Solid geometry is about three-dimensional shapes — boxes, cans, balls, pyramids — and the two questions the CSCA asks about them: how much space they fill (volume) and how much area covers their outside (surface area). It is a dependable source of marks, appearing on nearly every paper, often more than once.
This page handles the friendliest solids: the ones with straight sides that rise vertically — boxes, prisms, and cylinders. They all share one beautifully simple volume rule, and once you see it, half of solid geometry becomes a single idea repeated. We will build that rule from intuition, apply it to each shape, then handle surface area by a trick anyone can picture: unfold the solid flat.
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The volume idea: base × height
Picture a stack of identical coins. One coin has some small volume; stack them h units high and the total volume is just that one coin's face area multiplied by the height of the stack. Nothing about the stack's volume depends on the shape of the coin — only its area and how tall the stack is.
That is the whole principle for any solid with constant cross-section — a shape that looks the same all the way up. Its volume is the area of that cross-section (the base) times the height. Prisms (including boxes) and cylinders are exactly these "straight-up" solids.
Volume of a prism or cylinder. V = (base area) × height. The base is the cross-section that repeats all the way up; the height is how far it rises perpendicular to that base. This single rule covers boxes, triangular prisms, cylinders, and any uniform column.
A "straight-up" solid: its volume is the area of the repeating base multiplied by the height.
Boxes (cuboids)
A box — a cuboid — has a rectangular base of area length × width, so its volume is length × width × height. Its three dimensions multiply together.
Worked example · volume of a box
Find the volume of a box measuring 5 by 3 by 4.
The base is a 5 by 3 rectangle, with area 5 × 3 = 15.
Multiply by the height 4: 15 × 4 = 60.
So the volume is 60. For a box you can simply multiply all three dimensions in any order — 5 × 3 × 4 = 60 — since any face can serve as the base.
Try it! · Check-in question
The number of unit cubes (1 by 1 by 1) needed to fill a box measuring 2 by 3 by 6 is ( ).
The number of unit cubes is the volume. The bottom layer is 2 × 3 = 6 cubes, and there are 6 layers, so 6 × 6 = 36. The answer is C. Option A adds the dimensions (2 + 3 + 6 = 11). Option B is the surface area, 2(2×3 + 2×6 + 3×6) = 2(6 + 12 + 18) = 72; it counts unit squares on the outside, not cubes inside. Option D multiplies only two dimensions (3 × 6 = 18) and forgets the 2. Volume needs all three lengths multiplied together.
Cubes
A cube is the special box whose edges are all equal, length a. Its volume is a × a × a = a³ — which is exactly why "cubing" a number is named after it.
Try it! · Check-in question
A cube has edge length 4. Its volume is ( ).
A cube's volume is edge³ = 4 × 4 × 4 = 64. The answer is B. Option A multiplies the edge by the power (4 × 3 = 12) instead of using three factors of 4. Option C squares instead of cubing (4² = 16, the area of one face, not a volume). Option D squares and then multiplies by 3 (4² × 3 = 48) instead of by a third 4. Cubing means three factors of 4.
Other prisms
A prism need not have a rectangular base. A triangular prism has a triangle for its cross-section; the rule is unchanged — find the area of the triangular base, then multiply by the length the prism runs.
Worked example · triangular prism
A prism has a right-triangular base with legs 3 and 4, and it runs 10 units long. Find its volume.
Find the base area first. A right triangle's area is ½ × leg × leg = ½ × 3 × 4 = 6.
Multiply by the length: 6 × 10 = 60.
So the volume is 60. The "base" is whichever face is the repeating cross-section — here the triangle — and the "height" is how far that triangle is extruded, 10.
Try it! · Check-in question
A prism has a base of area 12 and a height of 7. Its volume is ( ).
Volume = base area × height = 12 × 7 = 84. The answer is D. Option A halves the product (½ × 84 = 42), as in a triangle's area. But 12 is already the base area, so there is no ½ left to use. Option B adds (12 + 7 = 19). Option C takes a third (⅓ × 84 = 28); the ⅓ belongs to pyramids and cones (next page), not prisms. For any prism it is simply base × height.
Cylinders
A cylinder is a prism whose base is a circle — a can standing upright. The volume rule does not change: base area times height. The only new ingredient is that the base is a circle, with area πr², so the volume is πr² × h.
Volume of a cylinder. V = πr²h, where r is the base radius and h is the height. It is "circle area × height" — the same base × height rule, with the circle's πr² as the base.
A cylinder is a circle extruded upward: volume = πr² × h.
Worked example · volume of a cylinder
Find the volume of a cylinder with radius 3 and height 5. Leave your answer in terms of π.
