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Cambridge IGCSE0580

Similarity

Mathematics 0580 Chapter Notes

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In this chapter

  1. 1Lengths of similar shapes
  2. 2Showing that triangles are similar
  3. 3Areas of similar shapes
  4. 4Volumes of similar solids
  5. 5Moving between lengths, areas and volumes

1. Lengths of similar shapes

In mathematically similar shapes, the angles in corresponding places are equal and corresponding lengths are in the same ratio. That ratio is the scale factor.

Find the scale factor by dividing a length on one shape by the corresponding length on the other. Then multiply the length you want by the scale factor. If you go from the larger shape to the smaller, the scale factor is less than 1.

Match the sides carefully. The longest side matches the longest side. Never find a difference and add it on. Keep the scale factor as a fraction, or to full accuracy, until the end.

k = (length on one shape)/(corresponding length on the other)

required length = k × given length

Examiner insight

“A frequent incorrect method of dealing with similar triangles was for candidates to write 18 – 12 = 6 so 27 – 12 = 15.” “The only common error was for candidates to lose accuracy by working with a two significant figure value as a scale factor.”

Worked example 12 marks

AD and BC are straight lines that intersect at X. Triangle ABX is similar to triangle DCX. CX = 5 cm, XD = 8 cm and AD = 20 cm. Find BX.

  1. 1

    AX = AD - XD = 20 - 8 = 12 cm [1]

  2. 2

    A ↔ D and B ↔ C, so BX/CX = AX/DX and BX = 5 × 12/8 = 7.5 cm [1]

2. Showing that triangles are similar

Two triangles are similar if their angles are equal in pairs. Show three pairs of equal angles. Name each pair, and give a geometric reason for each pair.

The usual reasons are corresponding angles, alternate angles (both on parallel lines), vertically opposite angles and a common angle.

End with a conclusion: the three angles are equal, so the triangles are similar. Saying only "they are similar" is not a reason.

Examiner insight

“A feature of the strongest solutions was the clarity of knowing that similar triangles would have 3 pairs of equal angles and in conclusion they stated this fact.” “Weaker solutions were not specific in the statements made, e.g. ‘they both have a right angle’, ‘they both share an angle’ without saying which pairs of angles they were referring to.”

Worked example 13 marks

In triangle ABC, the point P lies on AB and the point Q lies on AC. PQ is parallel to BC. Show that triangle APQ is similar to triangle ABC.

  1. 1

    angle APQ = angle ABC, corresponding angles (PQ is parallel to BC) [1]

  2. 2

    angle AQP = angle ACB, corresponding angles (PQ is parallel to BC) [1]

  3. 3

    angle A is common, so the three angles are equal and the triangles are similar [1]

3. Areas of similar shapes

If the length scale factor is k, the area scale factor is k². The same rule gives the surface area of similar solids.

Given two areas, take the square root of their ratio to get the length scale factor. Then use it on the lengths.

An area scale factor is not a length scale factor. Check which one each step needs.

area scale factor = k²

k = √((area of one)/(area of the other))

Examiner insight

“Many candidates did not recognise that area scale factor was the square of the linear scale factor.” “Candidates need to be aware that area scale factors are not the same as linear scale factors.”

Worked example 13 marks

The two cylinders, A and B, are mathematically similar. The surface area of A is 80 cm² and the surface area of B is 180 cm². The height of A is h cm and the height of B is 12 cm. Find the value of h.

  1. 1

    area ratio = 80/180 = 4/9 [1]

  2. 2

    length ratio = √(4/9) = 2/3 [1]

  3. 3

    h = 12 × 2/3 = 8 [1]

4. Volumes of similar solids

If the length scale factor is k, the volume scale factor is k³. Capacity scales in the same way as volume.

Given two volumes, take the cube root of their ratio to get the length scale factor. Then use it on the lengths.

The two volumes must be in the same units before you divide.

volume scale factor = k³

k = √[3](volume of one)/(volume of the other)

Examiner insight

“Approximately half of the candidates understood the requirement to cube root the volume scale factor to find the length scale factor and gave the correct answer.” “Candidates need to understand that a length scale has to be converted when working with a volume or area.”

