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MK · lesson 13 of 14 · 45 min

Area, Surface Area and Volume

Is the question asking about a boundary, a covering or a capacity? This advanced lesson opens with that distinction and explains why units come out squared or cubed. It then reviews flat area for rectangles, triangles, parallelograms, trapezoids, circles and annuli, along with composite and shaded regions and how area scales.

Reading · 45 minVideo · 13 minPractice questions (48)

Video lesson: Area, Surface Area and Volume

Video transcript

0:00 Hey there, if you're gearing up for the ASVAB and feeling, you know, maybe a little uneasy about the math section, you are in exactly the right place. I know math can be super intimidating, especially when you can't just reach for a calculator to do the heavy lifting for you. But honestly, don't worry. This explainer is basically your ultimate calculator free toolkit for completely conquering ASVAB measurement questions. We're going to break down area, surface area, and volume into really simple logical steps that'll make the mental math an absolute breeze. All right, let's jump straight into section one, the basics in flat area, focusing on boundary, covering, and capacity. Knowing the difference here is exactly how you'll always know which formula to pick. So what's really interesting here is how these flat 2D area formulas literally build the foundation for everything else we're going to do. You've got your rectangle, where area just equals length times width, for a square, it's just the side squared. Now, a triangle is simply half of a parallelogram. So its area is one half base times perpendicular height. We also have parallelograms, which are base times height, and trapezoids, where you average the two bases and then multiply by the height.

1:06 And then, of course, there's a circle, area equals pi r squared. Now, here is a massive mental math trick for the test. Just keep pi as a symbol. Seriously, do not try to multiply by 3.14 in your head, unless the multiple choice answers are actually given in decimals. Leaving it as the pi symbol is going to save you so much time. And for an annulus, which is just a flat ring, you literally just subtract the inner circles area from the outer circles area. Easy. 3D, volume, and surface area. We're stacking into the third dimension now, and you'll see how taking those flat shapes and just stretching them out gives us volume. Here is our lineup of 3D volume formulas. For a rectangular prism, or just a basic box, volume is simply length times width times height. For any prism or cylinder, the volume is just the area of the base. We call that big B times the height. You just find that flat base area first, and then multiply by how tall it is. Now here's a real game changer you'll want to memorize. Look at the pyramid and the cone. The crucial pattern here is that pointed solids always hold exactly one-third of their matching prism or cylinder.

2:10 No way around it. A pyramid is one-third big B times height, a cone is one-third pi r squared times height, and finally, a sphere's volume is four-thirds pi r cubed. I know, surface area looks super intimidating, but I promise it really isn't. Surface area is just unfolding the shape perfectly flat into a net and adding up the pieces. Think about a rectangular box. It just has three pairs of matching rectangle faces. You just add them up. A closed cylinder, unroll it, and you get two circular ends, and one big rectangle for the curved side. So the formula is just two pi r squared plus two pi r h. But, and this is a big but, read the question carefully. If it's an open-top container, like a cup, you have to leave out that missing top circle, making it just one pi r squared plus the side. Okay, section three, step-by-step worked examples. Let's actually see these rules in action. This is the safe space to see exactly how these work without touching a calculator. Let's imagine an L-shaped floor. It's a 16 by 10-foot rectangle, but it has a 6 by 4-foot corner completely missing. How do we find that area?

3:15 Well, you've got two great options here. First is the subtract method. You just pretend it's a full rectangle, 16 times 10 is 160, and then subtract the missing 6 by 4 chunk, which is 24, 160 minus 24, or you can use the split method. Break the L-shape into two separate rectangles, 16 by 6, and 10 by 4, which gives you 96 plus 40. Either way, you slice it, your total area comes out to exactly 136 square feet. Now, let's take a standard shipping box that's 8 by 5 by 4 inches. Let's find both the volume and the surface area to see the difference. For volume, it's super simple. Length times width times height. Eight times five times four. For surface area, we got to pair up the sides. Two times the quantity of 40 plus 32 plus 20. The volume comes out to 160 cubic inches, while the surface area is 184 square inches. Notice how the exact same box gives you completely different numbers and units, depending on whether you're measuring its capacity to hold things or the cardboard covering the outside. Picture a closed drum with radius of 5 feet and a height of 9 feet.

4:22 We need the total surface area. First, let's do the two flat circular ends. That's 2 times 25 pi, which gives us 50 pi. Then the curved side, or on the middle, is 2 pi times 5 times 9, which is 90 pi. Then you literally just add them together, 50 pi plus 90 pi equals 140 pi square feet. Do you see how easy that math becomes when you just let pi sit there as a symbol instead of trying to multiply out crazy decimals? Now what if we have an open top tank that's 8 feet long, 4 feet wide, and 3 feet high? Let's calculate those faces. The bottom floor is 8 times 4, which is 32. The two long walls are 2 times 8 times 3, which is 48. And the two short walls are 2 times 4 times 3, making 24. Add them all up. 32 plus 48 plus 24 equals 104 square feet. The key takeaway here, always picture the object physically in your mind. It's an open top, so we intentionally skip that top rectangle entirely. All right, let's look at a prism where the base is a right triangle with legs of 6 centimeters and 8 centimeters, and a hypotenuse of 10 centimeters.

