The shape of the lesson
By the end of the lesson, learners can
- Three scale factors: In the lab, enlarge a solid and watch three readouts: length, surface area and volume.
- Working backwards: When you are given two areas or two volumes, take the square root or cube root to get the length factor first.
- Capacity, mass and paint: In context, decide whether each quantity behaves like a length, an area or a volume.
- Identify what you are given: two lengths, two areas or two volumes.
- Reduce to \(k\): divide, then square root (areas) or cube root (volumes).
- Build up: \(k\) for lengths, \(k^2\) for areas, \(k^3\) for volumes, capacity and mass.
- Check direction: bigger shape, bigger answer.
- Units: 1 m2 = 10 000 cm2; 1 km2 = 1010 cm2.
One picture every idea returns to
If lengths are multiplied by \(k\), areas (including surface areas) are multiplied by \(k^2\) and volumes (and masses or capacities of the same material) by \(k^3\). Always get back to the LENGTH factor \(k\) first.
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, Square it, cube it.. Extended. 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: Three scale factors | In the lab, enlarge a solid and watch three readouts: length, surface area and volume. Project tab 3 of the lab (Similarity lab, area and volume) and let a student drive it while the class predicts. Model the worked example Area from a length ratio on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Working backwards | When you are given two areas or two volumes, take the square root or cube root to get the length factor first. Model the worked example Length from areas on the board, one line at a time. Your turn: Two similar cones have volumes 54 cm 3 and 250 cm 3. The smaller has height 6 cm. Find the height of the larger. | Pairs |
| 21 to 30 min | Teach and check: Capacity, mass and paint | In context, decide whether each quantity behaves like a length, an area or a volume. Model the worked example Similar bottles on the board, one line at a time. Your turn: A model car is made to a scale of 1: 20. The real car has a surface area of 12 m 2. Find the surface area of the model in cm 2. | Rally Coach |
| 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, 18 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 Capacity, mass and paint, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Multiplying an area by the LENGTH factor is the most common error: \(12 \times 2.5 = 30\) cm 2 is wrong. | Areas need \(k^2\). |
| Never go straight from an area ratio to a volume ratio by multiplying by the area factor. | Find \(k\) first: from areas 16: 25 the volume factor is \((5/4)^3\), not \((25/16)\) and not \((25/16)^2\). |
| Mass scales with volume only for solids made of the same material. | And a map scale of 1: 50 000 means AREAS shrink by \(50\,000^2\), not by 50 000. |
Areas and volumes of similar shapes: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Two similar solids have lengths in the ratio \(3 : 4\). Write down the ratio of (a) their surface areas (b) their volumes.
Two similar triangles have bases 6 cm and 15 cm. The smaller triangle has area 20 cm2. Find the area of the larger triangle.
Two similar jugs hold 250 ml and 2 litres. The smaller jug is 12 cm tall. How tall is the larger jug?
Two similar solids have surface areas 50 cm2 and 72 cm2. The smaller has volume 125 cm3. Find the volume of the larger.
A map has a scale of 1 : 50 000. A lake has an area of 6 cm2 on the map. Find the real area of the lake in km2.
Two paint tins are similar. The small tin is 10 cm tall and holds 1.25 litres; its label has area 150 cm2. The large tin is 16 cm tall.
(a) Find the capacity of the large tin. (b) Find the area of the large tin's label. (c) The small tin costs $4.50 and the large tin costs $16.00. Which is better value? Show your working.
Answer key and marking notes
- (a) \(3^2 : 4^2 = 9 : 16\)
- (b) \(3^3 : 4^3 = 27 : 64\)
- \(k = 15 \div 6 = 2.5\); area factor \(= 6.25\).
- \(20 \times 6.25 = 125\) cm2.
- 2 litres = 2000 ml, so the volume factor is \(2000 \div 250 = 8\).
- \(k = \sqrt[3]{8} = 2\).
- Height \(= 12 \times 2 = 24\) cm.
- Area factor \(= 72 \div 50 = 1.44\), so \(k = \sqrt{1.44} = 1.2\).
- Volume factor \(= 1.2^3 = 1.728\).
- \(125 \times 1.728 = 216\) cm3.
- Area factor \(= 50\,000^2 = 2.5 \times 10^{9}\).
- Real area \(= 6 \times 2.5 \times 10^{9} = 1.5 \times 10^{10}\) cm2.
- \(1\) km \(= 10^{5}\) cm, so \(1\) km2 \(= 10^{10}\) cm2. Area \(= 1.5\) km2.
- \(k = 16 \div 10 = 1.6\).
- (a) \(1.25 \times 1.6^3 = 1.25 \times 4.096 = 5.12\) litres.
- (b) \(150 \times 1.6^2 = 150 \times 2.56 = 384\) cm2.
