Area and volume formulas

Level: 101 · for anyone reviewing geometry before precalculus, or anyone who mixes up 4πr² and (4/3)πr³

One line: A formula's power of length is its dimension: scale every length by k and a perimeter grows by k, an area by k² and a volume by k³, so a misremembered formula can be caught before a single number goes in.

The idea

The book's second objective is a box of formulas to know: rectangle, triangle, circle, box, sphere, cylinder. Memorising eleven formulas is hard; memorising one check that tells them apart is easy. Every formula in the box multiplies lengths together:

• a perimeter or circumference is a length: no two lengths multiplied, 2l + 2w, 2πr;
• an area or surface area multiplies two lengths: lw, ½bh, πr², 4πr²;
• a volume multiplies three: lwh, πr²h, (4/3)πr³.

So count the lengths in each term. One is a length, measured in feet; two is an area, in square feet; three is a volume, in cubic feet. The constants in front, 2, ½, π, 4/3, are pure numbers with no units, and they do not count. If you double every length of a shape, its perimeter doubles, its area grows 2² = 4 times, and its volume 2³ = 8 times. A "volume" that only quadruples is really an area.

This page follows Section A.2, Geometry Essentials, objective 2, in Michael Sullivan's Precalculus: Example 4 and problems 2, 3, 6, 9 and 27–51.

The formulas

Shape	Measures	Formula	Lengths multiplied	Units
rectangle, length l, width w	perimeter	P = 2l + 2w	1	ft
	area	A = lw	2	ft²
triangle, base b, altitude h	area	A = ½bh	2	ft²
circle, radius r, diameter d = 2r	circumference	C = 2πr = πd	1	ft
	area	A = πr²	2	ft²
closed box, l × w × h	surface area	S = 2lh + 2wh + 2lw	2	ft²
	volume	V = lwh	3	ft³
sphere, radius r	surface area	S = 4πr²	2	ft²
	volume	V = (4/3)πr³	3	ft³
closed right circular cylinder, radius r, height h	surface area	S = 2πr² + 2πrh	2	ft²
	volume	V = πr²h	3	ft³

Three things in this table catch people out:

• The altitude of a triangle is not a side. It is the height measured at a right angle to the base, from the base to the opposite corner. For a slanted triangle it is shorter than the slanted sides, and it may even fall outside the triangle.
• Radius or diameter. The circle formulas use the radius. When a problem gives the diameter, halve it first, or use C = πd.
• "Closed". A closed box or cylinder has a top and a bottom, and the surface area counts them. For an open can, drop one πr².

Where the formulas come from

None of these is arbitrary. A rectangle l by w holds l · w unit squares. A triangle is half of a rectangle with the same base and height, so ½bh. A box l × w × h holds lwh unit cubes, and its surface is six rectangles in three matching pairs. A cylinder is a circle πr² stacked to height h, so its volume is πr²h, and its side, unrolled, is a rectangle 2πr wide and h tall, so its surface is the two circles plus 2πrh.

The sphere is the famous one. Archimedes proved that a sphere's surface equals the side of the cylinder that just fits around it: radius r, height 2r, side area 2πr · 2r = 4πr². And the volume and the surface are tied together: coat a sphere in a skin of thickness t, and the skin's volume is about the surface area times t. So the surface is the rate at which the volume grows as the radius grows, and 4πr² is what (4/3)πr³ gains per unit of radius. Section 3 of the program shows this. It is the same for a circle: a thin ring around it has area about 2πr · t, the circumference times the width. (In calculus words, each formula is the derivative of the next one up.)

What the program prints

Answers are kept exact, as a + bπ with fractions a and b, the way the book's answer key prints them, and rounded only at the end: 25π = 78.54, not 78.5 carried through.

Verified output of area_and_volume_formulas.py — regenerated by tools/run_examples.py, never hand-typed.

