Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 9: a square has side 4 cm and is enlarged by scale factor 3. What is the area of the new square?
Answer
144 cm². The side is 12 cm and 12² = 144, which is 9 times 16. - From the last lesson: a cone has a base area of 10 cm² and a height of 6 cm. What is its volume?
Answer
20 cm³, because ⅓ × 10 × 6 = 20. - From Year 9: what is the cube root of 27 ÷ 8?
Answer
3 ÷ 2, because 3 × 3 × 3 = 27 and 2 × 2 × 2 = 8.
Words used in this lesson
- similar solids
- solids with the same shape, one an enlargement of the other. For example: two cans in the ratio 3 : 5.
- scale factor
- the number every length is multiplied by. For example: k = 1.4 for an enlargement.
- corresponding
- matching in the two solids. For example: the two lids of two cans.
- frustum
- a cone with its top cut off by a plane parallel to the base. For example: a plant pot.
- ratio
- a comparison of two quantities by division. For example: 1 : 20 for a model.
The lantern that grew
A lantern-maker in Kyoto builds a small paper lantern, then a street lantern that is exactly four times as tall, with every dimension in the same ratio. Do the paper for the walls and the oil for the lamp go up in the same ratio as the heights?
Predict first: by what number do you think the paper needed is multiplied, and by what number the volume of oil? Use the figure to test your guesses on a cuboid, then think about the lantern.
Scale factor 1: the cuboid is 2 by 3 by 5 cm, with surface area 62 cm² and volume 30 cm³.
Show me the steps
- Set the scale factor to 2: every edge doubles. The surface area is 248 cm², which is 62 × 4, and the volume is 240 cm³, which is 30 × 8.
- Set the scale factor to 3: the surface area is 558 cm², which is 62 × 9, and the volume is 810 cm³, which is 30 × 27.
- Set the scale factor to 0.5: the surface area is 15.5 cm², which is 62 ÷ 4, and the volume is 3.75 cm³, which is 30 ÷ 8. Halving the lengths cuts the volume to an eighth.
Similar solids. Two solids are similar if one is an enlargement of the other: every pair of corresponding lengths is in the same ratio, k.
Three rules. Lengths are multiplied by k, areas by k², and volumes by k³. A face is made of length times length, so each of the two lengths is multiplied by k. A volume is length times length times length.
Going backwards. An area ratio is k², so k is its square root. A volume ratio is k³, so k is its cube root. If you know one ratio you can find the others: for example the volume ratio is the area ratio taken to the power 3 ÷ 2.
Percentage change. A scale factor k multiplies an area by k², so the percentage increase in area is (k² − 1) × 100%, and the same pattern holds for volume.
Similar means all lengths in the same ratio. A cylinder of radius 6 cm and height 10 cm is not an enlargement of one of radius 3 cm and height 10 cm, so the k³ rule does not apply: its volume is 4 times as big, not 8. And the area ratio is not the length ratio: a scale factor of 2 gives 4, not 2.
Worked examples
1. Two similar tins are 6 cm and 9 cm tall. The smaller has a label of area 80 cm² and holds 120 mL. Find the larger tin's label area and capacity.
- Scale factor: k = 9 ÷ 6 = 1.5.
- Area: 80 × 1.5² = 80 × 2.25 = 180 cm².
- Volume: 120 × 1.5³ = 120 × 3.375 = 405 mL.
2. Three similar vases A, B and C. The volumes of A and B are in the ratio 27 : 64. The surface areas of B and C are in the ratio 16 : 25. Find the ratio of the surface areas of A and C, and of their volumes.
- A to B: k³ = 27 : 64, so k = 3 : 4. B to C: k² = 16 : 25, so k = 4 : 5.
- Chain the scale factors: A to C is 3 ÷ 4 × 4 ÷ 5 = 3 ÷ 5.
- Areas: (3 : 5)² = 9 : 25. Volumes: (3 : 5)³ = 27 : 125.
3. By what percentage must the surface area of a solid increase to double its volume?
- Volume doubles, so k³ = 2 and k = ³√2 ≈ 1.26.
- Area is multiplied by k² = 1.26² ≈ 1.587.
- That is an increase of about 58.7%.
Your go. The first step is given. Two similar tins have surface areas 150 cm² and 600 cm². The smaller holds 250 cm³. Find the capacity of the larger.
- Area ratio: 600 ÷ 150 = 4, so the scale factor is k = √4.
- Now cube k and multiply by 250.
Check the answer
k = 2, so the volume is multiplied by 2³ = 8: 250 × 8 = 2 000 cm³.
With a GDC (optional)
A GDC finds a scale factor from a ratio: for a volume ratio, take the cube root with the root key or the power 1 ÷ 3, and for an area ratio the square root. Keep the unrounded k in memory, then apply it to the lengths, areas and volumes you need.
Rounding k early spoils a cube. Everything here can be done without a GDC: a scientific calculator's cube-root key gives a value such as ³√2.5, and where the ratio is a perfect cube or square, k is exact and the work is by hand.
Check before you go on
1. Two similar bottles are 10 cm and 15 cm tall. The smaller holds 200 mL. How much does the larger hold?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. Two similar jugs are 8 cm and 20 cm tall. The smaller has a surface area of 96 cm². Find the surface area of the larger, in square centimetres.
3. A 1 : 20 scale model of a lorry has a cargo space of 12 000 cm³. Find the volume of the real cargo space, in cubic metres. (1 m³ = 1 000 000 cm³.)
Level 2 · Building
4. A potter in Jingdezhen makes similar bowls. A bowl 8 cm tall needs clay costing 30 yuan and glaze costing 4 yuan. The clay cost follows volume and the glaze cost follows surface area. Find the total cost of clay and glaze for a similar bowl 20 cm tall, to the nearest yuan.
5. Two similar solids have surface areas of 45 cm² and 80 cm². The smaller has volume 135 cm³. Find the volume of the larger, in cubic centimetres.
6. A solid candle is a cone 12 cm tall with volume 1728 cm³. Two cuts parallel to the base are made 4 cm and 6 cm below the tip. Find the volume of the slice between the two cuts, in cubic centimetres.
Level 3 · Stretch
7. A larger tin is similar to a smaller one and holds 2.5 times as much. By what factor is its surface area larger? Give your answer to two decimal places.
8. A solid is enlarged so that its surface area increases by 44%. By what percentage does its volume increase? Give your answer to one decimal place.
Something to think about
A model animal is enlarged so that every length is 3 times as long. Its weight grows with its volume, and the strength of its leg bones grows with the area of their cross-section. By what factor does each grow, and what happens to the weight carried by each square centimetre of bone?
One way to think about it
Weight is multiplied by 3³ = 27 and the bone cross-section by 3² = 9, so each square centimetre of bone carries 27 ÷ 9 = 3 times as much. Enlarging by k makes the load on each square centimetre k times larger, which is why an elephant needs much thicker legs, for its size, than a gazelle, and why a giant ant could not stand up.
Exit ticket
9. Two similar solids have volumes in the ratio 8 : 125. What is the ratio of their surface areas?