5:28 The prism's total length is 15 centimeters. The big trick here is to find the flat base area first. One half of 6 times 8 is 24. Then for the volume, you just multiply that base area by the length. 24 times 15 is 360 cubic centimeters. Now for the surface area, the sides form a giant rectangle, equal to the perimeter, which is 24 times the length 15, giving us 360. Add in our two triangular bases, which are 48, to get a total surface area of 408 square centimeters. This next one is absolutely crucial. Imagine a cone with a radius of 6 inches, and a slant height of 10 inches. Remember, the slant height traces the tilted side, but volume needs the straight up vertical height. So using a right triangle, our vertical height h equals the square root of 10 squared minus 6 squared, which works out beautifully to 8. So volume is 1 third pi times 36 times 8, equaling 96 pi cubic inches. The surface area, however, does use that slant height. The curved surface is pi times 6 times 10, equaling 60 pi square inches.

6:36 Let's check out a square pyramid. This one has a 16 meter base and a vertical height of 6 meters. Its volume is 1 third times the base area, 256 times the height, 6. And here is a pro tip, divide before you multiply. 1 third of 6 is 2, and 256 times 2 is 512 cubic meters. See? Keeps the numbers super manageable for your brain. For surface area, the slant height spans half the base, which is 8, and the vertical height 6. It forms a triangle, giving us a slant height of 10. The total surface area is the base, 256, plus the four triangular faces, which total 320. That gives us 576 square meters altogether. What if we have two 5 centimeter cubes that are glued face to face, and we want to know the exposed surface area? Separately, two cubes would give us 12 faces, but when you glue them together, those two touching faces literally disappear from the surface area calculation. They're hidden. That leaves 10 exposed faces. Each square face has an area of 25 square centimeters. 10 times 25 is 250 square centimeters.

7:42 The rule of thumb? Always, always count only what is exposed on the outside. Here's a fantastic shortcut that will save you so much time on the test. Let's say a tank has a volume of 15 cubic meters, and a surface area of 40 square meters, and every single dimension is doubled. That means our scale factor is two. So, surface area always grows by that factor squared. Two squared is four, so 40 times four equals 160 square meters. Volume, on the other hand, always grows by that factor cubed. Two cubed is eight, so 15 times eight equals 120 cubic meters. It's a hard and fast rule. Lengths double, areas multiply by four, and volumes multiply by eight. What if the test works backward? Say they give you the volume and ask for a missing piece. Say a cylinder holds 245 pi cubic feet and is 5 feet tall, and you need to find the radius. Just work backward by plugging in what you already know. 245 pi equals pi times r squared times 5. To solve it, just divide both sides by 5 pi. That beautifully cancels out leaving r squared equals 49.

8:48 So, the radius r is 7 feet. It's really just a puzzle. We have a concrete slab 12 feet by 9 feet and 6 inches thick. Be super careful here. Notice the mixed units. 6 inches is half a foot or 0.5 feet. So volume equals 12 times 9 times 0.5, which is 54 cubic feet. But what if we need the answer in cubic yards? Remember, one cubic yard is 3 feet by 3 feet by 3 feet, which is 27 cubic feet. So 54 divided by 27 is exactly two cubic yards of concrete. The big lesson? Always convert your measurements to the exact same unit before you even start your calculation. Most common traps. Let's talk about how to dodge the favorite tricks that the test writers have up their sleeves. Pay close attention because these are the most common pitfalls. First up, slant height versus vertical height. Volume requires the vertical height. Surface area traces the slant height. Please don't mix them up. Second, using diameter instead of radius. Always, always halve the diameter before you square or cube anything.

9:54 Third, missing the one third. Pyramids and cones are pointy, meaning they hold exactly one third of a matching box or cylinder. Fourth, counting missing faces on open tops. If a box has no lid, don't calculate an area for a lid that isn't there. Fifth, converting by the linear factor. Remember, one square foot is not 12 square inches. It is 144. You have to square or cube the conversion. And lastly, using the wrong units entirely. Area is square units. Volume is cubic units. All right, section five, check your self questions. Time to put your mental math to the test. I'm going to read the question, give you a brief second to think about it, and let you guess the answer in your head before I reveal the solution. A triangular garden bed has a base of 12 meters and a perpendicular height of 7 meters. What is its area? Take a quick second and do the math in your head. Got it? OK, the answer is 42 square meters. Why? Well, because the area of a triangle is one half base times height. Take half of that 12 first to make it super easy on yourself. Half of 12 is 6, and 6 times 7 is 42.

