-sectional area.

1. At every height, the total cross-sectional area of solid \(B\) is \(12\,\text{cm}^2+23\,\text{cm}^2=35\,\text{cm}^2\). 2. Thus, solids \(A\) and \(B\) have equal cross-sectional areas at every corresponding height and have the same height. 3. By Cavalieri's principle, their volumes are equal. Using a prism with base area \(35\,\text{cm}^2\) and height \(9\,\text{cm}\), \(V_A=V_B=35\cdot9=315\,\text{cm}^3\).

Both solids have volume \(315\,\text{cm}^3\).

- Determine how much of each cross section is actually material. - Subtract the shaft area from the outer area. - Compare the result with a prism having a constant cross-sectional area.

1. The material area in each horizontal cross section is \(64\,\text{m}^2-16\,\text{m}^2=48\,\text{m}^2\). 2. A solid prism with base area \(48\,\text{m}^2\) and height \(20\,\text{m}\) has the same cross-sectional area at every corresponding height. 3. By Cavalieri's principle, the material volume is \(V=48\cdot20=960\,\text{m}^3\).

The volume of material is \(960\,\text{m}^3\).

- Replace the offset duct with a right comparison prism. - Use the fact that corresponding cross sections have equal areas. - Convert the volume only after calculating it.

1. The offset duct has the same volume as a right prism with base area \(0.08\,\text{m}^2\) and height \(2.5\,\text{m}\). 2. The volume is \(V=0.08\cdot2.5=0.2\,\text{m}^3\). 3. Since \(1\,\text{m}^3=1{,}000\,\text{dm}^3\), \(0.2\,\text{m}^3=200\,\text{dm}^3\).

The interior volume is \(200\,\text{dm}^3\).

- What happens to the shape and area of a prism's cross section when the cutting plane stays parallel to the ends? - State both conditions needed for a Cavalieri comparison with a right prism. - Calculate the right prism's volume only after the comparison is justified.

1. Every cross section parallel to the ends of the oblique prism is congruent to an end, so it has area \(1.8\,\text{m}^2\). 2. A right prism with the same end area has the same cross-sectional area at every corresponding height, and both prisms have perpendicular height \(6\,\text{m}\). 3. Cavalieri's principle therefore gives equal volumes. 4. The right prism's volume is \(V=Bh=1.8\cdot6=10.8\,\text{m}^3\), so the oblique container has the same capacity.

Corresponding parallel cross sections have area \(1.8\,\text{m}^2\) and the prisms have the same perpendicular height, so Cavalieri's principle gives equal volumes. The capacity is \(10.8\,\text{m}^3\).

- Use the stated standard solid as a comparison. - The sculpture's outer shape does not matter when corresponding cross-sectional areas are equal. - Calculate the comparison solid's volume.

1. The sculpture and the cone have the same height and equal cross-sectional areas at every corresponding height, so Cavalieri's principle gives them equal volumes. 2. The cone's volume is \(V=\frac{1}{3}\pi\cdot6^2\cdot15=180\pi\,\text{cm}^3\approx565.49\,\text{cm}^3\).

The volume is \(180\pi\,\text{cm}^3\approx565.49\,\text{cm}^3\).

- Decide whether the redesign changes the area of each layer. - Compare the original and new layers at the same height. - Calculate the percent change only after comparing the volumes.

1. The original and redesigned components have equal heights and equal corresponding cross-sectional areas, so Cavalieri's principle gives equal volumes. 2. The new volume is \(2.4\,\text{m}^3\). 3. The volume change is \(2.4-2.4=0\,\text{m}^3\), so the percent change is \(0\%\).

The new volume is \(2.4\,\text{m}^3\), and the percent change is \(0\%\).

- Read the dimensions of both cross sections from the diagram. - Compare their areas, not just their shapes. - Then use the common height of the solids.

1. For solid \(K\), each cross section is a rectangle with area \(A_K=8\cdot3=24\,\text{cm}^2\). 2. For solid \(L\), each cross section is a triangle with area \(A_L=\frac{1}{2}\cdot12\cdot4=24\,\text{cm}^2\). 3. The solids have equal heights and equal cross-sectional areas at every corresponding height, so Cavalieri's principle gives equal volumes. 4. Therefore, \(V_K=V_L=24\cdot15=360\,\text{cm}^3\).

