Pi Approximation Day puzzles for A Level
22 July is Pi Approximation Day: written day first, 22/7 is the famous fraction for π. A short summer puzzle set: a 5-minute starter, three 10-minute puzzles, an extension and full worked answers.
- Level
- A Level Maths (Year 12 and Year 13)
- Time
- 35 minutes, plus a 10-minute extension
- Topics
- Approximation and error; Continued fractions and recurrence; Radians and small-angle approximations
- Equipment
- The starter is non-calculator. A calculator is useful in the main activity.
Suggested timings
| Part | Time | What |
|---|---|---|
| Starter | 5 min | Quick questions on the board |
| Main: task A | 10 min | How good is 22/7? |
| Main: task B | 10 min | Continued fractions as a recurrence |
| Main: task C | 10 min | Small angles and π |
| Extension | 10 min | Fast finishers or homework |
Starter (5 minutes)
No calculator.
- Write 22/7 − 3 as a fraction.
- Work out 3 + 1/(7 + 1/15) as a single fraction.
- Convert 180° to radians and π/6 radians to degrees.
- Using sin θ ≈ θ, estimate sin 0.02.
Main activity (30 minutes)
Task A: How good is 22/7? (10 min)
Use your calculator’s value of π.
- Find the percentage error when 22/7 is used for π, to 3 significant figures.
- A satellite moves in a circle of radius 7000 km. By how many kilometres is one orbit overestimated if 22/7 is used for π? Give 3 significant figures.
- Which is the closer approximation to π: 22/7 or 355/113? By what factor, to the nearest whole number?
Task B: Continued fractions as a recurrence (10 min)
π = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + …)))). The approximations pn/qn follow pn = anpn−1 + pn−2 and qn = anqn−1 + qn−2, with a = 3, 7, 15, 1, 292, …
- Starting from 3/1 and 22/7, use the recurrence with a = 15 to find the next fraction.
- Use a = 1 to find the next fraction after that.
- Use a = 292 to find the next fraction.
Task C: Small angles and π (10 min)
Radians and small-angle approximations.
- Using sin θ ≈ θ, estimate 180 sin(1°) and compare it with π.
- Using cos θ ≈ 1 − θ2/2, estimate cos 0.1 and find the error in the approximation, to 2 significant figures.
- A wheel turns through 0.4 radians. Using π = 22/7, how many degrees is this, to the nearest degree?
Extension (10 minutes)
Archimedes’ bounds.
- Archimedes showed that 223/71 < π < 22/7. Find the width of the interval as a single fraction, and the midpoint to 5 decimal places.
For teachers
Teacher notes and full worked answers
- Pi Approximation Day is 22 July because 22/7, written day first, is the famous fraction for π.
- This is a short set for the last days of term or summer school: each puzzle stands alone.
- Task B uses the same kind of recurrence as the Fibonacci sequence: a nice link back to sequences.
Starter
- 1/7
- 22/7 − 21/7 = 1/7
- 333/106
- 7 + 1/15 = 106/15, so 3 + 15/106 = 333/106.
- π; 30°
- 180° = π radians, so π/6 = 30°.
- 0.02
- For small θ in radians, sin θ ≈ θ.
Task A: How good is 22/7?
- 0.0402%
- (22/7 − π) ÷ π × 100 = 0.04024…
- 17.7 km
- 2 × 7000 × (22/7 − π) = 17.70…
- 355/113, about 4740 times closer
- |22/7 − π| = 0.0012644…, |355/113 − π| = 0.00000026676…
- 0.0012644… ÷ 0.00000026676… = 4740.1…
Task B: Continued fractions as a recurrence
- 333/106
- p = 15 × 22 + 3 = 333, q = 15 × 7 + 1 = 106
- 355/113
- p = 1 × 333 + 22 = 355, q = 1 × 106 + 7 = 113
- 103993/33102
- p = 292 × 355 + 333 = 103 993, q = 292 × 113 + 106 = 33 102
Task C: Small angles and π
- π; 180 sin 1° = 3.1414 (4 d.p.)
- 1° = π/180 radians, so 180 sin(1°) ≈ 180 × π/180 = π.
- The calculator gives 180 sin 1° = 3.14143…
- 0.995; error 0.0000042 (4.2 × 10−6)
- 1 − 0.01/2 = 0.995
- cos 0.1 = 0.995004165…, so the error is 0.0000041652…
- 23°
- 0.4 × 180/π = 0.4 × 180 × 7/22 = 22.9…, so 23°.
Extension
- 1/497; 3.14185
- 22/7 − 223/71 = (1562 − 1561)/497 = 1/497
- Midpoint = (22/7 + 223/71) ÷ 2 = 3.141851…
The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.
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