Composite 3-D figures

1. Bottom block: V = lwh = 10 \times 10 \times 4 = 400\text{ in}^3.
2. Top block: V = 6 \times 6 \times 4 = 144\text{ in}^3.
3. The blocks do not overlap, so add: 400 + 144 = 544.

Answer: 544 in³

Check: The figure is smaller than one big 10 × 10 × 8 box (800\text{ in}^3) and bigger than the bottom block alone (400\text{ in}^3); 544 is between them ✓.

Tip: Composite volume = break the figure into familiar solids, find each volume, then add the pieces that are there (and subtract any holes).

1. Block: V = lwh = 10 \times 6 \times 4 = 240\text{ in}^3.
2. Hole: a cylinder with r = 1 and h = 4: V = \pi r^2 h = \pi(1)(4) = 4\pi \approx 12.57\text{ in}^3.
3. Subtract: 240 - 12.57 = 227.43.
4. Round: 227.4.

Answer: about 227.4 in³

Check: The hole is small, so the answer should be a little less than 240 ✓.

Tip: For holes, cut-outs, and hollow objects, subtract the missing part. The hole's height is the thickness it goes through (here 4 in).

1. Low box: V = lwh = 6 \times 3 \times 2 = 36\text{ ft}^3.
2. Tall box: V = 6 \times 3 \times 4 = 72\text{ ft}^3.
3. The boxes do not overlap, so add: 36 + 72 = 108.

Answer: 108 ft³

Check: Think of it another way: a 6 by 6 by 4 block (144 ft³) with a 6 by 3 by 2 corner missing (36 ft³): 144 - 36 = 108 ✓.

Tip: Split a composite solid into simple boxes that do not overlap, find each volume, and add. Write the pieces down so none is counted twice or missed.

1. Cube: V = s^3 = 4^3 = 64\text{ ft}^3.
2. Pyramid: base area B = 4^2 = 16, so V = \tfrac{1}{3}(16)(3) = 16\text{ ft}^3.
3. Total: 64 + 16 = 80.

Answer: 80 ft³

Check: The pyramid is one-third of a 4 by 4 by 3 box (48 \div 3 = 16) ✓, and 64 + 16 = 80 ✓.

Tip: In a "house" shape, the box part uses Bh and the roof uses \tfrac{1}{3}Bh. Both share the same base area B, so find B once and use it twice.

1. Box: V = 8 \times 8 \times 6 = 384\text{ ft}^3.
2. Roof: the base is the 8 × 8 top of the box, so B = 64. The height is 3 ft.
3. V = \tfrac{1}{3}Bh = \tfrac{1}{3}(64)(3) = 64\text{ ft}^3.
4. Add: 384 + 64 = 448.

Answer: 448.0 ft³

Check: The roof (64 ft³) is one-third of an 8 × 8 × 3 box (192 ft³) ✓.

Tip: Parts of a composite figure share measurements. The roof's base is the box's top, so you do not need a new base area. Use the slant height only for surface area.

1. Four walls, each 8 by 6: 4 \times (8 \times 6) = 4 \times 48 = 192\text{ ft}^2.
2. Floor: 8 \times 8 = 64\text{ ft}^2.
3. Roof (lateral area of the pyramid, using the slant height): \tfrac{1}{2}ps = \tfrac{1}{2}(32)(5) = 80\text{ ft}^2.
4. The top of the box is covered by the roof, so it is not counted.
5. Total: 192 + 64 + 80 = 336.

Answer: 336 ft²

Check: Each roof triangle is \tfrac{1}{2}(8)(5) = 20, and 4 \times 20 = 80 ✓.

Tip: Make a list of every face you can see from outside. Joined faces (box top, roof bottom) are inside the figure and do not count.

1. Whole (outer) cylinder: \pi(2^2)(6) = \pi(4)(6) = 24\pi.
2. Hole (inner) cylinder: \pi(1.5^2)(6) = \pi(2.25)(6) = 13.5\pi.
3. Subtract: 24\pi - 13.5\pi = 10.5\pi.
4. Evaluate: 10.5\pi \approx 32.99, which rounds to 33.0.

Answer: about 33.0 ft³

Check: The concrete ring is only 0.5 ft thick, so it should be much less than the solid cylinder (24\pi \approx 75.4) ✓.

Tip: Subtract the volumes (or the squared radii), not the radii: 2^2 - 1.5^2 = 1.75, but (2 - 1.5)^2 = 0.25.

1. Cylinder: V = \pi r^2 h = \pi(10^2)(20) = \pi(100)(20) = 2{,}000\pi.
2. Cone: V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(100)(6) = \tfrac{1}{3}\pi(600) = 200\pi.
3. Total: 2{,}000\pi + 200\pi = 2{,}200\pi.

Answer: 2{,}200\pi ft³ (about 6,912 ft³)

Check: The cone is one-third of a 6-ft-tall cylinder (600\pi), so 200\pi ✓. Total \approx 2{,}200 \times 3.1416 \approx 6{,}912 ✓.