The volumes are equal: \(V_K=V_L=360\,\text{cm}^3\).

- In b), pay attention to the dashed segment. - Compare the base and corresponding height of the two triangles. - Then use the fact that these cross sections occur at every height of the solids.

1. The triangles in a) and b) have the same base, \(10\,\text{cm}\), and the same corresponding height, \(6\,\text{cm}\). 2. Therefore, every cross section in each solid has area \(A=\frac{1}{2}\cdot10\cdot6=30\,\text{cm}^2\). 3. By Cavalieri's principle, \(V_a=V_b=30\cdot16=480\,\text{cm}^3\).

Both volumes are \(480\,\text{cm}^3\).

- Replace the column with a simple solid having the same cross sections. - Work backward from volume to area.

1. The column has the same volume as a prism with unknown base area \(B\) and height \(18\,\text{cm}\). 2. From \(18B=936\), \(B=52\,\text{cm}^2\).

Each cross section has area \(52\,\text{cm}^2\).

- Compare a horizontal slice of each solid at the same height. - Determine whether the slice areas change as the slicing level moves upward. - Once Cavalieri gives equal volumes, use the known prism volume relationship.

1. Every horizontal cross-section of the cylinder is a circle of area \(\pi r^2\). 2. Every horizontal cross-section of the right prism is congruent to its base, so it also has area \(\pi r^2\). 3. The two solids have the same height and equal corresponding cross-sectional areas. Cavalieri's principle therefore gives them equal volumes. 4. The prism's volume is base area times height, \((\pi r^2)h\). Hence the cylinder's volume is \(V=\pi r^2h\).

Corresponding horizontal sections of the cylinder and prism both have area \(\pi r^2\), and the solids have the same height. By Cavalieri's principle, their volumes are equal. Since the prism has volume \((\pi r^2)h\), the cylinder has volume \(V=\pi r^2h\).

- First calculate \(B_1\) and \(B_2\) from the square edge lengths. - Consider what shape results when the top face shrinks to a single point. - Substitute zero for \(B_2\) in the general formula.

1. When \(b=4\,\text{cm}\), \(B_1=8^2=64\,\text{cm}^2\) and \(B_2=4^2=16\,\text{cm}^2\). Therefore, \(V=\frac{12}{3}\left(64+\sqrt{64\cdot16}+16\right)=4\left(64+32+16\right)=448\,\text{cm}^3\). 2. When \(b=0\,\text{cm}\), \(B_2=0\). Then \(V=\frac{12}{3}\left(64+\sqrt{64\cdot0}+0\right)=4\cdot64=256\,\text{cm}^3\). 3. When the top area shrinks to zero, the solid ends at an apex and becomes a pyramid. In general, setting \(B_2=0\) gives \(V=\frac{h}{3}(B_1+0+0)=\frac{1}{3}B_1h\).

1. \(V=448\,\text{cm}^3\) 2. \(V=256\,\text{cm}^3\) 3. The solid becomes a pyramid, and the formula becomes \(V=\frac{1}{3}B_1h\).

- Use the proportional relationship between the radii and heights of the similar cones. - The large cone's height includes both the frustum and the small cone. - Use the cone volume formula for each cone before subtracting.

1. Similar triangles give \(\frac{x}{4}=\frac{x+6}{8}\). 2. Solving, \(8x=4x+24\), so \(x=6\,\text{cm}\). The large cone's height is \(12\,\text{cm}\). 3. The cone volumes are \(V_{\text{large}}=\frac{1}{3}\pi\cdot8^2\cdot12=256\pi\,\text{cm}^3\) and \(V_{\text{small}}=\frac{1}{3}\pi\cdot4^2\cdot6=32\pi\,\text{cm}^3\). 4. Therefore, the frustum's volume is \(256\pi-32\pi=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\). 5. Direct substitution gives \(V=\frac{6\pi}{3}\left(8^2+8\cdot4+4^2\right)=2\pi\left(64+32+16\right)=224\pi\,\text{cm}^3\), confirming the result.

a) \(x=6\,\text{cm}\) b) \(V=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\) c) Direct substitution also gives \(224\pi\,\text{cm}^3\).