Find the base (circle) area: πr² = π × 3² = 9π.
Multiply by the height: 9π × 5 = 45π.
So the volume is 45π. Square the radius before multiplying by the height — the most common slip is doing π × 3 × 5 and forgetting to square the 3.
Square the radius, not just multiply it. The base of a cylinder is πr², so the radius gets squared. Writing 2πrh or πr·h for the volume (forgetting the square) is the classic error. Volume is πr²h; the only place 2πr appears is the curved surface area, coming up next.
Try it! · Check-in question
A cylinder has a base diameter of 6 and a height of 10. Its volume is ( ).
The radius is half the diameter: r = 6 ÷ 2 = 3. Then V = πr²h = π × 3² × 10 = π × 9 × 10 = 90π. The answer is A. Option B uses the diameter 6 as the radius (π × 36 × 10 = 360π). Option C forgets to square the radius (π × 3 × 10 = 30π). Option D uses 2πrh = 2π × 3 × 10 = 60π, which is the curved surface area, not the volume. Halve the diameter first, then square.
Try it! · Check-in question
A cylinder has radius 5 and height 4. Its volume is ( ).
V = πr²h = π × 5² × 4 = π × 25 × 4 = 100π. The answer is C. Option A uses 2πrh = 2π × 5 × 4 = 40π, the curved surface formula. Option B swaps the radius and the height (π × 4² × 5 = 80π). Option D forgets to square the radius (π × 5 × 4 = 20π). The radius 5 squares to 25, then times 4 gives 100π.
Surface area: unfold it flat
Surface area is the total area of all the faces — the "skin" of the solid. The reliable way to find it is to imagine cutting the solid open and unfolding it flat into a 2-D shape called a net. Then you just add up the areas of the flat pieces.
A box
Unfold a box and you get six rectangles in three matching pairs: top and bottom (each lw), front and back (each lh), and the two sides (each wh). Adding all six gives the formula.
Surface area of a box. SA = 2(lw + lh + wh) — twice the sum of the three distinct face areas, since the faces come in three equal pairs.
Worked example · surface area of a box
Find the surface area of a box measuring 5 by 3 by 4.
Compute the three distinct face areas: lw = 5 × 3 = 15, lh = 5 × 4 = 20, wh = 3 × 4 = 12.
Add them: 15 + 20 + 12 = 47.
Double, because each face has a matching opposite: 2 × 47 = 94.
So the surface area is 94. Find the three different faces, add, then double — that captures all six faces without missing any.
Try it! · Check-in question
The surface area of a cube with edge length 3 is ( ).
A cube has 6 identical square faces, each of area edge² = 3² = 9, so SA = 6 × 9 = 54. The answer is C. Option A is the volume (3³ = 27), not an area. Option B counts only the 4 side faces (4 × 9 = 36) and leaves out the top and bottom. Option D counts only 2 faces (2 × 9 = 18). A cube always has 6 faces, giving 6a².
A cylinder
Unfolding a cylinder is the elegant one. The two circular ends give 2 × πr². The curved side, unrolled, is a rectangle: its height is the cylinder's height h, and its width is the distance around the circle — the circumference, 2πr. So the curved part has area 2πr × h.
A cylinder unrolled: two circles (2πr²) plus a rectangle whose width is the circumference 2πr and height is h, giving curved area 2πrh.
Surface area of a cylinder. Total SA = 2πr² + 2πrh — the two circular ends plus the curved side. The curved (lateral) part alone is 2πrh; add the two caps 2πr² for the full surface.
Worked example · surface area of a cylinder
Find the total surface area of a cylinder with radius 3 and height 5. Leave your answer in terms of π.
The two circular ends: 2πr² = 2 × π × 3² = 2 × 9π = 18π.
The curved side: 2πrh = 2 × π × 3 × 5 = 30π.
Add the pieces: 18π + 30π = 48π.
So the total surface area is 48π. Keep the two parts separate — caps 2πr² and side 2πrh — then add, so you neither double-count nor drop a piece.
Try it! · Check-in question
The curved (lateral) surface area of a cylinder with radius 4 and height 5 is ( ).
The curved surface alone is 2πrh = 2 × π × 4 × 5 = 40π. The answer is D. Option A is the volume, πr²h = π × 16 × 5 = 80π. Option B uses πrh = 20π: it drops the 2 from the circumference 2πr. Option C adds the two caps (2πr² = 32π) to get the total 72π, but the question asked only for the curved part. Curved area is 2πrh = 40π.
Try it! · Check-in question
A cup shaped like a cylinder has a bottom but no lid. Its radius is 2 and its height is 3. The area of material needed to make the cup is ( ).