Tip: Keep each part's own height. The cylinder uses 20 ft, the cone uses 6 ft — never add the heights and use one formula for the whole tank.

1. Cube: V = 6^3 = 216\text{ cm}^3.
2. Carved hemisphere: V = \tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\pi(2.5^3) = \tfrac{2}{3}\pi(15.625) \approx 32.72\text{ cm}^3.
3. Subtract: 216 - 32.72 = 183.28.
4. Round to the nearest tenth: 183.3.

Answer: about 183.3 cm³

Check: The hollow removes about 32.7 \div 216 \approx 15\% of the cube, and 216 \times 0.85 \approx 183.6 — close ✓.

Tip: "How much is left" → subtract the removed part. Be sure the removed part has the right shape: half a sphere is \tfrac{2}{3}\pi r^3, not \tfrac{4}{3}\pi r^3.

1. Large section: 8 \times 4 \times 1.5 = 48\text{ ft}^3.
2. Small section: 4 \times 4 \times 1.5 = 24\text{ ft}^3.
3. Total soil: 48 + 24 = 72\text{ ft}^3.
4. Cost: 72 \times \$2.50 = \$180.

Answer: $180

Check: The total floor area is 32 + 16 = 48\text{ ft}^2; times the depth, 48 \times 1.5 = 72\text{ ft}^3 ✓; 72 \times 2.5 = 180 ✓.

Tip: When all parts have the same depth, you can add the floor areas first and multiply by the depth once. Then apply the price to the total volume.

1. Cylinder base: radius = 6 \div 2 = 3, so V = \pi(3)^2(2).
2. Ball: radius = 4 \div 2 = 2, so V = \tfrac{4}{3}\pi(2)^3.
3. The parts are joined, so add: \pi(3)^2(2) + \tfrac{4}{3}\pi(2)^3.
4. Value: 18\pi + \tfrac{32}{3}\pi \approx 56.5 + 33.5 = 90.0\text{ in}^3.

Answer: \pi(3)^2(2) + \tfrac{4}{3}\pi(2)^3

Check: Each term matches one part: the first has r = 3 and a height (cylinder), the second has r = 2 cubed (sphere) ✓.

Tip: For "which expression" items, check three things in order: the right formula for each piece, radii (not diameters), and + vs. - (joined vs. removed).

1. Step 1 is correct: \pi(16)(10) = 160\pi.
2. Step 2 is wrong. The top is a hemisphere, half of a sphere: \tfrac{2}{3}\pi(4^3) = \tfrac{128}{3}\pi. Sam used a whole sphere.
3. Correct total: 160\pi + \tfrac{128}{3}\pi = \tfrac{608}{3}\pi \approx 636.7\text{ m}^3.

Answer: Step 2 is the first mistake — the top should be \tfrac{2}{3}\pi(4^3).

Check: Sam's total minus the correct total is 770.7 - 636.7 = 134.0 \approx \tfrac{128}{3}\pi — exactly one extra hemisphere ✓.

Tip: Sketch or name each part before choosing formulas: "cylinder + half sphere." The word "hemisphere" means you take half of \tfrac{4}{3}\pi r^3.

1. Main tank: 24 \times 12 \times 12 = 3{,}456\text{ in}^3.
2. Filter box: 6 \times 6 \times 12 = 432\text{ in}^3.
3. Total: 3{,}456 + 432 = 3{,}888\text{ in}^3.
4. Convert: 1\text{ ft}^3 = 12 \times 12 \times 12 = 1{,}728\text{ in}^3, so 3{,}888 \div 1{,}728 = 2.25.

Answer: 2.25 ft³

Check: Convert lengths first: main tank 2 \times 1 \times 1 = 2\text{ ft}^3; filter box 0.5 \times 0.5 \times 1 = 0.25\text{ ft}^3; total 2.25 ✓.

Tip: For cubic units, the conversion factor is cubed: 12^3 = 1{,}728 for in³ → ft³, and 3^3 = 27 for ft³ → yd³.

1. Cube: 6^3 = 216\text{ ft}^3.
2. Pyramid volume: 288 - 216 = 72\text{ ft}^3.
3. Pyramid base: B = 6^2 = 36\text{ ft}^2.
4. Solve: 72 = \tfrac{1}{3}(36)h = 12h, so h = 72 \div 12 = 6 ft.

Answer: 6 ft

Check: 216 + \tfrac{1}{3}(36)(6) = 216 + 72 = 288 ✓.

Tip: To find a missing part of a composite, first peel off the parts you know: total − known part = unknown part. Then work backwards with that part's formula only.

1. Original cube: 6 \times (4 \times 4) = 6 \times 16 = 96\text{ in}^2.
2. Cutting the corner removes three 1 in × 1 in squares from three outside faces: -3\text{ in}^2.
3. The notch exposes three new 1 in × 1 in faces inside the cut: +3\text{ in}^2.
4. New surface area: 96 - 3 + 3 = 96.

