Mathematics • Year 8 • Unit 3 • Lesson 8
Area of Circles — Mixed Challenge
Pull everything from Lesson 8 together: forward calculations, semicircles, quarter circles, reverse rearrangements and area-vs-circumference comparisons. Six mixed problems, one "find the mistake", and one open-ended donut design challenge.
1. Mixed problems — choose the right move
Each question pulls a different idea from Lesson 8. Decide whether you need r or d, full circle or fraction before you start writing. 3 marks each
1.1 r = 11 cm. Find A in exact form (n π).
1.2 d = 16 m. Find A to 2 d.p.
1.3 A semicircle has diameter 12 cm. Find its area to 2 d.p.
1.4 A quarter circle has radius 8 cm. Find its area to 2 d.p.
1.5 A circle has area A = 154 cm². Use π ≈ 22/7 to find the radius r exactly.
1.6 Compare: circle A has r = 4 cm, circle B has r = 8 cm. How many times bigger is the area of circle B compared to circle A?
2. Find the mistake
A Year 8 student has tried to find the area of a circle with radius r = 5 cm. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find A when r = 5 cm:
Line 1: A = π × r²
Line 2: A = (π × 5)²
Line 3: A = (15.708)² ≈ 246.74
Line 4: So the area is about 246.74 cm².
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Revisit lesson § Card 9 — "Squaring π instead of just r" is a common pitfall. A = π × (r²), not (πr)².3. Open-ended challenge — design a donut (annulus)
This question has more than one valid answer. 4 marks
3.1 A donut shape (mathematically called an annulus) is what you get when you cut a smaller circle out of a bigger circle. Its area is:
A_donut = π × R² − π × r² (R = outer radius, r = inner radius)
Your task: Design a donut with area exactly 75π cm² (exact form). You must:
(i) Choose whole-number values of R and r.
(ii) Show that π × R² − π × r² = 75π.
(iii) Sketch the donut, labelling R and r.
(iv) Find the decimal approximation of 75π to 2 d.p.
Bonus: find a second pair (R, r) that also works (different from your first pair).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — r = 11 (exact)
A = π × 11² = 121π cm² (exact). (≈ 380.13 cm².)
1.2 — d = 16 m
r = 16 ÷ 2 = 8 m. A = π × 8² = 64π ≈ 201.06 m².
1.3 — Semicircle d = 12
r = 6 cm. A = ½ × π × 6² = ½ × 36π = 18π ≈ 56.55 cm².
1.4 — Quarter circle r = 8
A = ¼ × π × 8² = ¼ × 64π = 16π ≈ 50.27 cm².
1.5 — Find r from A = 154 (π ≈ 22/7)
r² = A ÷ π = 154 ÷ (22/7) = 154 × (7/22) = 7 × 7 = 49. r = √49 = 7 cm exactly. Check: (22/7) × 49 = 22 × 7 = 154 ✓.
1.6 — Compare A and B (r = 4 vs r = 8)
A = π × 4² = 16π. B = π × 8² = 64π. Ratio = 64π ÷ 16π = 4 times bigger. (Doubling the radius multiplies area by 2² = 4.)
2 — Find the mistake
(a) The mistake is on Line 2 (carried into Line 3 and Line 4).
(b) The student wrote (π × 5)² instead of π × 5². The formula squares only the radius, not the whole expression. (π × 5)² = π² × 25, which has an extra factor of π.
(c) Corrected working:
A = π × r² = π × 5² = π × 25 = 25π ≈ 78.54 cm². ✓
Sanity check: the student's answer (246.74) is exactly π times bigger than the correct answer (78.54 × π ≈ 246.74) — that's the giveaway they squared π by accident.
3 — Donut design (sample solution)
Need R² − r² = 75. There are many valid integer solutions:
Solution 1: R = 10, r = 5. Check: π × 10² − π × 5² = 100π − 25π = 75π ✓. Decimal: 75 × 3.14159 ≈ 235.62 cm².
Solution 2: R = 14, r = 11. Check: π × 14² − π × 11² = 196π − 121π = 75π ✓.
Solution 3: R = 38, r = 37. Check: π × 38² − π × 37² = 1444π − 1369π = 75π ✓ (very thin donut!).
Marking: 1 mark for choosing valid integer R and r; 1 mark for showing R² − r² = 75 with working; 1 mark for sketch labelled with R and r; 1 mark for decimal 75π ≈ 235.62 cm². Bonus mark for a second valid pair.