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Cambridge IGCSE0607

Similarity

International Mathematics 0607 Chapter Notes

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In this chapter

  1. 1Similar shapes and the scale factor
  2. 2Finding a missing length
  3. 3Areas of similar shapes
  4. 4Volumes of similar solids
  5. 5Lengths, surface areas and volumes together
  6. 6Showing that triangles are similar

1. Similar shapes and the scale factor

Mathematically similar shapes are enlargements of each other. Each length on one shape is the same number of times longer than the corresponding length on the other. That number is the scale factor.

Find it by dividing. Never subtract.

scale factor = (length on the new shape)/(corresponding length on the original)

new length = original length × scale factor

Examiner insight

“The most common error was to subtract 5 from 20 to give the increase in length rather than the scale factor.” “Scale factors are always found by division or multiplication and never by subtraction or addition.”

Worked example 13 marks

Rectangle A and rectangle B are mathematically similar. Find the perimeter of rectangle A.

  1. 1

    scale factor from A to B = 21 ÷ 7 = 3 [1]

  2. 2

    height of A = 15 ÷ 3 = 5 cm

  3. 3

    perimeter = 2(7 + 5) [1]

  4. 4

    = 24 cm [1]

2. Finding a missing length

Pair each side with its corresponding side on the other shape. Use a pair where both lengths are known to find the scale factor. Then multiply or divide the side you want.

Check the size of the answer. A length on the smaller shape must be smaller than its partner on the larger one.

(side on shape 1)/(its match on shape 2) = (another side on shape 1)/(its match on shape 2)

Examiner insight

“This comes from the totally wrong method of noting that 6 is 2 less than 8 so x must be 2 less than 20.” “Similar triangles are enlargements of each other so one way to gain a mark is to find the scale factor between the two or to show the full method.”

Worked example 12 marks

Parallelogram ABCD is mathematically similar to parallelogram EFGH. Find the length of BC.

  1. 1

    AB matches EF, so scale factor = 15 ÷ 3 = 5 [1]

  2. 2

    BC matches FG, so BC = 2 × 5 = 10 cm [1]

3. Areas of similar shapes

If the lengths of two similar shapes are in the ratio 1 : k, their areas are in the ratio 1 : k². To go from a length scale factor to an area scale factor, square it.

To go the other way, take the square root of the area scale factor. Decide first whether the ratio you hold is of lengths or of areas.

(Area of A)/(Area of B) = ((Length of A)/(Length of B))²

Examiner insight

“The most common incorrect approach was to square the ratio of the areas to obtain the length scale factor instead of square rooting, or to multiply 24 by the reciprocal of the correct ratio.” “Given that XYZ was a smaller triangle, candidates are reminded to check their answer against what they are expecting to find, so the length of YZ should be less than the equivalent length (24 cm) in triangle ABC.”

Worked example 12 marks

Trapezium U is mathematically similar to trapezium T. The area of trapezium U is 80 cm². Find the area of trapezium T.

  1. 1

    T is the larger shape, so its area must be more than 80 cm².

  2. 2

    length scale factor from U to T = 32 ÷ 20, so area scale factor = (32/20)² = 2.56 [1]

  3. 3

    area of T = 80 × 2.56 = 204.8 cm² [1]

4. Volumes of similar solids

If the lengths of two similar solids are in the ratio 1 : k, their volumes are in the ratio 1 : k³. To go from a length scale factor to a volume scale factor, cube it.

To go the other way, take the cube root. On a graphic display calculator, enter the scale factor to the power 1/3, for example `2^(1/3)`, and read off the value.

(Volume of A)/(Volume of B) = ((Length of A)/(Length of B))³

Examiner insight

“Almost all the candidates fell into one of two groups: those who used the volume scale factor as the linear scale factor producing the incorrect answer 12.5, and those using the correct cube root of the volume scale factor and reaching the correct answer.”

Worked example 12 marks

The two solids are mathematically similar. The volume of solid B is 2 times the volume of solid A. The height of solid A is 8 cm. The height of solid B is h cm. Find the value of h.

  1. 1

    B is the larger solid, so h must be more than 8.

  2. 2

    h/8 = √[3]2 [1]

  3. 3

    h = 8 × 2^(1/3) = 10.07… = 10.1 to 3 significant figures [1]

5. Lengths, surface areas and volumes together

Lengths go with k, surface areas with k² and volumes with k³. A surface area is an area, so it is squared, not cubed.