1. DOUBLE EVERY LENGTH, AND EACH FORMULA SAYS ITS OWN DIMENSION
   shape     formula                                  at size 1           at size 2  ratio
   rectangle perimeter 2l + 2w                          14.0000             28.0000      2
   rectangle area lw                                    12.0000             48.0000      4
   triangle  area bh/2                                  12.0000             48.0000      4
   circle    circumference 2 pi r                       12.5664             25.1327      2
   circle    area pi r^2                                12.5664             50.2655      4
   box       surface 2lh + 2wh + 2lw                    52.0000            208.0000      4
   box       volume lwh                                 24.0000            192.0000      8
   sphere    surface 4 pi r^2                           50.2655            201.0619      4
   sphere    volume (4/3) pi r^3                        33.5103            268.0826      8
   cylinder  surface 2 pi r^2 + 2 pi r h                87.9646            351.8584      4
   cylinder  volume pi r^2 h                            62.8319            502.6548      8
   Perimeters and circumferences double, areas and surfaces grow 4 = 2^2,
   volumes grow 8 = 2^3. The ratio is 2 raised to the number of lengths
   multiplied together, whatever the constant in front.

2. WHAT THE TEST CATCHES, AND WHAT IT CANNOT: PROBLEMS 6 AND 9
   'The surface area of a sphere of radius r is (4/3) pi r^2.'
   At r = 2 it gives 16/3 pi = 16.76; at r = 4, 64/3 pi = 67.02.
   Ratio 4: it has the shape of an area, so far so good.
   The dimension test cannot see a constant. The right one is 4, not 4/3:
   a sphere of radius 1 fits in a cylinder of radius 1 and height 2, whose
   side has area 2 pi r h = 4 pi = 12.57, and Archimedes showed that the
   sphere's surface is exactly that. False.
   For the volume, 4 pi r^2 and 4 pi r^3 (choices d and c of problem 6): the first
   is an area, so the test rejects it. The second passes the test, but a sphere
   inside a cube of side 2r must hold less than the cube's 8 r^3,
   and 4 pi = 12.57 is more than 8. Only (4/3) pi = 4.19 fits.

3. THE SURFACE IS WHAT THE VOLUME GAINS PER UNIT OF RADIUS
   Put a skin of thickness t on a sphere of radius 3 and weigh the skin:
     t = 1       skin volume / t =   154.9852
     t = 1/10    skin volume / t =   116.9091
     t = 1/100   skin volume / t =   113.4747
     t = 1/1000  skin volume / t =   113.1350
   The skin per unit thickness settles on 4 pi 3^2 = 113.0973, the surface area.
   The same for a circle: a ring of width t has area about 2 pi r t,
   the circumference times the width. Each formula is the rate of the next one up.

4. THE BOOK'S PROBLEMS 27-38, EXACT FIRST, ROUNDED LAST
     27  rectangle 6 by 7 in, area        42
     28  rectangle 9 by 4 cm, area        36
     29  triangle h 14, b 4 in, area      28
     30  triangle h 9, b 4 cm, area       18
     31  circle r 5 m, area               25 pi = 78.54
     31  circle r 5 m, circumference      10 pi = 31.42
     32  circle r 2 ft, area              4 pi = 12.57
     32  circle r 2 ft, circumference     4 pi = 12.57
     33  box 6 x 8 x 5 ft, volume         240
     33  box 6 x 8 x 5 ft, surface        236
     34  box 9 x 4 x 8 in, volume         288
     34  box 9 x 4 x 8 in, surface        280
     35  sphere r 5 cm, volume            500/3 pi = 523.60
     35  sphere r 5 cm, surface           100 pi = 314.16
     36  sphere r 3 ft, volume            36 pi = 113.10
     36  sphere r 3 ft, surface           36 pi = 113.10
     37  cylinder r 9, h 8 in, volume     648 pi = 2035.75
     37  cylinder r 9, h 8 in, surface    306 pi = 961.33
     38  cylinder r 8, h 9 in, volume     576 pi = 1809.56
     38  cylinder r 8, h 9 in, surface    272 pi = 854.51
   Problem 32 is a coincidence worth noticing: at r = 2 the area 4 pi and the
   circumference 4 pi are the same number, in different units. Area/circumference
   is r/2, so it happens only at r = 2, and only because units were dropped.
   37 and 38 swap r and h: the volumes differ (648 pi against 576 pi), because
   r is squared and h is not.

5. SHADED REGIONS: PROBLEMS 39-42, A SQUARE OF SIDE 2 AND A CIRCLE
   Inscribed circle: radius 1. Circumscribed circle: radius = half the diagonal,
   and the diagonal of a 2 by 2 square is sqrt(2^2 + 2^2) = sqrt(8), so r^2 = 2.
     39  circle inside the square             pi = 3.14
     40  square minus the circle inside it    4 - pi = 0.86
     41  circle around the square             2 pi = 6.28
     42  circle minus the square inside it    2 pi - 4 = 2.28
   The whole region is always a sum or difference of pieces with formulas.