11:00 Last one, a storage box is six inches long, four inches wide, and five inches high. What is its volume? Go ahead, do the math in your head right now. Did you get it? The answer is 120 cubic inches. We need capacity here, so we use volume equals length times width times height. Six times four is 24, and 24 times five is 120. You are now officially ready to tackle these ASVAB questions with total confidence. You don't need a calculator, you just need the structure we talked about today. But I want to leave you with this final thought. The next time you look at a building, a water tower, or even just a simple box of cereal in your pantry, ask yourself, are you looking at its perimeter, its surface area, or its volume? Start seeing the math in the real world around you, and this test will honestly feel like a breeze.

Contents · 48 sections
  1. Lesson Goal and MK Context
  2. Three Different Questions: Boundary, Covering, Capacity
  3. Why Units Are Squared or Cubed
  4. Rectangles and Squares
  5. Triangles: Base and Perpendicular Height
  6. Parallelograms
  7. Trapezoids
  8. Circle Area as a Prior Skill
  9. Annuli and Holes in Flat Regions
  10. Composite 2D Shapes: Split or Subtract
  11. Shaded Areas and Multiple Regions
  12. Scale Factors for Flat Area
  13. What Counts as a Three-Dimensional Solid
  14. Rectangular Prisms: Volume
  15. Prisms: Base Area × Height
  16. Cylinders: Volume
  17. Pyramids: One-Third of a Prism
  18. Cones: One-Third of a Cylinder
  19. Spheres: Volume
  20. Surface Area From Nets
  21. Rectangular Prism and Cube Surface Area
  22. Right Prisms: Lateral Area and Total Surface Area
  23. Cylinders: Curved and Total Surface Area
  24. Open-Top and Open-Bottom Containers
  25. Pyramids: Surface Area and Slant Height
  26. Cones: Curved and Total Surface Area
  27. Spheres: Surface Area
  28. Composite Solids: Add and Subtract Volume
  29. Composite Surface Area: Count Only Exposed Faces
  30. Unit Conversion: Linear, Square and Cubic
  31. Capacity, Fill Level and Displacement
  32. Scaling Three-Dimensional Solids
  33. Reverse Formulas and Solving for a Dimension
  34. No-Calculator Area and Volume Routine
  35. Formula Recognition and Fast Mental Arithmetic
  36. Worked Example 1 — Composite L-Shaped Floor
  37. Worked Example 2 — Box Volume Versus Surface Area
  38. Worked Example 3 — Closed Cylinder
  39. Worked Example 4 — Open-Top Tank
  40. Worked Example 5 — Triangular Prism
  41. Worked Example 6 — Cone Height Versus Slant Height
  42. Worked Example 7 — Square Pyramid
  43. Worked Example 8 — Two Joined Cubes
  44. Worked Example 9 — Scale Change
  45. Worked Example 10 — Reverse Cylinder Dimension
  46. Worked Example 11 — Cubic Unit Conversion
  47. Most Common Traps
  48. Final MK Review and Self-Check
Visual lesson

Lesson slides

15 slides
Area, Surface Area and Volume – slide 1 of 15

Lesson Goal and MK Context

This lesson pulls together the measurement formulas the Mathematics Knowledge test uses most. You will find areas of flat shapes, including composite regions and shapes with holes, and then move to solids: prisms, cylinders, pyramids, cones and spheres.

For each solid you will find volume and surface area, handle open containers and joined shapes, solve a formula backward for a missing dimension, and convert square and cubic units correctly. All work is set up for mental or pencil arithmetic.

Three Different Questions: Boundary, Covering, Capacity

Most measurement problems ask one of three things. How long is the edge? That is perimeter, in plain units. How much material covers the surface? That is area or surface area, in square units. How much fits inside? That is volume, in cubic units.

Picture a raised garden bed: the fence around it is a perimeter, the plastic liner on its sides and bottom is a surface area, and the soil that fills it is a volume. Decide which question you face before choosing a formula.

Raised garden bed: the fence along its top edge is a perimeter (ft), the soil filling it is a volume (ft³), and the liner on its sides and bottom is a surface area (ft²).

Why Units Are Squared or Cubed

Area multiplies two lengths, so its unit is a length times a length, such as square feet. Volume multiplies three lengths, giving cubic units such as cubic centimeters.

A square foot is a 1 ft by 1 ft tile; a cubic foot is a 1 ft by 1 ft by 1 ft block. The unit tells you what kind of quantity an answer is.

A square foot as a 1 ft by 1 ft tile next to a cubic foot as a 1 ft by 1 ft by 1 ft block drawn in oblique view.

Rectangles and Squares

For a rectangle of length l and width w, and a square of side s:

A cutting board 8 inches by 5 inches covers 40 square inches. A square window 7 feet on a side has area 49 square feet. Both lengths must be in the same unit.

Cutting board 8 in by 5 in drawn as an 8 by 5 grid of 1-square-inch squares; A = 8 × 5 = 40 in².

