Mathematics • Year 10 • Unit 1 • Lesson 20
Pyramids, Cones and Spheres — Mixed Challenge
Pull together every idea from Lesson 20: pyramid V = ⅓·Abase·h, cone V = ⅓πr²h, sphere V = (4/3)πr³, the one-third rule, hemispheres, and composite solids combining tapered + straight pieces. Choose the right tool for each problem, spot another student's "forgot the ⅓" blunder, then design two desserts with the same volume but different shapes.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 20. Decide which formula applies before you start writing. Give exact answers in π unless told otherwise. 3 marks each
1.1 A cone has radius 5 cm and perpendicular height 9 cm. Find the volume in exact form.
1.2 A square-based pyramid has base side 12 cm and perpendicular height 9 cm. Find the volume.
1.3 A sphere has radius 9 cm. Find the volume in exact form and to 1 d.p.
1.4 A hemisphere has radius 7 cm. Find its volume in exact form. (Hint: hemisphere = ½ sphere.)
1.5 A composite solid is a cone of radius 6 cm and perpendicular height 8 cm with a hemisphere of the same radius (6 cm) sealed on top of the cone's base. Find the total volume in exact form.
1.6 A rectangular-based pyramid has base 10 m × 8 m and perpendicular height 6 m. (a) Find its volume. (b) How many times more volume does a rectangular prism with the same base and height hold?
2. Find the mistake
Another Year 10 student has tried to find the volume of a cone with radius 6 cm and perpendicular height 10 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 the volume of a cone with r = 6 cm, h = 10 cm:
Line 1: Abase = πr² = π(6)² = 36π cm²
Line 2: V = Abase × h = 36π × 10 = 360π cm³
Line 3: So the cone's volume is 360π ≈ 1131.0 cm³.
(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? A cone's volume is one-third of a cylinder's — the student has used the cylinder formula πr²h and forgotten the ⅓.3. Open-ended challenge — design two desserts with the same volume
This question has many valid answers. Be creative but show every number. 4 marks
3.1 A dessert café wants two new menu items that hold exactly the same volume of mousse: about 500 mL (= 500 cm³, within ±25 cm³). One must be a spherical scoop in a flat dish, the other must be a conical pot (radius and perpendicular height of your choice).
For each dessert:
(a) Choose its dimensions (sphere: radius r; cone: radius r and perpendicular height h).
(b) Calculate the volume in cm³ and confirm it is within ±25 cm³ of 500.
(c) Convert to mL and check both desserts match to within 25 mL.
Constraints: sphere radius between 4 cm and 8 cm; cone radius between 4 cm and 8 cm; cone height between 5 cm and 20 cm.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cone (r = 5, h = 9)
V = ⅓π(5)²(9) = ⅓π(25)(9) = ⅓π(225) = 75π cm³ (≈ 235.6 cm³).
1.2 — Square pyramid (base 12, h = 9)
Abase = 144 cm². V = ⅓ × 144 × 9 = ⅓ × 1296 = 432 cm³.
1.3 — Sphere (r = 9)
V = (4/3)π(9)³ = (4/3)π(729) = 4 × 243 × π = 972π cm³ ≈ 3053.6 cm³ (1 d.p.).
1.4 — Hemisphere (r = 7)
V = ½ × (4/3)π(7)³ = (2/3)π(343) = 686π/3 cm³ (≈ 718.4 cm³).
1.5 — Cone (r = 6, h = 8) + hemisphere (r = 6) on top
Vcone = ⅓π(6)²(8) = ⅓π(288) = 96π cm³.
Vhem = ½ × (4/3)π(6)³ = ½ × (4/3)π(216) = ½ × 288π = 144π cm³.
Total = 96π + 144π = 240π cm³ (≈ 754.0 cm³).
1.6 — Rectangular pyramid (10 × 8 × h = 6)
(a) Abase = 80 m². V = ⅓ × 80 × 6 = 160 m³.
(b) A prism with the same base and height has V = 80 × 6 = 480 m³ — that is 3 times the pyramid's volume (the one-third rule).
2 — Find the mistake
(a) The mistake is on Line 2.
(b) A cone's volume is V = ⅓πr²h, not πr²h. The student has computed the matching cylinder's volume and forgotten the one-third factor. The lesson's misconception card calls this out: "Forgetting the ⅓ factor for pyramids and cones."
(c) Corrected working:
Abase = π(6)² = 36π cm² ✓
V = ⅓ × 36π × 10 = ⅓ × 360π = 120π cm³ ≈ 377.0 cm³.
The student's answer of 360π would be a cylinder of the same radius and height — exactly three times the cone.
3 — Open-ended challenge (sample solutions)
We need a spherical scoop and a conical pot, both holding ≈ 500 cm³.
Dessert A — spherical scoop, r = 5 cm
V = (4/3)π(5)³ = (4/3)π(125) = 500π/3 ≈ 523.6 cm³ ✓ (within ±25 of 500).
≈ 523.6 mL.
Dessert B — conical pot, r = 5 cm, h = 20 cm (then check; if too big, scale down)
V = ⅓π(5)²(20) = ⅓π(500) = 500π/3 ≈ 523.6 cm³ ✓.
≈ 523.6 mL — exact match to the sphere!
This works because cone V = ⅓π r²h and sphere V = (4/3)π r³, so they're equal when h = 4r. With r = 5, h = 20 gives an exact volume match.
Marking: 1 for sphere dimensions and correct volume; 1 for cone dimensions and correct volume; 1 for both volumes within ±25 cm³ of 500; 1 for valid mL conversion and a match within 25 mL. Other valid dimension sets accepted — e.g. sphere r = 5, cone r = 7, h ≈ 10.2 cm gives V ≈ 522 cm³.