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Composite 3-D Figures

Find the volume and surface area of solids built from prisms, cylinders, pyramids, cones, and spheres.

Practice this skill 25 min Composite 3-D figures Leer en español
Key idea

What you'll learn

Real objects are rarely one simple shape. A grain silo is a cylinder with a dome on top. A pencil is a cylinder with a cone. A house is a box with a pyramid roof. A pipe is a cylinder with a cylinder-shaped hole. These are composite figures — solids built from two or more basic solids.

In this lesson you will learn to:

  • break a composite 3-D figure into prisms, cylinders, pyramids, cones, and spheres,
  • find its volume by adding parts (and subtracting holes),
  • find its surface area by counting only the surfaces on the outside,
  • work backwards to find a missing dimension of one part.

Why it matters on the GED: Composite figures are the GED's favorite way to make a 3-D question harder. You already know every formula you need — they are all on the formula sheet. The challenge is planning: which parts, add or subtract, and which faces are hidden.

Real life: How much grain fits in a silo? How much concrete is in a pipe? How much paint covers a shed with a roof? Same thinking.

Words to know

Key words

  • composite figure — a solid made by joining or cutting basic solids (box + pyramid roof).
  • decompose — break a figure into parts you know (a capsule = cylinder + 2 hemispheres).
  • hemisphere — half of a sphere (a dome): volume \tfrac{2}{3}\pi r^3, curved area 2\pi r^2.
  • shared (common) dimension — a length two parts have in common (a cone on a cylinder shares the radius).
  • hidden face (joint) — a surface where two parts are glued together; it is inside the figure.
  • lateral area — the area of the sides only, without bases (cylinder side 2\pi r h, cone side \pi r s).
  • hollow / with a hole — a solid with a part removed; its volume = outer solid − removed solid.
  • volume — space inside (cubic units); surface area — total outside area (square units).

All Math words in the glossary

Step 1: Break the figure into basic solids

Every composite problem starts the same way: name the parts. Look for lines where one shape ends and another begins.

FigureParts
silocylinder + hemisphere
pencil / rocketcylinder + cone
house or shedrectangular prism + pyramid
ice-cream conecone + hemisphere
capsule / pillcylinder + 2 hemispheres (= 1 sphere)
pipe / washer / block with a holebig solid − cylinder
stacked blocks / stepsprism + prism

Then label each part with its own dimensions. Watch for shared dimensions:

  • A cone on a cylinder uses the same radius.
  • A dome on a silo has the same radius as the silo, and its height equals that radius.
  • A pyramid roof's base is the top of the box below it.
  • A ball that fits exactly in a box has a diameter equal to the box's edge.

Example: A rocket is 14 cm tall: a 10 cm cylinder and a cone. The cone's height must be 14 - 10 = 4 cm.

Rocket made of a cylinder 10 cm tall with a cone on top. The whole rocket is 14 cm tall, so the cone is 14 minus 10 = 4 cm tall. The cone and cylinder share the same radius. 10 cm ? = 4 cm 14 cm cone height 14 − 10 = 4 cm cylinder

Step 2: Volume — add the parts, subtract the holes

V_{\text{total}} = V_{\text{part 1}} + V_{\text{part 2}} + \dots - V_{\text{holes}}

Example (add): a cube 6 in on a side with a square pyramid on top, 4 in tall. Cube: 6^3 = 216. Pyramid: \tfrac{1}{3}(36)(4) = 48. Total: 216 + 48 = 264\text{ in}^3.

A cube with 6-inch edges with a square pyramid 4 inches tall on top; the pyramid base is the top of the cube. Volume = 216 + 48 = 264 cubic inches. 6 in 6 in 4 in pyramid: ⅓(36)(4) = 48 cube: 6³ = 216 total: 216 + 48 = 264 in³

Example (subtract): a pipe 6 ft long with outer radius 2 ft and inner radius 1.5 ft. Outer: \pi(4)(6) = 24\pi. Hole: \pi(2.25)(6) = 13.5\pi. Concrete: 24\pi - 13.5\pi = 10.5\pi \approx 33.0\text{ ft}^3.

Concrete pipe 6 feet long with outer radius 2 feet and inner radius 1.5 feet. Concrete volume = outer cylinder 24 pi minus hole 13.5 pi = 10.5 pi, about 33.0 cubic feet. 2 ft 1.5 ft 6 ft outer: π(2²)(6) = 24π hole: π(1.5²)(6) = 13.5π concrete: 24π − 13.5π = 10.5π ≈ 33.0 ft³

Keep \pi as a label. If every part has a \pi, add or subtract the numbers in front of it: 540\pi + 144\pi = 684\pi. Turn it into a decimal once, at the end. This avoids rounding errors.