- Compare cross sections taken the same distance from each apex. - Consider how cross-sectional area scales in a pyramid. - Once volume equality is established, calculate the volume only once.

1. At the same fractional distance from the apex, corresponding cross sections of all three pyramids have the same area scale factor. 2. Because the pyramids have equal base areas and equal heights, corresponding cross sections have equal areas. By Cavalieri's principle, the volumes are equal. 3. The common volume is \(V=\frac{1}{3}Bh=\frac{1}{3}\cdot48\cdot9=144\,\text{cm}^3\).

All three pyramids have volume \(144\,\text{cm}^3\).

- Think about how a parallel cross section of a cone is scaled relative to its base. - At the same fraction of the perpendicular height, compare the scale factors in the oblique and right cones. - Only after establishing equal corresponding areas should you use the right cone's volume formula.

1. In either cone, a cross section parallel to the base is similar to the base. Its radius is determined by the same linear scale factor from the apex because the two cones have the same base radius and perpendicular height. 2. Therefore, at every corresponding height, the two circular cross sections have equal radii and equal areas. 3. The cones have the same perpendicular height, so Cavalieri's principle gives them equal volumes. 4. The comparison right cone has volume \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(4.5)^2(12)=81\pi\,\text{cm}^3\approx254.47\,\text{cm}^3\).

Similarity makes the corresponding parallel cross sections equal in area at every height, so Cavalieri's principle gives equal volumes. The oblique cone's volume is \(81\pi\,\text{cm}^3\approx254.47\,\text{cm}^3\).

- Compare cross sections near the base and near the top for each pair. - Determine whether cross-sectional area changes with height. - Equal base areas alone are not sufficient.

1. For a), all corresponding cross sections have equal areas, so the volumes are equal. 2. For b), prism cross sections remain constant while pyramid cross sections decrease toward the apex, so Cavalieri's principle does not establish equal volume. 3. For c), corresponding cross sections of the right and oblique cones have equal areas, so the volumes are equal. 4. For d), cylinder and cone cross sections do not have equal areas at every height, so the principle does not establish equal volume.

a) yes b) no c) yes d) no

- Simplify and compare the area expressions pairwise. - Look for a cross-sectional area difference that remains constant. - Interpret a constant area difference as the cross section of an added prism.

1. For every height, \(A_K(z)=A_L(z)\), so Cavalieri's principle gives \(V_K=V_L\). 2. Also, \(A_M(z)-A_K(z)=1\,\text{cm}^2\) and \(A_M(z)-A_L(z)=1\,\text{cm}^2\) at every height. 3. This constant cross-sectional area difference over \(10\,\text{cm}\) corresponds to a volume difference of \(1\cdot10=10\,\text{cm}^3\). 4. Therefore, \(V_M\) is \(10\,\text{cm}^3\) greater than each of \(V_K\) and \(V_L\).

\(V_K=V_L\). The volume \(V_M\) is \(10\,\text{cm}^3\) greater than each of them.

- Subtract the inner circular area from the outer circular area for one parallel cross section. - Ask whether that annular area changes as the cross section moves through the oblique pipe. - Match both cross-sectional area and perpendicular height to a right hollow cylinder before calculating volume.

1. Every cross section parallel to the ends is an annulus with area \(A=\pi(5^2-3^2)=16\pi\,\text{cm}^2\). 2. A right hollow cylinder with outer radius \(5\,\text{cm}\), inner radius \(3\,\text{cm}\), and height \(18\,\text{cm}\) has the same annular area in every corresponding cross section. 3. The two solids have the same perpendicular height, so Cavalieri's principle gives equal volumes. 4. Thus, \(V=(16\pi)(18)=288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\).

Each parallel cross section has area \(16\pi\,\text{cm}^2\). These areas match those of a right hollow cylinder at every corresponding height, so Cavalieri's principle applies. The material volume is \(288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\).

- Identify a familiar solid with the same cross-sectional area at every height. - The changing cross-sectional shape does not matter if the area remains equal. - Calculate in one unit first, then convert at the end.