Unfold the cup: one circle for the bottom and one rectangle for the side. Bottom: πr² = π × 2² = 4π. Side: 2πrh = 2π × 2 × 3 = 12π. Total: 4π + 12π = 16π. The answer is B. Option A also counts a lid (two circles, 8π + 12π = 20π), but the cup has no lid. Option C is the side alone and forgets the bottom. Option D uses πrh = 6π for the side (it drops the 2 from 2πr): 4π + 6π = 10π. Count only the pieces the cup really has.
Your turn! · Practice questions
Practice · Question 1
Box P measures 2 by 3 by 4. Box Q has every edge twice as long as box P. The volume of Q is how many times the volume of P? ( )
Volume of P: 2 × 3 × 4 = 24. Box Q measures 4 by 6 by 8, so its volume is 4 × 6 × 8 = 192. Then 192 ÷ 24 = 8. The answer is A. Each of the three lengths doubles, so the volume doubles three times: 2 × 2 × 2 = 8. Option B assumes the volume only doubles, like the edges. Option C uses 2 × 2 = 4, which is the factor for an area (a face has only two lengths). Option D adds the three doublings (2 + 2 + 2 = 6) instead of multiplying them.
Practice · Question 2
A cylinder has radius 3 and volume 72π. Its height is ( ).
Use V = πr²h and solve for h: π × 3² × h = 72π, so 9πh = 72π. Divide both sides by 9π: h = 8. The answer is D. Option A forgets to square the radius (3h = 72, so h = 24). Option B divides 72π by 9 but forgets that the π cancels; a length has no π here. Option C uses the curved surface formula 2πrh = 72π, so 6πh = 72π and h = 12.
Practice · Question 3
A triangular prism has volume 60 and length 5. Its triangular end has a base of 6. The height of the triangle is ( ).
Work backwards. Volume = base area × length, so base area = 60 ÷ 5 = 12. The base is a triangle: ½ × 6 × h = 12, so 3h = 12 and h = 4. The answer is B. Option A forgets the ½ in the triangle's area (6h = 12, so h = 2). Option C stops at the base area, 12, which is an area, not the height. Option D skips the division by the length (½ × 6 × h = 60, so h = 20).
Practice · Question 4
A cube of edge 4 is cut into two equal boxes, each 4 by 4 by 2. The total surface area of the two boxes is greater than the surface area of the cube by ( ).
Cube: SA = 6 × 4² = 96. Each box: SA = 2(4×4 + 4×2 + 4×2) = 2(16 + 8 + 8) = 64, so the two boxes have 2 × 64 = 128. The increase is 128 − 96 = 32. The answer is C. You can see it directly: the cut makes two new 4 by 4 faces, one on each piece, and 2 × 16 = 32. Option A assumes cutting does not change the area. Option B counts only one new face (16). Option D is the total area of the two boxes, not the increase.
Practice · Question 5
A cylinder has height 4, and its curved surface area is 24π. Its volume is ( ).
First find the radius from the curved surface: 2πrh = 24π, so 2πr × 4 = 24π, 8πr = 24π, and r = 3. Then V = πr²h = π × 9 × 4 = 36π. The answer is A. Option B forgets the 2 in 2πr (πr × 4 = 24π gives r = 6, then π × 36 × 4 = 144π). Option C forgets to square the radius in the volume (π × 3 × 4 = 12π). Option D swaps the radius and the height in the volume (π × 4² × 3 = 48π).
Practice · Question 6
A cube has volume 64. Its surface area is ( ).
Volume 64 means edge³ = 64, so the edge is 4 (4 × 4 × 4 = 64). Each face has area 4² = 16, and there are 6 faces: SA = 6 × 16 = 96. The answer is B. Option A multiplies the volume by 6 (6 × 64 = 384); the 6 multiplies the area of one face, not the volume. Option C is one face only. Option D counts only 3 faces (3 × 16 = 48) and forgets that each face has an opposite partner. Find the edge first, then use 6a².
Things to remember
Concept
What to know
Volume principle
For any "straight-up" solid (prism or cylinder): V = base area × height. One rule covers boxes, prisms, and cylinders.
Box / cuboid
V = l × w × h. Cube: V = a³.
Triangular prism
V = (½ × base × height of triangle) × length. Don't forget the triangle's ½.
Cylinder volume
V = πr²h. Square the radius first — never πrh or 2πrh.
Box surface area
SA = 2(lw + lh + wh). Cube: SA = 6a² (six equal faces).
Cylinder surface area
Total = 2πr² + 2πrh: two circular caps plus the curved side. Curved (lateral) part alone = 2πrh.
Surface area method
Unfold into a net and add the flat pieces. The cylinder's side unrolls to a rectangle of width 2πr (the circumference).