Do not use the outer box. A shape that is 10 in wide at the bottom and 6 in wide at the top is not a 10-in box all the way up. Find each piece separately.

Two stacked blocks: a bottom block 10 by 10 by 4 and a centered top block 6 by 6 by 4. Volume = 400 + 144 = 544 cubic units. The dashed red outline is a 10 by 10 by 8 box (800) that is NOT the figure. 10 10 4 4 6 bottom: 10 × 10 × 4 = 400 top: 6 × 6 × 4 = 144 total = 544 not the dashed box 10 × 10 × 8 = 800

Pipe trap: subtract the volumes, not the radii. (2 - 1.5)^2 = 0.25 is wrong; 2^2 - 1.5^2 = 1.75 is right.

Step 3: Surface area — only what is on the outside

Surface area is trickier, because the place where two parts are joined is hidden. You do not add the full surface areas of the parts.

Method: list every surface you could touch or paint from outside, find each one, and add.

Example: a cylinder (r = 3, h = 10) with a cone on top (s = 5), bottom included.

SurfaceFormulaArea
bottom circle\pi r^29\pi
side of cylinder2\pi r h60\pi
side of cone\pi r s15\pi
top of cylinder / base of conehidden0
total84\pi \approx 263.9
Exploded view of a cylinder with radius 3 and height 10 and a cone on top with slant height 5. Outside surfaces: bottom circle 9 pi, cylinder side 60 pi, cone side 15 pi. The top of the cylinder and the base of the cone are glued together, so they are hidden and not counted. Total 84 pi. s = 5 r = 3 h = 10 hidden joint cone side: π(3)(5) = 15π joint: hidden = 0 cyl. side: 2π(3)(10) = 60π bottom: π(3²) = 9π total = 84π ≈ 263.9

Example: a shed, box 8 × 8 × 6 with a pyramid roof (slant height 5), floor included: walls 4(8 \times 6) = 192, floor 64, roof \tfrac{1}{2}(32)(5) = 80. Total 336\text{ ft}^2. The box's top is hidden.

Shed: a box 8 by 8 by 6 feet with a square pyramid roof whose slant height is 5 feet. Walls 4 times 8 times 6 = 192, floor 64, roof one-half times 32 times 5 = 80; total 336 square feet. The top of the box is hidden under the roof. s = 5 8 ft 8 ft 6 ft roof: ½(32)(5) = 80 box top: hidden walls: 4(8 × 6) = 192 floor: 8 × 8 = 64 total = 336 ft²

Holes add surface. Drilling a hole of radius 1 in through a 4-in block removes two circles (2\pi) but creates an inside wall (2\pi(1)(4) = 8\pi). Surface area goes up by 6\pi even though volume goes down.

A block 4 inches tall with a hole of radius 1 inch drilled straight through. Drilling removes two circles from the top and bottom (2 pi in total) but adds an inside wall of 2 pi times 1 times 4 = 8 pi, so the surface area goes up by 6 pi. r = 1 4 in lose 2 circles: 2 × π(1²) = 2π gain inside wall: 2π(1)(4) = 8π SA goes up 6π

Useful pieces: cylinder side 2\pi r h; hemisphere curved part 2\pi r^2; two hemispheres make one sphere, 4\pi r^2.

Working backwards with composite figures

Sometimes the total is given and one dimension of one part is missing.

  1. Find the volume (or area) of every part you can compute.
  2. Subtract it from the total to get the unknown part's volume.
  3. Solve that part's formula for the missing length.
  4. Reread the question — does it want that length, or a total length?

Example: a tank is a cylinder (radius 3 ft) with a hemisphere on top. Total volume 72\pi. Hemisphere: \tfrac{2}{3}\pi(27) = 18\pi. Cylinder: 72\pi - 18\pi = 54\pi = 9\pi h, so h = 6. Total height = 6 + 3 = 9 ft.

Tank: a cylinder with radius 3 feet and unknown height h, topped by a hemisphere with radius 3. Total volume 72 pi. Hemisphere 18 pi, so the cylinder is 54 pi = 9 pi h and h = 6; total height 6 + 3 = 9 feet. r = 3 h = ? 3 ft total ? total V = 72π dome: ⅔π(27) = 18π cylinder: 72π − 18π = 54π = 9πh → h = 6 total height 6 + 3 = 9 ft
Formula

Formulas you combine (all on the GED formula sheet)

SolidVolumeSurface area
rectangular prismV = lwhSA = 2lw + 2lh + 2wh
cylinderV = \pi r^2 hSA = 2\pi r h + 2\pi r^2
pyramidV = \tfrac{1}{3}BhSA = \tfrac{1}{2}ps + B
coneV = \tfrac{1}{3}\pi r^2 hSA = \pi r s + \pi r^2
sphereV = \tfrac{4}{3}\pi r^3SA = 4\pi r^2

All of these are on the GED formula sheet. There is no single "composite formula" — you build one from these. Hemisphere (\tfrac{2}{3}\pi r^3, 2\pi r^2) and "lateral only" pieces (2\pi r h, \pi r s, \tfrac{1}{2}ps) come from these formulas by taking half or dropping bases.