1. The sculpture has the same volume as a cylinder with radius \(6\,\text{cm}\) and height \(20\,\text{cm}\). 2. Its volume is \(V=\pi\cdot6^2\cdot20=720\pi\,\text{cm}^3\). 3. Since \(1{,}000\,\text{cm}^3=1\,\text{dm}^3\), \(720\pi\,\text{cm}^3=0.72\pi\,\text{dm}^3\approx2.26\,\text{dm}^3\).

The volume is \(0.72\pi\,\text{dm}^3\approx2.26\,\text{dm}^3\).

- Compare cross sections at the same perpendicular distance from the base. - Decide whether shifting the apex changes the area scale factor of those cross sections. - Calculate the common volume only once.

1. At the same perpendicular distance from the base, a cross section parallel to the base has the same area in both pyramids. Shifting the apex sideways changes the position of the cross section, but not its area scale factor. 2. The pyramids have equal heights and equal corresponding cross-sectional areas, so Cavalieri's principle shows that their volumes are equal. 3. Their common volume is \(V=\frac{1}{3}\cdot80\cdot9=240\,\text{cm}^3\).

Both pyramids have volume \(240\,\text{cm}^3\).

- Compare cross sections near the base and near the top. - Determine whether the cross-sectional area stays constant in both solids. - Equal base area and equal height alone are not sufficient.

1. The corresponding cross-sectional areas are not equal at every height. A prism's cross-sectional area remains constant, while a pyramid's cross-sectional area decreases toward the apex. 2. Therefore, equal base area and equal height do not satisfy the cross-section condition of Cavalieri's principle for these two solids. 3. The prism's volume is \(V_{\text{prism}}=54\cdot10=540\,\text{cm}^3\). 4. The pyramid's volume is \(V_{\text{pyramid}}=\frac{1}{3}\cdot54\cdot10=180\,\text{cm}^3\).

Nora's reasoning is incorrect. The prism has volume \(540\,\text{cm}^3\), and the pyramid has volume \(180\,\text{cm}^3\).

- Treat the difference between the two cross sections as a separate simple solid. - This area difference remains constant over the entire height. - Add the resulting extra volume to the known volume.

1. The constant cross-sectional area difference can be represented by an added prism with base area \(7\,\text{cm}^2\) and height \(12\,\text{cm}\). 2. The added volume is \(7\cdot12=84\,\text{cm}^3\). 3. Therefore, \(V_A=420+84=504\,\text{cm}^3\).

The volume of solid \(A\) is \(504\,\text{cm}^3\).

- Compare the constant cross-section scale factor with the volume scale factor. - Write an equation for the unknown value. - Check that the resulting factor is consistent with the larger volume.

1. Because the corresponding cross-sectional areas differ by a constant factor, the volumes differ by the same factor. 2. Therefore, \(\frac{x}{5}=1.4\). 3. Multiplying by \(5\) gives \(x=7\).

\(x=7\).

- Use all the dimensions shown in the diagram. - Compare the areas of the two different shapes. - Determine what equal cross-sectional areas at every height imply.

1. The trapezoidal cross section in a) has area \(A_a=\frac{10+6}{2}\cdot4=32\,\text{cm}^2\). 2. The rectangular cross section in b) has area \(A_b=8\cdot4=32\,\text{cm}^2\). 3. The solids have equal heights and equal cross-sectional areas at every height, so Cavalieri's principle gives \(V_a=V_b=32\cdot9=288\,\text{cm}^3\).

Both volumes are \(288\,\text{cm}^3\).

- Distinguish the material from the hollow region in the diagram. - Find the total material area of each cross section. - Then use the common height.

1. In a), the material area is \(12\cdot8-8\cdot4=96-32=64\,\text{cm}^2\). 2. In b), the material area is \(8\cdot8=64\,\text{cm}^2\). 3. The corresponding material cross sections have equal areas at every height, so both components contain \(64\cdot20=1{,}280\,\text{cm}^3\) of material.

Both components contain \(1{,}280\,\text{cm}^3\) of material.

- Find the three cross-sectional areas one at a time. - Because the solids have equal heights, compare their cross-sectional areas directly. - Check which solid has the largest cross section.