Triangles: Base and Perpendicular Height

Every triangle is half of a parallelogram with the same base and height:

The height must meet the base, or the line containing it, at a right angle. In a right triangle the two legs serve as base and height. In an obtuse triangle the height may fall outside the shape. A triangle with base 10 and height 6 has area 30.

Triangle with base b = 10 and dashed perpendicular height h = 6, area ½ × 10 × 6 = 30, beside an obtuse triangle whose height falls outside the shape.

Parallelograms

A parallelogram's area is base times perpendicular height, A = bh. Slicing off a triangle from one end and moving it to the other end makes a rectangle with the same base and height.

The slanted side is not the height. A parallelogram with base 12 and height 4 covers 48 square units, whatever its side length.

Parallelogram with base 12 and dashed perpendicular height 4; the triangle cut from the left end moves to the right end to form a 12 by 4 rectangle of area 48.

Trapezoids

A trapezoid's area uses the average of its two parallel bases times the perpendicular height:

With bases 5 and 9 and height 6, the average base is 7, so the area is 42.

Trapezoid with parallel bases 5 and 9 and dashed height 6; a dashed midline of length 7 shows the average base.

Circle Area as a Prior Skill

Circle area is needed for cylinders, cones and many composite shapes:

A disk of radius 4 has area 16 pi. If a diameter of 10 is given, use radius 5, for 25 pi. Keep pi in the answer unless the choices are decimals. The circle's boundary is 2 pi r, a length, not an area.

Circle with diameter d = 10 drawn across it and radius r = 5 from the centre; use r = 5, so A = 25π.

Annuli and Holes in Flat Regions

For a flat ring, or any shape with a hole, subtract the hole from the whole. For a ring with outer radius R and inner radius r:

A washer with R of 7 and r of 2 has area pi times 45, or 45 pi.

Ring with outer radius R = 7 and inner radius r = 2, R labelled beside its radius line; ring area 49π − 4π = 45π.

Composite 2D Shapes: Split or Subtract

A composite region can be found two ways. Split it into rectangles, triangles or other basic pieces that do not overlap, then add. Or surround it with a simple outer shape and subtract the missing pieces.

Example: a 14 by 10 foot room has a 4 by 3 foot corner taken by a closet. The floor area is 140 minus 12, or 128 square feet. Use whichever method needs fewer pieces.

A 14 ft by 10 ft room with a 4 ft by 3 ft closet corner cut out; floor area 140 − 12 = 128 ft².

Shaded Areas and Multiple Regions

For shaded regions, find the large area and subtract every unshaded piece. Combine identical pieces first.

Example: a square of side 8 has a quarter circle of radius 4 drawn in each corner. Four quarter circles make one full circle, 16 pi. The shaded area left is

or about 13.7 square units.

Square of side 8 with a quarter circle of radius 4 at each corner; the shaded region left is 64 − 16π, about 13.7.

Scale Factors for Flat Area

If every length of a flat figure is multiplied by k, its area is multiplied by k squared.

A 4 by 6 inch photo has area 24. Enlarged 3 times, it is 12 by 18, with area 216, which is 9 times 24. Lengths triple, area multiplies by 9.

A 6 in by 4 in photo (area 24) enlarged 3 times to 18 in by 12 in, which holds 9 copies of the small photo: 9 × 24 = 216.

What Counts as a Three-Dimensional Solid

A solid takes up space in three directions: length, width and height. Polyhedra have flat polygon faces meeting along edges and corners, as in prisms and pyramids.

Other solids have curved surfaces: a cylinder has two circular bases and a curved side, a cone has one circular base rising to a point, and a sphere is every point a fixed distance from a center. A prism has two matching parallel bases; a pyramid has one base and a tip called the apex.

Five solids: prism with two matching parallel triangular bases, pyramid with a square base and an apex, cylinder with two circular bases, cone with one circular base and an apex, and sphere with centre and radius r.

Rectangular Prisms: Volume

A rectangular box holds length times width times height unit cubes:

A planter 10 by 4 by 3 feet holds 120 cubic feet of soil. A cube with edge s has volume s cubed.

Rectangular box 10 ft by 4 ft by 3 ft drawn in isometric view with unit-cube grid lines and hidden edges dashed; it holds 120 unit cubes.

Prisms: Base Area × Height

Every prism, whatever the shape of its base, has volume equal to base area times height: V = Bh. Here B is the area of one base and h is the distance between the two bases.

Find B with the matching flat formula first. A prism with a hexagonal base of area 20 square inches and a height of 7 inches holds 140 cubic inches.

Upright hexagonal prism with its top base shaded and labelled B = 20 in² and height h = 7 in; V = Bh = 20 × 7 = 140 in³.

Cylinders: Volume

A cylinder is like a prism with a circular base, so its volume is base area times height:

A can with radius 5 cm and height 4 cm holds 25 pi times 4, or 100 pi cubic centimeters. Use the radius, not the diameter.