6. APPLICATIONS: EXAMPLE 4 AND PROBLEMS 47-51
     Ex 4  ornament: triangle b 4, h 6, plus semicircle r 2   12 + 2 pi = 18.28 cm^2
     47    wheel, diameter 16 in, 4 turns: 4 * 16 pi          64 pi = 201.06 in
     48    disk, diameter 4 ft, rolled 20 ft: 20 / (4 pi)     1.59 turns
     49    border: outer square 6 + 2 + 2 = 10, minus inner   64 ft^2
     50    squares of area 100 and 16: legs 10 - 4 = 6 and 4   12 ft^2
     51    Norman window 4 by 6 plus semicircle r 2, area      24 + 2 pi = 30.28 ft^2
     51    its frame: 6 + 4 + 6 plus the arc, half of 4 pi    16 + 2 pi = 22.28 ft
   A turn of a wheel moves it one circumference, pi d, not 2 pi d: the
   diameter is given, so the circumference is pi d.

Section 4's problem 32 is worth a second look. At r = 2 the area is 4π and the circumference is 4π, the same number. That is a coincidence of units: one is 4π square feet and the other 4π feet, and they can no more be equal than 4 pounds can equal 4 miles. Change the unit to inches and the two numbers part company.

Questions

Try each one before opening the answer.

1. Fill in (problems 2 and 3): the area of a triangle with base b and altitude h is ; the circumference of a circle of radius r is .

Answer

A = ½bh and C = 2πr.

2. Problem 6: which is the volume of a sphere of radius r: (a) (4/3)πr², (b) (4/3)πr³, (c) 4πr³, (d) 4πr²? Rule out as many as you can without remembering.

Answer

(b). The dimension test rules out (a) and (d): r² makes an area. Between (b) and (c), the sphere fits inside a cube of side 2r, whose volume is 8r³, and 4π ≈ 12.6 is more than 8, so (c) is too big. (4/3)π ≈ 4.19, about half the cube, which is believable.

3. True or false (problem 9): the surface area of a sphere of radius r is (4/3)πr².

Answer

False; it is 4πr². This one the dimension test cannot catch, because (4/3)πr² *is* an area in shape. Only the constant is wrong, which is why it is a favourite trick question.

4. Every side of a box is tripled. What happens to its volume and to its surface area?

Answer

The volume grows 3³ = 27 times and the surface area 3² = 9 times. This is why big animals have thick legs and small ones lose heat fast: weight follows volume, strength and skin follow area.

5. A pizza of diameter 16 inches costs twice as much as one of diameter 10 inches. Which is the better deal?

Answer

Area goes with the square of the diameter: (16/10)² = 2.56. The big one has 2.56 times the pizza for 2 times the price, so the big one wins. Diameter alone suggests only 1.6 times as much, which is the common mistake.

6. Problems 37 and 38 swap the radius and the height of a cylinder (9 and 8). Do the volumes match? Why not?

Answer

No: π·9²·8 = 648π against π·8²·9 = 576π. The radius is squared and the height is not, so the radius counts for more.

7. Problem 41: a square of side 2 is inscribed in a circle. Find the area of the circle.

Answer

The diagonal of the square is a diameter. By Pythagoras, the diagonal is √(2² + 2²) = √8, so r = √8/2 and r² = 8/4 = 2. Area πr² = 2π ≈ 6.28. Note that r² was needed, not r, so √8 never had to be simplified.

8. Problem 47: how far does a wheel of diameter 16 inches travel in four revolutions?

Answer

One turn rolls out one circumference, πd = 16π inches. Four turns: 64π ≈ 201.06 inches, about 16.8 feet. The trap is 2πd: the 2 in 2πr is there because r is half of d.

9. Problem 49: a square of side 6 ft with a border 2 ft wide all round. Find the area of the border.

Answer

The outer square has side 6 + 2 + 2 = 10, not 6 + 2 = 8: the border is on both sides. 100 − 36 = 64 ft².