1. The cross-sectional areas are \(A_a=5\cdot5=25\,\text{cm}^2\), \(A_b=10\cdot2.5=25\,\text{cm}^2\), and \(A_c=\frac{1}{2}\cdot12\cdot5=30\,\text{cm}^2\). 2. Since a) and b) have equal cross-sectional areas at every height and the same height, Cavalieri's principle gives equal volumes. 3. The volumes are \(V_a=V_b=25\cdot12=300\,\text{cm}^3\) and \(V_c=30\cdot12=360\,\text{cm}^3\).

Solids a) and b) have equal volumes of \(300\,\text{cm}^3\) each. Solid c) has volume \(360\,\text{cm}^3\).

- Distinguish the outer region from the hollow space. - Compare the resulting areas symbolically. - Keep the factor \(\pi\) until the final approximation.

1. The annular material area in a) is \(\pi\left(5^2-3^2\right)=16\pi\,\text{cm}^2\). 2. The circular material area in b) is \(\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The material cross sections have equal areas at every height, so Cavalieri's principle gives \(V=(16\pi)(11)=176\pi\,\text{cm}^3\approx552.92\,\text{cm}^3\).

Both material volumes are \(176\pi\,\text{cm}^3\approx552.92\,\text{cm}^3\).

- In b), add the areas of the two separate regions. - Compare that total with the area in a). - Separating the regions does not change their total area.

1. In a), the triangular area is \(A_A=\frac{1}{2}\cdot12\cdot8=48\,\text{cm}^2\). 2. In b), each triangle has area \(\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\), so the total cross-sectional area is \(48\,\text{cm}^2\). 3. The solids have equal total cross-sectional areas at every height and equal heights. By Cavalieri's principle, \(V_A=V_B=48\cdot10=480\,\text{cm}^3\).

Both solids have volume \(480\,\text{cm}^3\).

- First find the fully dimensioned cross-sectional area. - Use the equal volumes together with the common height. - Then calculate the common volume only once.

1. Because the solids have equal heights, equal volumes, and constant cross-sectional areas, those constant areas must be equal. 2. The triangular cross section in b) has area \(\frac{1}{2}\cdot12\cdot5=30\,\text{cm}^2\). 3. For a), \(6x=30\), so \(x=5\,\text{cm}\). 4. The common volume is \(V=30\cdot13=390\,\text{cm}^3\).

\(x=5\,\text{cm}\), and the common volume is \(390\,\text{cm}^3\).

- Calculate both cross-sectional areas. - Simplify the resulting ratio completely. - For equal heights, the cross-sectional area ratio equals the volume ratio.

1. From the diagram, \(A_A=9\cdot4=36\,\text{cm}^2\) and \(A_B=\frac{1}{2}\cdot12\cdot9=54\,\text{cm}^2\). 2. Thus, \(\frac{V_A}{V_B}=\frac{A_A}{A_B}=\frac{36}{54}=\frac{2}{3}\) because the solids have the same height. 3. The volumes are \(V_A=36\cdot8=288\,\text{cm}^3\) and \(V_B=54\cdot8=432\,\text{cm}^3\).

The ratio of \(V_A\) to \(V_B\) is \(2\) to \(3\). Also, \(V_A=288\,\text{cm}^3\) and \(V_B=432\,\text{cm}^3\).

- Test the claim by calculating all three cross-sectional areas. - For the parallelogram, use the dashed height. - Compare the volumes only after comparing the areas.

1. The cross-sectional areas are \(A_a=8\cdot5=40\,\text{cm}^2\), \(A_b=10\cdot4=40\,\text{cm}^2\), and \(A_c=\frac{1}{2}\cdot10\cdot8=40\,\text{cm}^2\). 2. Ben's claim is false. The triangle's greater height offsets the factor \(\frac{1}{2}\) in its area formula. 3. All three solids have equal cross-sectional areas at every height and equal heights, so Cavalieri's principle gives \(V=40\cdot7=280\,\text{cm}^3\) for each solid.

Ben's claim is false. All three volumes are \(280\,\text{cm}^3\).

- Calculate the reference cross-sectional area first. - Test each candidate separately. - Because the heights are equal, the cross-sectional areas determine the comparison.