Cylinder with radius 5 cm and height 4 cm; the hidden back of the bottom circle is dashed.

Pyramids: One-Third of a Prism

A pyramid holds exactly one third of the prism with the same base and height:

A pyramid with a 9 by 9 base and height 5 has volume 81 times 5 over 3, or 135 cubic units.

Square pyramid with a 9 by 9 base and height 5 inside a see-through prism with the same base and height; the pyramid is ⅓ of the prism: ⅓ × 81 × 5 = 135.

Cones: One-Third of a Cylinder

A cone holds one third of the cylinder with the same base and height:

Here h is the straight-up height from base to tip. A cone with radius 6 and height 5 holds one third of 180 pi, or 60 pi.

Cone with radius 6 and height 5 drawn inside the cylinder with the same base and height; the cone is ⅓ of the cylinder: ⅓ of 180π = 60π.

Spheres: Volume

A sphere of radius r has volume

Cube the radius first. For r of 6, r cubed is 216, and four thirds of 216 is 288, so V is 288 pi.

Sphere with radius 6 from the center; r³ = 216 and V = 288π.

Surface Area From Nets

Surface area is the total area of every outside face, as if you wrapped the solid in paper. A net is the solid unfolded flat into connected faces.

Unfolding a box gives six rectangles in matching pairs. Unrolling a cylinder gives two circles and one rectangle. Find each face's area and add them, counting every face exactly once.

Net of a 5 by 3 by 2 box: six rectangles in a cross, matching pairs in the same colour (two 15s, two 10s, two 6s), with the folded box beside it.

Rectangular Prism and Cube Surface Area

A box has three pairs of matching faces, so SA = 2(lw + lh + wh). A box 5 by 3 by 2 has surface area 2 times (15 plus 10 plus 6), or 62. A cube has six equal squares, so SA = 6s squared.

Right Prisms: Lateral Area and Total Surface Area

The lateral area of a right prism is the area of its side faces only. Laid flat, the sides form one rectangle whose width is the base perimeter, so lateral area equals perimeter times height.

Total surface area adds the two bases. A prism with a 3-4-5 right triangle base and length 10 has lateral area 12 times 10, or 120. Each base is 6, so the total is 132.

Prism with a 3-4-5 right triangle base and length 10, and its net: side faces 3, 4 and 5 wide form one 12 by 10 rectangle (area 120), plus two triangle bases of area 6.

Cylinders: Curved and Total Surface Area

Unrolled, a cylinder's side is a rectangle whose width is the circumference. So the curved area is 2 pi r h. A closed cylinder adds two circular ends:

With r of 3 and h of 7, the side is 42 pi and the total is 60 pi.

Net of a closed cylinder with r = 3 and h = 7: two circles of area 9π each and a rectangle 6π wide (2πr) and 7 tall with area 42π.

Open-Top and Open-Bottom Containers

Leave out any face that is missing. An open-top can has one circular base plus the curved side:

A cup with r of 3 and h of 10 needs 9 pi plus 60 pi, or 69 pi. A pipe open at both ends has only the curved side, 2 pi r h. An open box drops its top face.

Open-top cup with radius 3 and height 10: the top is an opening showing the inside wall, bottom circle 9π and curved side 60π, total 9π + 60π = 69π.

Pyramids: Surface Area and Slant Height

A square pyramid's surface is its base plus four triangles. Each triangle uses the slant height l, measured up the middle of a face, not the vertical height h.

The slant height comes from h and half the base: a base of 10 and height 12 give l of 13, so SA is 100 plus 260, or 360.

Square pyramid with base edge 10 labelled on the front base edge, vertical height h = 12 and slant height l = 13 up the middle of a face; h, half the base 5 and l form a right triangle; SA = 100 + 260 = 360.

Cones: Curved and Total Surface Area

A cone's curved surface uses its slant height l, the distance from the tip down the side to the base edge:

If r is 5 and the height is 12, then l is 13. The curved part is 65 pi and the total is 90 pi.

Cone with radius 5 and height 12; the slant height l = 13 runs from the tip down the side; curved area 65π, total 90π.

Spheres: Surface Area

A sphere's surface area is four times the area of its largest circle:

A ball of radius 7 has surface area 196 pi.

Ball of radius 7 with its largest circle marked, equal to 4 times a disk of area 49π; SA = 4 × 49π = 196π.

Composite Solids: Add and Subtract Volume

Break a composite solid into basic solids. Add volumes of pieces joined together and subtract volumes that are hollowed out.

Example: a grain silo is a cylinder of radius 3 and height 10 topped with a half sphere of radius 3. The cylinder holds 90 pi. The half sphere holds half of 36 pi, or 18 pi. Together they hold 108 pi cubic units.

Grain silo: a cylinder of radius 3 and height 10 (90π) topped by a half sphere of radius 3 (18π), total 108π.

Composite Surface Area: Count Only Exposed Faces

When solids are joined, the touching faces are hidden and must be removed from both pieces.