10. A circle's area and its circumference are the same number. What is its radius, and why is the question a little dishonest?

Answer

πr² = 2πr gives r = 2. It is dishonest because the two have different units, square units and units, so "equal" depends on the unit chosen. In inches instead of feet the numbers would differ.

11. A triangle has base 10 and slanted sides 13 and 13. Why is its area not ½ · 10 · 13?

Answer

13 is a side, not the altitude. The altitude drops from the top to the middle of the base, making two right triangles with hypotenuse 13 and leg 5, so the altitude is √(169 − 25) = 12, and the area is ½ · 10 · 12 = 60.

12. Why is the surface area of a sphere "what the volume gains per unit of radius"?

Answer

Growing the radius by a small t adds a thin skin, and the skin is roughly the surface spread to thickness t: volume ≈ 4πr² · t. So the gain per unit of t tends to 4πr². The program's section 3 watches the ratio settle as t shrinks.

How to practise

1. The dimension check first, every time. Before using a formula, count the lengths multiplied in each term and say the unit out loud: "cubic feet". A wrong unit means a wrong formula.
2. Problems 27–51 on paper, then compare with the output: exact answers first (25π), decimals last.
3. Sketch before you compute. The shaded-region problems 39–42 and 49–51 are always a sum or a difference of shapes that have formulas; say which before writing numbers.
4. The cards, a few minutes a day. The trap cards are the ones worth failing on purpose.

Flashcards

The page as a deck of Anki cards: area_and_volume_formulas.txt. Import with File → Import; the header sets everything. Tags: formula, dimension, problems, trap.

Where this goes next

Congruent and similar triangles uses the scaling rule of section 1: similar triangles with sides in the ratio k have areas in the ratio k². And the circle comes back in coordinates in circles, at the end of the book's first chapter, where "radius r" becomes an equation, (x − h)² + (y − k)² = r², which is the distance formula held fixed.

Po polsku, w skrócie

Wzorów na pola i objętości jest jedenaście, ale wystarczy jedna kontrola, żeby ich nie mylić: policz, ile długości mnoży się w każdym wyrazie. Obwód to jedna długość (2πr), pole dwie (πr², 4πr²), objętość trzy (πr²h, (4/3)πr³). Stałe z przodu, takie jak 2, ½, π, 4/3, to czyste liczby i się nie liczą. Jeśli podwoimy wszystkie wymiary, obwód rośnie dwa razy, pole cztery, objętość osiem. „Objętość", która rośnie tylko czterokrotnie, jest w rzeczywistości polem.

Ta kontrola ma granice: (4/3)πr² ma kształt pola, a jednak nie jest polem powierzchni kuli, bo zła jest stała; poprawnie 4πr². Pułapki z zadań: wysokość trójkąta to nie jego bok, tylko odcinek prostopadły do podstawy; koło podane średnicą trzeba najpierw podzielić na pół; ramka szerokości 2 wokół kwadratu 6 daje kwadrat 10, nie 8. Program liczy odpowiedzi dokładnie, jako a + bπ, i zaokrągla dopiero na końcu, a także pokazuje, że pole powierzchni kuli to przyrost objętości na jednostkę promienia.

Auf Deutsch: Stichwörter

Rechteck, Dreieck, Kreis, Quader, Kugel, Zylinder; die Potenz der Länge in einer Formel ist ihre Dimension, Streckung um k vervielfacht um k² und k³.

Stichwörter: Flächenformel, Volumenformel, Umfang, Oberfläche, Dimension, Streckfaktor k, k² und k³, Kreiszahl π, Kugel, Zylinder.

See also

• The Pythagorean theorem and its converse — the diagonal of a square in problems 41–42, and the altitude in question 11
• Congruent and similar triangles — scale factor k, areas k²
• Significant figures — why 25π is kept exact and rounded only at the end
• Exact vs approximate — π is exact as a symbol, approximate as 3.14
• Precalculus: a reading guide — where Appendix A.2 sits in the course
• Area ↗ and Volume ↗ — Wikipedia, with the derivations
• On the Sphere and Cylinder ↗ — Archimedes' proof that the sphere's surface is 4πr²
• Pole powierzchni ↗ — Wikipedia po polsku
• Michael Sullivan, Precalculus (Pearson), Appendix A, Section A.2 Geometry Essentials, objective 2; this page follows it