1. The reference cross-sectional area is \(A_R=7\cdot4=28\,\text{cm}^2\). 2. Candidate a) has area \(\frac{1}{2}\cdot14\cdot4=28\,\text{cm}^2\), candidate b) has area \(\frac{9+5}{2}\cdot4=28\,\text{cm}^2\), and candidate c) has area \(5\cdot5=25\,\text{cm}^2\). 3. Therefore, a) and b) have the same volume as \(R\): \(V_R=V_a=V_b=28\cdot9=252\,\text{cm}^3\). 4. Candidate c) has volume \(V_c=25\cdot9=225\,\text{cm}^3\).

Candidates a) and b) have the same volume as \(R\). \(V_R=V_a=V_b=252\,\text{cm}^3\), and \(V_c=225\,\text{cm}^3\).

- Use the diagram to find both cross-sectional areas. - Calculate the known volume first. - Then work backward to find the unknown height.

1. The cross-sectional areas are \(A_A=9\cdot4=36\,\text{cm}^2\) and \(A_B=\frac{1}{2}\cdot10\cdot6=30\,\text{cm}^2\). 2. By Cavalieri's principle, each solid has the same volume as a right prism with the same constant cross-sectional area and height. Thus, \(V_A=36\cdot10=360\,\text{cm}^3\) and \(V_B=30h\). 3. For equal volumes, \(30h=360\). Therefore, \(h=12\,\text{cm}\).

Solid \(B\) must be \(12\,\text{cm}\) tall.

- Start with the known volume of the whole prism. - Use the fact that the dissection gives three equal-volume pieces. - For a general polygonal base, think about dividing the base into triangles while keeping the same apex and height.

1. The triangular prism has volume \(Bh\). 2. The dissection partitions that volume into \(3\) equal-volume pyramids, so each pyramid has volume \(\frac{1}{3}Bh\). 3. For a pyramid with any polygonal base, divide the base into triangles with areas \(B_1,B_2,\ldots\). This divides the pyramid into triangular pyramids that all have the same perpendicular height \(h\). 4. Their volumes add to \(\frac{1}{3}B_1h+\frac{1}{3}B_2h+\cdots=\frac{1}{3}(B_1+B_2+\cdots)h=\frac{1}{3}Bh\). 5. Therefore, the pyramid volume formula is \(V=\frac{1}{3}Bh\).

The prism's volume \(Bh\) is split into three equal-volume pyramids, so each has volume \(\frac{1}{3}Bh\). Triangulating any polygonal base and adding the resulting triangular-pyramid volumes extends the formula to every polygonal pyramid: \(V=\frac{1}{3}Bh\).

- Compare cross-sections taken at the same relative height in the two solids. - When all linear dimensions scale by a factor, determine how area scales. - After establishing equal corresponding slice areas, connect the cone to the known pyramid volume formula.

1. At a fraction \(t\) of the distance from the apex to the base, linear dimensions in each cross-section are scaled by \(t\) from the corresponding base dimensions. 2. Therefore, each cross-sectional area is scaled by \(t^2\). Because the cone and pyramid have the same base area \(B\), their corresponding cross-sections both have area \(t^2B\). 3. The solids have the same height and equal corresponding cross-sectional areas, so Cavalieri's principle gives them equal volumes. 4. The pyramid has volume \(\frac{1}{3}Bh\). Therefore the cone also has volume \(\frac{1}{3}Bh\). 5. Since \(B=\pi r^2\), the cone formula is \(V=\frac{1}{3}\pi r^2h\).

Similarity makes every pair of corresponding cross-sections have the same area \(t^2B\). By Cavalieri's principle, the cone and matching pyramid have equal volume, so \(V=\frac{1}{3}Bh=\frac{1}{3}\pi r^2h\).

- Match each square base area with the corresponding circular base area. - Think about how a side length or radius changes linearly between the two ends of a frustum. - Compare cross-sectional areas at the same relative height before transferring a volume formula from one solid to the other.