Worked example

Example 1 (easy): cube with a pyramid on top

A cube with 6-in edges has a square pyramid on top. The pyramid's base matches the top of the cube, and the pyramid is 4 in tall. Find the total volume.

  1. Cube: 6 \times 6 \times 6 = 216\text{ in}^3.
  2. Pyramid: B = 6 \times 6 = 36; V = \tfrac{1}{3}(36)(4) = 48\text{ in}^3.
  3. Add: 216 + 48 = 264.
Example 1, pulled apart: a cube with 6-inch edges (216 cubic inches) and a square pyramid with a 6 by 6 base and height 4 inches (48 cubic inches). Together 264 cubic inches. h = 4 in 6 in 6 in 6 in pyramid B = 6 × 6 = 36 ⅓(36)(4) = 48 cube 6 × 6 × 6 = 216 216 + 48 = 264 in³

Answer: 264 in³

Check: the pyramid is one-third of a 6 × 6 × 4 box (144 \div 3 = 48) ✓.

Worked example

Example 2 (real world): capacity of a silo

A silo is a cylinder with radius 10 ft and height 30 ft, topped by a hemisphere. How many cubic feet does it hold?

  1. Cylinder: \pi(10^2)(30) = 3{,}000\pi.
  2. Hemisphere: \tfrac{2}{3}\pi(10^3) = \tfrac{2{,}000}{3}\pi \approx 666.67\pi.
  3. Total: 3{,}000\pi + 666.67\pi = 3{,}666.67\pi \approx 11{,}519.2.
Example 2: silo made of a cylinder with radius 10 feet and height 30 feet, topped by a hemisphere with radius 10 feet. Cylinder 3,000 pi plus hemisphere about 666.67 pi gives about 11,519 cubic feet. r = 10 10 30 ft hemisphere: ⅔π(10³) ≈ 666.67π cylinder: π(10²)(30) = 3,000π total ≈ 3,666.67π ≈ 11,519 ft³

Answer: about 11,519 ft³

Check: exact form \tfrac{11{,}000}{3}\pi; 11{,}000 \div 3 \times 3.14159 \approx 11{,}519.2 ✓.

Worked example

Example 3 (subtract): block with a hole

A wooden block is 12 in by 8 in by 5 in. A hole with radius 2 in is drilled through the 5-in height. How much wood is left?

  1. Block: 12 \times 8 \times 5 = 480\text{ in}^3.
  2. Hole: \pi(2^2)(5) = 20\pi \approx 62.83\text{ in}^3.
  3. Wood: 480 - 62.83 = 417.17.
Example 3: wooden block 12 by 8 by 5 inches with a hole of radius 2 inches drilled through the 5-inch height. Block 480 cubic inches minus hole 20 pi (about 62.83) leaves about 417.2 cubic inches of wood. r = 2 12 in 8 in 5 block: 12 × 8 × 5 = 480 hole: π(2²)(5) = 20π ≈ 62.83 wood ≈ 417.2 in³

Answer: about 417.2 in³

Check: a small hole should remove a small fraction: 62.8 \div 480 \approx 13\% ✓.

Worked example

Example 4: paint for a rocket-shaped bin

A storage bin is a cylinder (radius 4 ft, height 10 ft) with a cone on top that is 3 ft tall. The whole outside, including the bottom, will be painted. Find the area.

  1. Cone slant height: s^2 = 3^2 + 4^2 = 25, so s = 5 ft.
  2. Bottom: \pi(4^2) = 16\pi.
  3. Cylinder side: 2\pi(4)(10) = 80\pi.
  4. Cone side: \pi(4)(5) = 20\pi.
  5. The joint between cylinder and cone is hidden. Total: 16\pi + 80\pi + 20\pi = 116\pi \approx 364.4.
Example 4: storage bin made of a cylinder with radius 4 feet and height 10 feet and a cone on top 3 feet tall. The cone slant height is 5 feet (3-4-5 triangle). Painted area: bottom 16 pi + cylinder side 80 pi + cone side 20 pi = 116 pi, about 364.4 square feet; the joint is hidden. r = 4 s = 5 3 ft 10 ft cone side: π(4)(5) = 20π joint: hidden cyl. side: 2π(4)(10) = 80π bottom: π(4²) = 16π total = 116π ≈ 364.4 ft²

Answer: about 364.4 ft²

Check: three surfaces, each counted once; the cone used s = 5, not h = 3 ✓.