Answer: 96 in² (unchanged)

Check: Look at the block from the front, top and side: each view still shows a full 4 by 4 outline (16 in²), and each of the 6 directions still adds up to 16 in² ✓.

Tip: Removing material does not always reduce surface area. Count the faces that disappear and the new faces that appear. (The volume does drop: 64 - 1 = 63\text{ in}^3.)

1. Cylinder: \pi(1.5^2)(6) = \pi(2.25)(6) = 13.5\pi\text{ ft}^3.
2. Two hemispheres = one sphere: \tfrac{4}{3}\pi(1.5^3) = \tfrac{4}{3}\pi(3.375) = 4.5\pi\text{ ft}^3.
3. Total: 13.5\pi + 4.5\pi = 18\pi \approx 56.55\text{ ft}^3.
4. Convert: 56.55 \times 7.48 \approx 422.98, about 423 gallons.

Answer: about 423 gallons

Check: Estimate: 18 \times 3 = 54 ft³ and 54 \times 7.5 = 405; the real value is a bit more since \pi > 3 ✓.

Tip: A capsule shape (cylinder with a hemisphere on both ends) is "cylinder + one full sphere." That shortcut saves a step and prevents counting only one end.

1. Step 1 is correct: the cylinder's curved side is 2\pi r h = 24\pi.
2. Step 2 is correct: the bottom of the cylinder is exposed.
3. Step 3 is wrong: the top circle of the cylinder is covered by the hemisphere, so it is not on the outside.
4. Step 4 is correct: the hemisphere's curved surface is 2\pi r^2 = 8\pi.
5. Correct total: 24\pi + 4\pi + 8\pi = 36\pi \approx 113.1\text{ in}^2.

Answer: Step 3

Check: Rae's total 40\pi is exactly one circle (4\pi) more than 36\pi — the hidden face ✓.

Tip: For composite surface area, list only the faces you can touch from outside. Where two parts join, both touching faces disappear.

1. Step 1 is a true fact: the wall is 5 - 4 = 1 in thick.
2. Step 2 is wrong. The wall is a ring around the hole, not a thin cylinder of radius 1. The right method is outer cylinder − inner cylinder.
3. Outer: \pi(5^2)(20) = 500\pi. Inner: \pi(4^2)(20) = 320\pi.
4. Concrete: 500\pi - 320\pi = 180\pi \approx 565.5\text{ in}^3.

Answer: Step 2

Check: The ring's area is \pi(25 - 16) = 9\pi, not \pi(1^2) = \pi; times 20 in gives 180\pi ✓. Tom's answer is 9 times too small.

Tip: Never subtract the radii first. Square each radius, subtract the areas (or volumes), and only then multiply by the length.

1. Hemisphere: \tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\pi(27) = 18\pi.
2. Cylinder volume: 72\pi - 18\pi = 54\pi.
3. Cylinder height: \pi(3^2)h = 54\pi → 9h = 54 → h = 6 ft.
4. The dome is as tall as its radius, 3 ft.
5. Total height: 6 + 3 = 9 ft.

Answer: 9 ft

Check: \pi(9)(6) + \tfrac{2}{3}\pi(27) = 54\pi + 18\pi = 72\pi ✓.

Tip: In "work backwards" composite problems, subtract the part you can compute, solve for the unknown part, and then reread what the question wants — here the total height.

1. Radius: r = 5 \div 2 = 2.5 cm (for both parts).
2. Cone: \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(2.5)^2(11).
3. Hemisphere: half of \tfrac{4}{3}\pi r^3, which is \tfrac{2}{3}\pi(2.5)^3.
4. Total: \tfrac{1}{3}\pi(2.5)^2(11) + \tfrac{2}{3}\pi(2.5)^3.
5. Value: \approx 72.0 + 32.7 = 104.7\text{ cm}^3.

Answer: \tfrac{1}{3}\pi(2.5)^2(11) + \tfrac{2}{3}\pi(2.5)^3

Check: Each term has the right power: the cone has r^2 times a height, the hemisphere has r^3 ✓, and both use r = 2.5 ✓.

Tip: Eliminate wrong expressions one feature at a time: radius vs. diameter, \tfrac{1}{3} on the cone, \tfrac{2}{3} (half sphere) vs. \tfrac{4}{3} (whole sphere).

1. Cylinder's curved side: 2\pi r h = 2\pi(5)(12) = 120\pi\text{ ft}^2.
2. Hemisphere's curved top: 2\pi r^2 = 2\pi(25) = 50\pi\text{ ft}^2.
3. Leave out the bottom (on the ground) and the cylinder's top circle (covered by the hemisphere).
4. Total to paint: 120\pi + 50\pi = 170\pi \approx 534.1\text{ ft}^2.
5. Gallons: 534.1 \div 200 \approx 2.67. Two gallons are not enough, so buy 3.

Answer: 3 gallons

Check: 3 gallons cover 600\text{ ft}^2 \ge 534.1 ✓, but 2 gallons cover only 400\text{ ft}^2 ✗.

Tip: When you must buy enough of something, always round up — even 2.1 gallons means buying 3.