1. Equal lower-base areas give \(a_1^2=\pi r_1^2\), so \(r_1=\frac{a_1}{\sqrt{\pi}}\). Equal upper-base areas give \(a_2^2=\pi r_2^2\), so \(r_2=\frac{a_2}{\sqrt{\pi}}\). 2. At fraction \(t\) of the height from the lower base, the square cross-section has side length \((1-t)a_1+ta_2\), while the circular cross-section has radius \((1-t)r_1+tr_2\). 3. Since \(a_1=\sqrt{\pi}r_1\) and \(a_2=\sqrt{\pi}r_2\), the square cross-sectional area is \(\left((1-t)a_1+ta_2\right)^2=\pi\left((1-t)r_1+tr_2\right)^2\), exactly the area of the circular cross-section. 4. The solids have the same height and equal cross-sectional areas at every height, so Cavalieri's principle shows that their volumes are equal. 5. For the square pyramidal frustum, \(B_1=a_1^2=\pi r_1^2\) and \(B_2=a_2^2=\pi r_2^2\). Substituting gives \(V=\frac{h}{3}\left(\pi r_1^2+\sqrt{\pi^2r_1^2r_2^2}+\pi r_2^2\right)\). 6. Because the radii are nonnegative, \(\sqrt{\pi^2r_1^2r_2^2}=\pi r_1r_2\). Therefore, \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

a) \(r_1=\frac{a_1}{\sqrt{\pi}}\) and \(r_2=\frac{a_2}{\sqrt{\pi}}\) b) Corresponding parallel cross-sections have equal area, so Cavalieri's principle gives equal volumes. Hence \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

- Rewrite both volume formulas in terms of \(r_1\), \(r_2\), and \(h\). - Compare the two expressions by studying their difference. - Determine when that difference can equal zero.

1. Substituting \(r_1=r_2=r\) gives \(V=\frac{1}{3}\pi h(r^2+r^2+r^2)=\pi r^2h\), the cylinder volume formula. 2. The proposed formula becomes \(V_{\text{estimate}}=\pi h\left(\frac{r_1+r_2}{2}\right)^2=\frac{\pi h}{4}(r_1^2+2r_1r_2+r_2^2)\). 3. Subtracting the estimate from the exact formula gives \(V-V_{\text{estimate}}=\frac{\pi h}{12}\left(4r_1^2+4r_1r_2+4r_2^2-3r_1^2-6r_1r_2-3r_2^2\right)\). 4. Therefore, \(V-V_{\text{estimate}}=\frac{\pi h}{12}(r_1-r_2)^2\). This difference is positive when \(r_1\ne r_2\), so the proposed cylinder formula is not exact and underestimates the frustum volume.

1. The formula simplifies to \(V=\pi r^2h\). 2. The proposed formula is not exact when \(r_1\ne r_2\). It underestimates the volume by \(\frac{\pi h}{12}(r_1-r_2)^2\).

- Identify the segment in the cube that becomes a tetrahedron edge. - Use the general volume formula for a pyramid. - Determine the shape and area of a corner pyramid's base. - Solve the edge-length relationship for \(a\) before substituting.

1. Each tetrahedron edge is a face diagonal of the cube. By the Pythagorean theorem, \(b^2=a^2+a^2=2a^2\), so \(b=a\sqrt{2}\). 2. A corner pyramid has a right-triangular base with area \(\frac{1}{2}a^2\) and perpendicular height \(a\). Thus, \(V_{\text{corner}}=\frac{1}{3}\cdot\frac{1}{2}a^2\cdot a=\frac{1}{6}a^3\). 3. Subtracting the four corner pyramids from the cube gives \(V_{\text{tetrahedron}}=a^3-4\cdot\frac{1}{6}a^3=\frac{1}{3}a^3\). 4. From \(b=a\sqrt{2}\), \(a=\frac{b}{\sqrt{2}}\). Therefore, \(V=\frac{1}{3}\left(\frac{b}{\sqrt{2}}\right)^3=\frac{b^3}{6\sqrt{2}}=\frac{\sqrt{2}}{12}b^3\).

a) \(b=a\sqrt{2}\) b) \(V_{\text{corner}}=\frac{1}{6}a^3\) and \(V_{\text{tetrahedron}}=\frac{1}{3}a^3\) c) \(V=\frac{\sqrt{2}}{12}b^3\)

- Consider how a height changes when a solid is turned upside down. - Convert complementary heights into corresponding heights. - Changing a solid's orientation does not change its volume.