Worked example

Example 5 (GED-hard, work backwards): length of a capsule tank

A fuel tank is a cylinder with radius 3 m and a hemisphere on each end. Its total volume is 99\pi m³. How long is the whole tank, end to end?

  1. Two hemispheres = one sphere: \tfrac{4}{3}\pi(27) = 36\pi.
  2. Cylinder: 99\pi - 36\pi = 63\pi.
  3. \pi(3^2)h = 63\pi → 9h = 63 → h = 7 m.
  4. Each end adds its radius, 3 m. Total length: 3 + 7 + 3 = 13 m.
Example 5: capsule-shaped fuel tank lying on its side: a cylinder of radius 3 m and unknown length h with a hemisphere on each end. Total volume 99 pi; the two hemispheres make one sphere, 36 pi, so the cylinder is 63 pi and h = 7 m. End to end: 3 + 7 + 3 = 13 m. 3 3 3 h = 7 3 total length = 3 + 7 + 3 = 13 m

Answer: 13 m

Check: 9\pi(7) + 36\pi = 63\pi + 36\pi = 99\pi ✓.

Calculator tip

On the TI-30XS MultiView

  • Add the \pi-coefficients first, then press π once. Silo: 540 + 144 = 684, then 684 × π enter → 684\pi; press toggle → 2148.849…
  • Or use parentheses for the whole setup: π × 6 x² × 15 + 2 n/d 3 ▶ × π × 6 ^ 3 enter.
  • Square roots for slant heights: 2nd x² then the expression, e.g. 3 x² + 4 x² → 5.
  • Use the toggle key to switch an answer like \tfrac{40}{3}\pi between fraction and decimal.
  • Write each part on your erasable note board before typing. The calculator cannot tell you which surfaces are hidden — that is the part the GED is really testing.
  • Answer choices with \pi (like 102\pi or 248 + 6\pi) usually need no calculator at all.
Common mistake

Common mistakes

  • Using one big box for the whole figure. Stacked blocks of different sizes must be found separately: 400 + 144 = 544, not 10 \times 10 \times 8 = 800.
  • Counting hidden faces in surface area. The joint between a cone and a cylinder (or a roof and a box) is inside. Adding each part's full SA double-counts two circles or squares.
  • Adding instead of subtracting a hole (or the reverse). Holes and hollow centers are subtracted for volume.
  • Subtracting radii in a pipe. (2 - 1.5)^2 \ne 2^2 - 1.5^2. Find each volume first.
  • Using a whole sphere when the figure has a hemisphere, or forgetting that two hemispheres make one sphere.
  • Mixing height and slant height in the cone or pyramid part. Volume uses h; surface area uses s.
Test strategy

GED test strategy

  • Sketch and label each part on your erasable note board. Write the formula for each part before plugging in numbers.
  • Decide add or subtract for every part, and for surface area, cross out hidden faces.
  • Look for shared dimensions — they are often not labeled twice (the dome's radius is the silo's radius).
  • Estimate: the total should be bigger than the largest part (when adding) and smaller than the outer solid (when subtracting).
  • Expression answers: if the choices are setups like \pi(3)^2(10) + \tfrac{1}{3}\pi(3)^2(4), write your own setup first, then match it term by term.
  • Time: composite problems take 2–3 minutes. If stuck, flag it and come back.

Check yourself

Composite 3-D figures

Two rectangular wooden blocks are glued together as shown.

Two rectangular blocks stacked. The bottom block is 10 in long, 10 in wide and 4 in tall. The top block, sitting on the back-left corner of the bottom block, is 6 in long, 6 in wide and 4 in tall. 10 in 10 in 4 in 6 in 6 in 4 in

What is the total volume of the figure?

Answer choices
Hint

Find the volume of each block separately, then combine them.

Check yourself

Composite 3-D figuresFill in

A solid ice-cream cone model is made of a cone topped with a hemisphere. The radius of both parts is 2 in, and the cone is 6 in tall.

Ice-cream cone model: a cone pointing down with a hemisphere on top. The cone and the hemisphere share a circular edge with radius 2 in. The cone is 6 in tall from that circle to its tip. 2 in 6 in

What is the volume of the model, to the nearest tenth of a cubic inch?

in³

Type your answer in the box. Fractions like 3/4 are fine.

Hint

Volume = cone + half of a sphere. Both use r = 2.

Summary

  • Break the figure into basic solids and label each part's dimensions, including shared ones.
  • Volume: add the parts and subtract any holes. Keep \pi until the end.
  • Surface area: add only the outside surfaces; joints where parts meet are hidden. Holes can add surface.
  • Use the formula-sheet formulas for each part; hemisphere = half a sphere, two hemispheres = one sphere.
  • Working backwards: subtract the known parts from the total, solve for the missing length, then reread what is asked.
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