Example: a 4 cm cube sits on top of a 10 by 10 by 2 cm slab. The slab's area is 280 and the cube's is 96. The cube's bottom and the patch of slab it covers are each 16, so the exposed area is 376 minus 32, or 344.

A 4 cm cube lifted above a 10 by 10 by 2 cm slab to show the 16 cm² patch it covers; slab 280, cube 96, touching faces 16 + 16 = 32, exposed area 376 − 32 = 344 cm².

Unit Conversion: Linear, Square and Cubic

A length factor is squared for area and cubed for volume. Since 1 foot is 12 inches, 1 square foot is 144 square inches and 1 cubic foot is 1728 cubic inches.

Since 1 yard is 3 feet, 1 cubic yard is 27 cubic feet, so 54 cubic feet is 2 cubic yards.

A 1 yd cube with 3 ft edges split into 3 × 3 × 3 = 27 one-foot cubes, and a 1 ft square split into a 12 by 12 grid: 1 ft² = 144 in².

Capacity, Fill Level and Displacement

Capacity is the volume a container can hold. In metric units, 1 liter equals 1000 cubic centimeters.

So a tank 50 by 30 by 40 cm holds 60 liters, and filled three fourths of the way it holds 45 liters.

A submerged object pushes the water up by its own volume. If a rock raises the water in that tank by 2 cm, its volume is 50 times 30 times 2, or 3000 cubic centimeters.

Tank 50 cm by 30 cm by 40 cm with a rock on the bottom; the water rises 2 cm and the risen layer, 50 × 30 × 2 = 3000 cm³, is shaded.

Scaling Three-Dimensional Solids

When every length of a solid is multiplied by k, surface area is multiplied by k squared and volume by k cubed:

Doubling all dimensions of a box gives 4 times the paint and 8 times the space. Halving them leaves one eighth of the volume.

A cube with edge 1 beside a cube with edge 2 built from 2 × 2 × 2 = 8 small cubes; one face holds 4 small squares (area × 4), the volume is 8 cubes (volume × 8).

Reverse Formulas and Solving for a Dimension

To find a missing dimension, write the formula, substitute what you know, and solve.

Example: a box has volume 180 cubic inches with a 6 by 5 inch base.

Box with a 6 in by 5 in base and unknown height h = ?; V = 180 in³.

So 30h equals 180 and h is 6 inches.

No-Calculator Area and Volume Routine

First, name the shape and decide whether the question asks for length, area or volume. Second, convert everything to one unit and change diameters to radii. Third, write the formula symbolically, for example

then substitute. Simplify the one-third or one-half with a factor that divides evenly before multiplying. Keep pi as a symbol. Last, check that the units are square for area and cubic for volume, and that the size makes sense.

Formula Recognition and Fast Mental Arithmetic

Spot the pattern: pointed solids, pyramids and cones, carry a one third; spheres carry four thirds for volume and 4 for surface area.

Divide first when you can. Learn squares through 15 and cubes through 6 to save time.

Worked Example 1 — Composite L-Shaped Floor

Problem: A floor is a 16 by 10 foot rectangle with a 6 by 4 foot corner missing. Find its area.

L-shaped floor: a 16 ft by 10 ft rectangle with a 6 ft by 4 ft corner missing (dashed); Area = ?.

Step 1 (subtract method): Find the area of the full rectangle as if nothing were missing.

Step 2: Find the area of the missing corner.

Step 3: Take the corner away from the full rectangle.

Answer: The floor is 136 square feet.

Check (split method): Cut the L into two rectangles instead. One strip is 16 by 6, and the leftover piece is 10 by 4 (because

10 − 4 = 6
and
16 − 6 = 10

).

Both methods give 136 square feet, so the answer is confirmed.

L-shaped floor two ways: a 16 by 10 rectangle minus a 6 by 4 corner (160 − 24 = 136), and split into a 16 by 6 strip and a 10 by 4 piece (96 + 40 = 136).

Worked Example 2 — Box Volume Versus Surface Area

Problem: A shipping box is 8 by 5 by 4 inches. Find how much it holds (volume) and how much cardboard covers it (surface area).

Shipping box 8 in by 5 in by 4 in with its three face types shaded in three colours (8 × 5, 8 × 4, 5 × 4); V = ? and SA = ?.

Step 1: Volume is length × width × height. Multiply two numbers at a time:

8 × 5 = 40
then
40 × 4 = 160

Step 2: For surface area, find the three different face sizes. A box has two of each.

Step 3: Add the three faces, then double for the matching opposite faces.

Answer: The box holds 160 cubic inches and is covered by 184 square inches of cardboard.

Check: Volume uses cubic units (in³) and surface area uses square units (in²). Different questions, different units. Recount faces:

2 + 2 + 2 = 6

the faces of a box.

Worked Example 3 — Closed Cylinder

Problem: A closed drum (cylinder) has radius 5 ft and height 9 ft. Find its total surface area.