- (c) Small: \(4.50 \div 1.25 = 3.60\), so $3.60 per litre. Large: \(16.00 \div 5.12 = 3.125\), about $3.13 per litre.
- The large tin is better value.
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.
Two similar cones have heights 6 cm and 9 cm. The curved surface area of the smaller cone is \(40\text{ cm}^2\). Find the curved surface area of the larger cone.
- A\(90\text{ cm}^2\)
- B\(60\text{ cm}^2\)
- C\(135\text{ cm}^2\)
- D\(43\text{ cm}^2\)
Two similar jars have volumes in the ratio \(8 : 27\). The smaller jar is 10 cm tall. How tall is the larger jar?
- A15 cm
- B33.75 cm
- C18.4 cm
- D22.5 cm
Two similar solids have surface areas \(49\text{ cm}^2\) and \(64\text{ cm}^2\). The smaller solid has a volume of \(343\text{ cm}^3\). Find the volume of the larger solid.
- A\(512\text{ cm}^3\)
- B\(448\text{ cm}^3\)
- C\(392\text{ cm}^3\)
- D\(585\text{ cm}^3\)
A map has a scale of 1 : 25 000. A forest covers \(3\text{ cm}^2\) on the map. Find the real area of the forest in \(\text{km}^2\).
- A\(0.1875\text{ km}^2\)
- B\(18\,750\text{ km}^2\)
- C\(0.0000075\text{ km}^2\)
- D\(1875\text{ km}^2\)
Two similar ornaments are made of the same glass. Their masses are 1.6 kg and 5.4 kg. The smaller ornament has a surface area of \(80\text{ cm}^2\). Find the surface area of the larger ornament.
- A\(180\text{ cm}^2\)
- B\(270\text{ cm}^2\)
- C\(120\text{ cm}^2\)
- D\(911\text{ cm}^2\)
A solid cone has volume \(160\text{ cm}^3\). It is cut by a plane parallel to its base, halfway up its height. The top part is a small cone. Find the volume of the bottom part (the frustum).
- A\(140\text{ cm}^3\)
- B\(80\text{ cm}^3\)
- C\(120\text{ cm}^3\)
- D\(20\text{ cm}^3\)
Quiz key
Length factor \(k = 9 \div 6 = 1.5\), so the area factor is \(k^2 = 2.25\) and the larger area is \(40 \times 2.25 = 90\text{ cm}^2\). The tempting 60 multiplies the area by the LENGTH factor. 135 uses the volume factor \(1.5^3\), and 43 adds the 3 cm difference in height.
Volume factor \(= \tfrac{27}{8}\), so the length factor is \(k = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}\) and the height is \(10 \times 1.5 = 15\) cm. The tempting 33.75 cm multiplies the height by the VOLUME factor \(\tfrac{27}{8}\). 18.4 cm takes a square root instead of a cube root, and 22.5 cm multiplies by \(k^2 = 2.25\).
Area factor \(= \tfrac{64}{49}\), so \(k = \sqrt{\tfrac{64}{49}} = \tfrac{8}{7}\) and the volume factor is \(\left(\tfrac{8}{7}\right)^3 = \tfrac{512}{343}\). Volume \(= 343 \times \tfrac{512}{343} = 512\text{ cm}^3\). The tempting 448 multiplies the volume by the AREA factor \(\tfrac{64}{49}\). 392 uses \(k\) alone, and 585 squares the area factor, which is \(k^4\), not \(k^3\).
Real area \(= 3 \times 25\,000^2 = 1\,875\,000\,000\text{ cm}^2\). Since \(1\text{ km} = 100\,000\) cm, \(1\text{ km}^2 = 100\,000^2 = 10^{10}\text{ cm}^2\), so the area is \(1.875 \times 10^9 \div 10^{10} = 0.1875\text{ km}^2\). The tempting 18 750 divides by \(100\,000\) only once, converting the area as if it were a length. 0.0000075 multiplies by 25 000 instead of \(25\,000^2\), and 1875 divides by \(10^6\), the number of \(\text{m}^2\) in a \(\text{km}^2\).
Volume (mass) factor \(= 5.4 \div 1.6 = \tfrac{27}{8}\), so \(k = \sqrt[3]{\tfrac{27}{8}} = 1.5\) and the area factor is \(1.5^2 = 2.25\). Surface area \(= 80 \times 2.25 = 180\text{ cm}^2\). The tempting 270 multiplies by the mass ratio 3.375 directly. 120 uses \(k\), and 911 squares the mass ratio.
The small cone has length factor \(\tfrac{1}{2}\), so its volume is \(160 \times \left(\tfrac{1}{2}\right)^3 = 20\text{ cm}^3\). The frustum is \(160 - 20 = 140\text{ cm}^3\). The tempting 80 halves the volume, using the length factor. 120 uses the area factor \(\tfrac{1}{4}\) for the small cone (\(160 - 40\)), and 20 is the small cone, not the frustum.
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.