The shape of the lesson
By the end of the lesson, learners can
- Pyramids: one third: In the lab, press Pour three times: it takes exactly three full pyramids (or cones) to fill the prism (or cylinder) with the same base and height.
- Cones and the slant height: A cone has two different heights (Fig. 1).
- Spheres and hemispheres: In the lab, change the radius of the sphere: doubling the radius makes the surface area 4 times bigger and the volume 8 times bigger.
- Pyramid and cone volume: \(\tfrac{1}{3}\) of base area \(\times\) vertical height.
- Cone curved surface: \(\pi rl\) with slant height \(l = \sqrt{r^2 + h^2}\).
- Sphere: \(V = \tfrac{4}{3}\pi r^3\), \(A = 4\pi r^2\).
- Solid shapes: add flat faces (cone base \(\pi r^2\), hemisphere face \(\pi r^2\)).
- Melting and recasting: the volume stays the same.
One picture every idea returns to
Pyramid \(V = \tfrac{1}{3}Ah\); cone \(V = \tfrac{1}{3}\pi r^2 h\) and curved surface \(\pi r l\); sphere \(V = \tfrac{4}{3}\pi r^3\) and surface \(4\pi r^2\). All are in the List of formulas. The cone's curved surface uses the SLANT height \(l\), where \(l^2 = r^2 + h^2\); that relationship is not given.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Points and curves.. A pyramid or cone holds exactly one third of the prism it fits inside. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Pyramids: one third | In the lab, press Pour three times: it takes exactly three full pyramids (or cones) to fill the prism (or cylinder) with the same base and height. Project tab 3 of the lab (Surface area and volume lab, pyramids and cones) and let a student drive it while the class predicts. Model the worked example Square-based pyramid on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Cones and the slant height | A cone has two different heights (Fig. 1). Model the worked example Volume and curved surface on the board, one line at a time. Your turn: A cone has radius 3 cm and vertical height 4 cm. Find its curved surface area in terms of \(\pi\). | Pairs |
| 21 to 30 min | Teach and check: Spheres and hemispheres | In the lab, change the radius of the sphere: doubling the radius makes the surface area 4 times bigger and the volume 8 times bigger. Project tab 4 of the lab (Surface area and volume lab, spheres) and let a student drive it while the class predicts. Model the worked example Sphere on the board, one line at a time. Your turn: A sphere has surface area \(100\text{ cm}^2\). Find its radius. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Spheres and hemispheres, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The \(h\) in \(\tfrac{1}{3}Ah\) is the VERTICAL height from the base to the apex, not the sloping height of a face. | Forgetting the \(\tfrac{1}{3}\) gives the volume of the prism instead. |
| Putting the vertical height into \(\pi rl\): for the cone above, \(\pi \times 5 \times 12\) is wrong. | The curved surface needs the slant height 13. |
| The total surface area of a solid hemisphere is \(2\pi r^2 + \pi r^2 = 3\pi r^2\). | Halving \(4\pi r^2\) and stopping forgets the flat face. |
Surface area and volume: cones, pyramids and spheres: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
A pyramid has a square base of side 8 cm and vertical height 9 cm. Find its volume.
A sphere has diameter 14 cm. Calculate (a) its volume (b) its surface area.
A cone has radius 4 cm and vertical height 9 cm. Calculate (a) its volume (b) its curved surface area.
Without a calculator, find the volume and the total surface area of a solid hemisphere of radius 3 cm. Give your answers in terms of \(\pi\).
A ball is a sphere with volume \(500\text{ cm}^3\).
Calculate (a) its radius (b) its surface area.
A solid metal sphere of radius 6 cm is melted down. All the metal is used to make solid cones, each with radius 2 cm and vertical height 3 cm.
How many cones are made?
Answer key and marking notes
- Base area \(= 64\text{ cm}^2\).
- \(V = \tfrac{1}{3} \times 64 \times 9 = 192\text{ cm}^3\)
- \(r = 7\) cm.
- (a) \(\tfrac{4}{3}\pi \times 343 = 1436.7... = 1440\text{ cm}^3\) (3 s.f.)
- (b) \(4\pi \times 49 = 615.7... = 616\text{ cm}^2\)
- (a) \(\tfrac{1}{3}\pi \times 16 \times 9 = 48\pi = 150.7... = 151\text{ cm}^3\)
- (b) \(l = \sqrt{16 + 81} = \sqrt{97} = 9.848...\) cm
- Curved surface \(= \pi \times 4 \times 9.848... = 123.7... = 124\text{ cm}^2\)
- \(V = \tfrac{1}{2} \times \tfrac{4}{3}\pi \times 27 = 18\pi\text{ cm}^3\)
- Surface \(= \tfrac{1}{2} \times 4\pi \times 9 + \pi \times 9 = 18\pi + 9\pi = 27\pi\text{ cm}^2\)
- (a) \(\tfrac{4}{3}\pi r^3 = 500\), so \(r^3 = \dfrac{1500}{4\pi} = 119.3...\) and \(r = 4.923... = 4.92\) cm
- (b) \(4\pi \times 4.923...^2 = 304.6... = 305\text{ cm}^2\) (use the unrounded radius)
- Sphere: \(\tfrac{4}{3}\pi \times 216 = 288\pi\text{ cm}^3\).