Worked example 13 marks

Two bottles are mathematically similar. The larger bottle has a capacity of 1.35 litres and a height of 24 cm. The smaller bottle has a capacity of 0.4 litres. Calculate the height of the smaller bottle.

  1. 1

    volume ratio = 0.4/1.35 = 40/135 = 8/27 [1]

  2. 2

    length ratio = √[3]8/27 = 2/3 [1]

  3. 3

    height = 24 × 2/3 = 16 cm [1]

5. Moving between lengths, areas and volumes

Always pass through the length scale factor. From an area ratio, take the square root to get k. From a volume ratio, take the cube root to get k.

To go from areas to volumes, find k first, then cube it. Do not use the area ratio as a volume ratio.

Write k as a fraction where you can. It keeps the working exact.

length : k area : k² volume : k³

Examiner insight

“Others confused squaring, square rooting, cubing and cube rooting.”

Worked example 14 marks

Do not use a calculator. Two mathematically similar tins have surface areas 16 cm² and 36 cm². The smaller tin has a volume of 40 cm³. Find the volume of the larger tin.

  1. 1

    area ratio, larger to smaller = 36/16 = 9/4 [1]

  2. 2

    k = √(9/4) = 3/2 [1]

  3. 3

    volume factor = k³ = 27/8 [1]

  4. 4

    volume = 40 × 27/8 = 135 cm³ [1]

Recap

  • Similar shapes: corresponding lengths are in the same ratio, the scale factor k.
  • Find k by dividing corresponding lengths. Then multiply.
  • Extended: show three pairs of equal angles, each with a geometric reason.
  • Extended: area scale factor = k². Surface area follows the same rule.
  • Extended: volume scale factor = k³. Capacity follows the same rule.
  • Extended: go through k. Square root for areas, cube root for volumes.

End-of-chapter exercise

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

  1. Do not use a calculator. Two rectangles are mathematically similar. The smaller rectangle is 4 cm by 6 cm. The longer side of the larger rectangle is 15 cm. Find the shorter side of the larger rectangle.2 marks
  2. Triangle ABC is mathematically similar to triangle PQR. AB = 6 cm, BC = 9 cm and PQ = 8 cm. Calculate the length of QR.2 marks
  3. Do not use a calculator. A poster is 40 cm wide and 60 cm high. A mathematically similar poster is 1.2 m wide. Find the height of the larger poster in centimetres.3 marks
  4. Triangle DEF is mathematically similar to triangle XYZ. DE = 7 cm, EF = 9.2 cm and XY = 11 cm. Calculate the length of YZ. Give your answer correct to 3 significant figures.3 marks
  5. Do not use a calculator. Two mathematically similar triangles have corresponding sides of 3 cm and 12 cm. The area of the smaller triangle is 5 cm². Find the area of the larger triangle.2 marks
  6. Two mathematically similar picture frames have areas 18 cm² and 50 cm². The height of the smaller frame is 6 cm. Find the height of the larger frame.3 marks
  7. Do not use a calculator. Two mathematically similar bottles have heights 10 cm and 15 cm. The smaller bottle holds 200 ml. Find the capacity of the larger bottle.3 marks
  8. Lines AB and CD are parallel. The lines AD and BC cross at X. Give geometric reasons why triangle ABX is similar to triangle DCX.3 marks
  9. Do not use a calculator. Two mathematically similar water tanks have capacities 1.6 litres and 5.4 litres. The surface area of the larger tank is 1800 cm². The height of the smaller tank is 30 cm. (a) Find the surface area of the smaller tank. (b) Find the height of the larger tank.5 marks
  10. A solid cone has a height of 20 cm. A cut parallel to the base, 8 cm above the base, removes a small cone from the top. The small cone is mathematically similar to the original cone. The solid that remains has a volume of 1176 cm³. (a) Calculate the volume of the original cone. (b) The curved surface area of the original cone is 575 cm². Calculate the curved surface area of the small cone.6 marks

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