To get from a volume ratio to an area ratio, go through the lengths: cube root the volume ratio, then square the result.

length ratio = 1 : k area ratio = 1 : k² volume ratio = 1 : k³

Examiner insight

“The most common error was to find that the volume scale factor was 8 and just multiply the total surface area of the small solid by 8.” “Candidates were expected to realise that the information given related to volumes and that they needed to cube root the given values to find the ratio of lengths.”

Worked example 14 marks

Do not use a calculator. Two mathematically similar solids have volumes 24 cm³ and 81 cm³. The surface area of the smaller solid is 100 cm². Find the surface area of the larger solid.

  1. 1

    volume ratio = 24 : 81 = 8 : 27 [1]

  2. 2

    length ratio = √[3]8 : √[3]27 = 2 : 3 [1]

  3. 3

    area ratio = 2² : 3² = 4 : 9 [1]

  4. 4

    surface area of larger solid = 100 × 9/4 = 225 cm² [1]

6. Showing that triangles are similar

Similar triangles are enlargements of each other. Their equivalent angles are the same and their equivalent sides are in the same ratio.

To show two triangles are similar, show that their angles are equal in pairs and give a geometric reason for each pair. The angle sum of a triangle is one such reason. Then use the ratio of corresponding sides.

Examiner insight

“The reason needed for the two angles being equal was some reference to the angle sum of a triangle.” “Some earned one mark by giving two correct pairs of angles without reasons.”

Worked example 14 marks

In triangle PQR, angle P = 50°, angle Q = 70°, PQ = 6 cm and QR = 8 cm. In triangle STU, angle S = 50°, angle U = 60° and ST = 9 cm. Show that the triangles are similar, then find TU.

  1. 1

    angle R = 180° - 50° - 70° = 60° (angle sum of a triangle) [1]

  2. 2

    angle P = angle S = 50° and angle R = angle U = 60°, so angle Q = angle T too (angle sum of a triangle) and the triangles are similar [1]

  3. 3

    ST matches PQ, as each is opposite the 60° angle, so scale factor = 9 ÷ 6 = 1.5 [1]

  4. 4

    TU matches QR, as each is opposite the 50° angle, so TU = 8 × 1.5 = 12 cm [1]

Recap

  • Similar shapes are enlargements of each other: corresponding lengths have the same scale factor.
  • Scale factor = new length ÷ corresponding original length. Never subtract.
  • Pair corresponding sides first, then multiply or divide. Check the answer is smaller or larger as it should be.
  • Extended: length ratio 1 : k, area ratio 1 : k², volume ratio 1 : k³.
  • Extended: from an area ratio take the square root. From a volume ratio take the cube root.
  • Extended: to show triangles are similar, show equal angles in pairs and give a geometric reason, such as the angle sum of a triangle.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State how the scale factor is found from two corresponding lengths.1 mark
  2. Rectangle P is 4 cm by 6 cm. Rectangle Q is mathematically similar to P and its shorter side is 10 cm. Find the longer side of Q.2 marks
  3. Do not use a calculator. Triangle ABC is mathematically similar to triangle PQR. AB = 12 cm, BC = 18 cm and PQ = 8 cm. Find QR.2 marks
  4. A photograph is 15 cm wide and 10 cm high. An enlargement of it is mathematically similar and 24 cm wide. Find the perimeter of the enlargement.3 marks
  5. A model of a ship is mathematically similar to the real ship. The model is 45 cm long and its mast is 30 cm tall. The mast on the real ship is 12 m tall. Find the length of the real ship in metres.4 marks
  6. Do not use a calculator. The lengths of two similar shapes are in the ratio 2 : 5. Find the ratio of their areas.2 marks
  7. Do not use a calculator. Two mathematically similar triangles have areas 18 cm² and 50 cm². The shortest side of the smaller triangle is 6 cm. Find the shortest side of the larger triangle.3 marks
  8. Two mathematically similar bottles have heights 10 cm and 15 cm. The smaller bottle has volume 200 cm³. Find the volume of the larger bottle.3 marks
  9. Do not use a calculator. A drink is sold in mathematically similar cartons. The small carton holds 250 ml and the large carton holds 2 litres. A label covers the whole front face of each carton. The label on the large carton has area 360 cm². (a) Find the area of the label on the small carton. (b) A third similar carton has a label of area 810 cm². Find, in litres, how much it holds.6 marks
  10. Do not use a calculator. In triangle ABC, point D lies on side AB and point E lies on side AC. DE is parallel to BC. AD = 6 cm and DB = 4 cm. The area of triangle ADE is 27 cm². (a) Show that triangles ADE and ABC are similar. (b) Find the area of quadrilateral DBCE.6 marks

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