- One cone: \(\tfrac{1}{3}\pi \times 4 \times 3 = 4\pi\text{ cm}^3\).
- Number of cones \(= 288\pi \div 4\pi = 72\).
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A pyramid has a rectangular base measuring 8 cm by 6 cm. Its vertical height is 15 cm. Find its volume.
- A\(240\text{ cm}^3\)
- B\(720\text{ cm}^3\)
- C\(360\text{ cm}^3\)
- D\(16\text{ cm}^3\)
A ball is a sphere of diameter 9 cm. Find its volume, correct to 3 significant figures.
- A\(382\text{ cm}^3\)
- B\(3050\text{ cm}^3\)
- C\(254\text{ cm}^3\)
- D\(286\text{ cm}^3\)
A cone has base radius 8 cm and vertical height 6 cm. Find its curved surface area, in terms of \(\pi\).
- A\(80\pi\text{ cm}^2\)
- B\(48\pi\text{ cm}^2\)
- C\(144\pi\text{ cm}^2\)
- D\(128\pi\text{ cm}^2\)
Find the total surface area of a solid hemisphere of radius 5 cm, correct to 3 significant figures.
- A\(236\text{ cm}^2\)
- B\(157\text{ cm}^2\)
- C\(314\text{ cm}^2\)
- D\(262\text{ cm}^2\)
A sphere has volume \(500\text{ cm}^3\). Find its radius, correct to 3 significant figures.
- A4.92 cm
- B10.9 cm
- C5.42 cm
- D6.31 cm
A cone has volume \(300\text{ cm}^3\) and base radius 6 cm. Find its slant height, correct to 3 significant figures.
- A9.97 cm
- B7.96 cm
- C6.56 cm
- D14.0 cm
Quiz key
Base area \(= 8 \times 6 = 48\text{ cm}^2\). \(V = \tfrac{1}{3} \times 48 \times 15 = 240\text{ cm}^3\). The tempting 720 forgets the \(\tfrac{1}{3}\), giving the volume of the cuboid the pyramid fits in. 360 uses \(\tfrac{1}{2}\) as if it were a triangle, and 16 is \(\tfrac{1}{3}\) of the base area with no height.
\(r = 4.5\) cm, so \(V = \tfrac{4}{3}\pi \times 4.5^3 = 121.5\pi = 381.7... = 382\text{ cm}^3\). The tempting 3050 uses the diameter 9 as the radius. 254 is the surface area \(4\pi r^2\), and 286 forgets the \(\tfrac{4}{3}\).
\(l = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\) cm, so curved surface \(= \pi \times 8 \times 10 = 80\pi\text{ cm}^2\). The tempting \(48\pi\) puts the vertical height into \(\pi rl\). \(144\pi\) also adds the base (\(64\pi\)), which is not part of the curved surface, and \(128\pi\) is the volume \(\tfrac{1}{3}\pi \times 64 \times 6\).
Curved part \(= \tfrac{1}{2} \times 4\pi \times 5^2 = 50\pi\). Flat face \(= \pi \times 5^2 = 25\pi\). Total \(= 75\pi = 235.6... = 236\text{ cm}^2\). The tempting 157 is the curved part alone, forgetting the flat face. 314 is a whole sphere, and 262 is the volume \(\tfrac{2}{3}\pi \times 5^3\).
\(r^3 = \dfrac{3 \times 500}{4\pi} = 119.3...\), so \(r = \sqrt[3]{119.3...} = 4.92\) cm. The tempting 10.9 cm takes the square root instead of the cube root. 5.42 cm forgets the \(\tfrac{4}{3}\), and 6.31 cm uses the surface area formula \(4\pi r^2 = 500\).
\(\tfrac{1}{3}\pi \times 6^2 \times h = 300\), so \(12\pi h = 300\) and \(h = 7.957...\) cm. Then \(l = \sqrt{6^2 + 7.957...^2} = 9.97\) cm. The tempting 7.96 cm is the vertical height, not the slant height. 6.56 cm forgets the \(\tfrac{1}{3}\) when finding \(h\), and 14.0 cm adds \(r + h\) instead of using Pythagoras.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.