1. Turn solid \(L\) upside down. A cross section that was originally at height \(12-z\) is then at height \(z\). 2. Therefore, at every corresponding height \(z\), solid \(K\) and the inverted solid \(L\) have equal cross-sectional areas. 3. Turning a solid over does not change its volume, so Cavalieri's principle gives \(V_K=V_L\).

After solid \(L\) is turned upside down, all corresponding cross-sectional areas are equal. Therefore, \(V_K=V_L\).

- Identify which length measures the distance between the base planes. - Compare the oblique cylinder with a right cylinder having the same perpendicular height. - Find the error amount only after calculating the correct volume.

1. Tim used the generator length instead of the perpendicular height between the bases. The corresponding right cylinder has height \(12\,\text{cm}\). 2. The correct volume is \(V=\pi\cdot5^2\cdot12=300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\). 3. Tim's result is \(325\pi\,\text{cm}^3\). The difference is \(325\pi-300\pi=25\pi\,\text{cm}^3\approx78.54\,\text{cm}^3\).

The correct volume is \(300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\). Tim's result is too large by \(25\pi\,\text{cm}^3\approx78.54\,\text{cm}^3\).

- Add the two cross-sectional areas at the same height. - Check whether the height-dependent terms cancel. - Interpret the resulting total cross-sectional area using a comparison solid.

1. At each height, the sum of the cross-sectional areas is \(A_A(z)+A_B(z)=10+2z+28-2z=38\,\text{cm}^2\). 2. Together, the two solids have the same total cross-sectional area at every height as a prism with base area \(38\,\text{cm}^2\) and height \(9\,\text{cm}\). 3. Therefore, \(V_A+V_B=38\cdot9=342\,\text{cm}^3\).

\(V_A+V_B=342\,\text{cm}^3\).

- Calculate the two volumes independently. - Then compare at least one pair of corresponding cross sections. - A true implication does not necessarily remain true when reversed.

1. The prism's volume is \(V_{\text{prism}}=30\cdot10=300\,\text{cm}^3\). 2. The pyramid's volume is \(V_{\text{pyramid}}=\frac{1}{3}\cdot90\cdot10=300\,\text{cm}^3\). 3. However, their base cross-sectional areas are already different: \(30\,\text{cm}^2\) for the prism and \(90\,\text{cm}^2\) for the pyramid. 4. Thus, equal height and equal volume do not imply equal cross-sectional areas at every corresponding height. The converse is false.

The claim is false. Both volumes are \(300\,\text{cm}^3\), but the corresponding cross-sectional areas are not equal at every height.

- In the sphere's axial circle, relate the slice radius, the distance \(y\), and the sphere radius with a right triangle. - In the comparison solid, subtract the cone's circular slice from the cylinder's circular slice. - Establish equality of corresponding slice areas before comparing total volumes. - The two removed cones are congruent and together occupy both halves of the cylinder.

1. In the sphere, the radius \(x\) of the circular slice satisfies \(x^2+y^2=r^2\), so its area is \(\pi x^2=\pi(r^2-y^2)\). 2. At the same height in the comparison solid, the cylinder slice has area \(\pi r^2\). The removed cone at that level has slice radius \(|y|\), so its slice area is \(\pi y^2\). 3. The remaining cross-sectional area is therefore \(\pi r^2-\pi y^2=\pi(r^2-y^2)\), exactly matching the sphere slice. 4. The solids have equal height and equal corresponding cross-sectional areas, so Cavalieri's principle gives them equal volumes. 5. The cylinder has volume \(\pi r^2(2r)=2\pi r^3\). The two removed cones have total volume \(2\left(\frac{1}{3}\pi r^2(r)\right)=\frac{2}{3}\pi r^3\). 6. Hence the sphere volume is \(2\pi r^3-\frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3\).

At height \(y\), both solids have cross-sectional area \(\pi(r^2-y^2)\). Cavalieri's principle therefore makes their volumes equal, giving \(V=2\pi r^3-\frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3\).