Closed drum: a cylinder with radius 5 ft and height 9 ft, both round ends marked; SA = ?.

Step 1: The two round ends are circles. Each has area πr², and

5 × 5 = 25

Step 2: The curved side unrolls into a rectangle. Its width is the circumference 2πr and its length is the height.

Multiply the numbers:

2 × 5 = 10
then
10 × 9 = 90

Step 3: Add the ends and the side.

Answer: The surface area is 140π ft², about 440 ft².

Check: Use the formula SA = 2πr(r + h) = 2π ×

5 × (5 + 9) = 10

π × 14 = 140π. Same result.

Worked Example 4 — Open-Top Tank

Problem: An open-top tank is 8 ft long, 4 ft wide and 3 ft high. How much sheet metal makes the bottom and four walls?

Open-top tank 8 ft long, 4 ft wide and 3 ft high, drawn without a lid so the inside shows; sheet metal = ?.

Step 1: Find the bottom, length × width.

Step 2: Find the two long walls. Each is length × height.

Step 3: Find the two short walls. Each is width × height.

Step 4: Add the pieces. There is no top, so do not count it.

Answer: The tank needs 104 square feet of sheet metal.

Check: A closed box would add a top of 32, for 136 square feet. Subtracting that top again gives

136 − 32 = 104

Worked Example 5 — Triangular Prism

Problem: A prism has a right triangle base with legs 6 cm and 8 cm and a hypotenuse of 10 cm. Its length is 15 cm. Find its volume and surface area.

Triangular prism with a right triangle end of legs 6 cm and 8 cm and hypotenuse 10 cm, length 15 cm, drawn in 3D with hidden edges dashed.

Step 1: Find the triangle's area. In a right triangle the legs are the base and height, so take half of 6 × 8.

Step 2: Volume is base area × length.

24 × 15 = 24 × 10 + 24 × 5 = 240 + 120

Step 3: The three rectangular sides unroll into one long rectangle. Its width is the triangle's perimeter.

Step 4: Add the two triangle ends,

2 × 24 = 48

Answer: The volume is 360 cm³ and the surface area is 408 cm².

Check: Add the sides one at a time:

6 × 15 = 90
8 × 15 = 120
10 × 15 = 150
and
90 + 120 + 150 = 360

Same side area.

Worked Example 6 — Cone Height Versus Slant Height

Problem: A cone has radius 6 in and slant height 10 in. Find its volume.

Cone with radius 6 in and slant height l = 10 in; the unknown vertical height h = ? is dashed, forming a right triangle with r and l; V = ?.

Step 1: The volume formula needs the vertical height h, not the slant height. The radius, height and slant height form a right triangle, with the slant height as the hypotenuse.

Step 2: Solve for h. 10² = 100 and 6² = 36.

Step 3: Use

V = 1/3

πr²h with r² = 36 and h = 8.

Divide 36 by 3 first (12), then

12 × 8 = 96

Answer: The volume is 96π cubic inches, about 301 in³.

Check: 6-8-10 is a well-known right triangle (double of 3-4-5). The slant height is used for the surface area instead: the curved side is π ×

6 × 10 = 60

π.

Worked Example 7 — Square Pyramid

Problem: A square pyramid has a 16 m base edge and a vertical height of 6 m. Find its volume and surface area.

Square pyramid with base edge 16 m and vertical height 6 m; half the base, 8, runs from the centre to the middle of a front edge, and the slant height s = ? is dashed, forming a right triangle with legs 8 and 6.

Step 1: Find the base area.

16 × 16 = 256

Step 2: Volume is one third of base area × height. Divide 6 by 3 first to get 2.

Step 3: Each triangular face needs the slant height. It is the hypotenuse of a right triangle with legs 8 (half the base) and 6 (the height).

Step 4: Each face is a triangle with base 16 and height 10. There are 4 faces.

Step 5: Add the square base.

Answer: The volume is 512 m³ and the surface area is 576 m².

Check: 6-8-10 is double the 3-4-5 right triangle, so a slant height of 10 makes sense.

Worked Example 8 — Two Joined Cubes

Problem: Two cubes with 5 cm edges are glued face to face. Find the exposed surface area.

Two 5 cm cubes glued face to face into one 10 by 5 by 5 cm block, drawn in oblique view; the glued face is shaded orange with its hidden edges dashed; exposed SA = ?.

Step 1: Count faces. Each cube has 6 faces, so two cubes have 12.

Step 2: Gluing hides one face on each cube, so 2 faces disappear.

Step 3: Each face is a 5 by 5 square.

Step 4: Multiply the number of exposed faces by the area of one face.

Answer: The exposed surface area is 250 cm².

Check: Together the cubes make a 10 by 5 by 5 box. Its faces are

10 × 5 = 50
10 × 5 = 50
and
5 × 5 = 25

Worked Example 9 — Scale Change

Problem: A storage tank has volume 15 cubic meters and surface area 40 square meters. Every dimension is doubled. Find the new values.

Step 1: The scale factor is 2, because every length is multiplied by 2.

Step 2: Area uses two lengths multiplied together, so it grows by the scale factor squared.

Step 3: Volume uses three lengths, so it grows by the scale factor cubed.

Answer: The new surface area is 160 m² and the new volume is 120 m³.

Check: Try a 1 m cube. Doubled, it becomes a 2 m cube: each face goes from 1 to 4 square meters (4 times), and the volume goes from 1 to 8 cubic meters (8 times).

Worked Example 10 — Reverse Cylinder Dimension

Problem: A cylinder holds 245π cubic feet and is 5 ft tall. Find its radius.

Cylinder 5 ft tall with its radius marked r = ?; V = 245π ft³.

Step 1: Write the volume formula V = πr²h and fill in what you know.

Step 2: Divide both sides by π. It cancels on both sides.

Step 3: Divide both sides by 5 to get r² alone.

Step 4: Take the square root.

7 × 7 = 49

Answer: The radius is 7 feet.

Check: π × 7² × 5 = π ×

49 × 5 = 245

π. It matches the given volume.

Worked Example 11 — Cubic Unit Conversion

Problem: A slab is 12 ft by 9 ft and 6 inches thick. How many cubic yards of concrete does it need?

Concrete slab 12 ft by 9 ft and 6 in thick; cubic yards = ?.

Step 1: Put all lengths in feet. There are 12 inches in a foot, and 6 is half of 12.

Step 2: Volume is length × width × thickness.

12 × 9 = 108

and half of 108 is 54.

Step 3: Convert to cubic yards. 1 yard = 3 feet, so 1 cubic yard =

3 × 3 × 3 = 27 cubic feet

Divide by 27.

Answer: The slab needs 2 cubic yards of concrete.

Check: Work in yards from the start: 12 ft = 4 yd, 9 ft = 3 yd, 6 in = 1/6 yd. Then

4 × 3 × 1/6 = 12/6 = 2

cubic yards.

Most Common Traps

Slant height versus height. Volume of a cone or pyramid uses the vertical height; their surface area uses the slant height.

Diameter for radius. Halve the diameter before squaring or cubing.

Missing the one-third. Pointed solids hold one third of the matching prism or cylinder.

Counting faces that are not there. Open tops and glued faces are not part of the surface.

Converting by the linear factor. Square feet to square inches is times 144, and

Scaling volume by k. Volume scales by k cubed, area by k squared.

Wrong units. Area answers are square units; volume answers are cubic.

Final MK Review and Self-Check

Flat area: rectangle lw, triangle half bh, parallelogram bh, trapezoid half the base sum times h, circle pi r squared.

Volume: prism and cylinder are base area times height; pyramid and cone are one third of that.

Surface area adds every exposed face. Scaling by k multiplies area by k squared and volume by k cubed.

Self-check: did you use radius, the right height, consistent units, and the correct unit power?

Quick review

Trapezoid area: what is the formula?
: average the two parallel bases and multiply by the perpendicular height h. Bases 5 and 9 with height 6 give 42 square units.
Volume of any prism or cylinder?
, where B is the area of one base and h is the distance between the bases. For a cylinder , so . Answer in cubic units.
How does the volume of a pyramid or cone compare with a prism or cylinder of the same base and height?
It is one third: for a pyramid and for a cone. Pointed solids carry the one-third.
Volume and surface area of a sphere of radius r?
(cubic units) and (square units). For r = 6, .
Total surface area of a closed cylinder?
: two circular ends plus the curved side (a rectangle whose width is the circumference ). Open top: drop one .
Cone or pyramid: when do you use the slant height l and when the vertical height h?
Volume uses the vertical height h; surface area uses the slant height l. They are linked by a right triangle, e.g. for a cone .
If every length of a figure is multiplied by k, what happens to area and volume?
Area (and surface area) is multiplied by ; volume by . Doubling all dimensions gives 4 times the surface and 8 times the volume.
Key square and cubic unit conversions for feet, inches and yards?
Square the length factor for area and cube it for volume: , , . Also 1 liter = 1000 cubic centimeters.
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About this lesson

The three-dimensional part covers volume for rectangular prisms, general prisms, cylinders, pyramids, cones and spheres, then surface area from nets for prisms, cylinders, open-top containers, pyramids with slant height, cones and spheres. You also practice composite solids, counting only exposed faces, linear-to-square-to-cubic conversion, fill level and displacement, scaling solids and solving formulas in reverse for a missing dimension. Eleven worked examples range from an L-shaped floor and an open-top tank to two joined cubes and a cubic unit conversion.

Budget about 45 minutes. These multi-step measurement problems are a recurring part of Mathematics Knowledge, an AFQT subtest, so the closing self-check and list of common traps are worth a careful read. Then take the lesson test.

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