MathematicsFoundation20 min read

Similar Figures

Same shape, different size — and what that does to areas and volumes

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01

Congruent and similar are not the same word

Definition

Similar figures — Two polygons are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. For triangles, either condition on its own is enough to force the other.

Two figures are congruent if one can be placed exactly on top of the other: same shape and same size. They are similar if they have the same shape but not necessarily the same size — an enlargement or reduction of one another.

Congruence is the special case of similarity where the scale factor happens to be 1. Every congruent pair is similar; almost no similar pair is congruent.

TestApplies toMeaning
AAA (or AA)trianglestwo pairs of equal angles force the third, so the triangles are similar
SSStrianglesall three pairs of sides in the same ratio
SAStrianglestwo pairs of sides in ratio with the included angles equal
Angles + ratiosother polygonsboth conditions must be checked — equal angles alone is not enough

Equal angles is not enough outside triangles

A square and a 2 cm × 5 cm rectangle both have four right angles, but they are not similar. Triangles are special: fixing the angles fixes the shape, which is why AA works for them and only for them.

02

The scale factor, and the two powers of it

If every length in a figure is multiplied by k, then every area is multiplied by and every volume by . This is not a separate rule to memorise; it falls straight out of the fact that area is two lengths multiplied and volume is three.

A 3 cm cube has surface area 54 cm² and volume 27 cm³. Double every edge and you get a 6 cm cube with surface area 216 cm² (four times as much) and volume 216 cm³ (eight times as much). Lengths ×2, areas ×4, volumes ×8.

length ratio = karea ratio = k²volume ratio = k³if a question gives you an area ratio, take the square root to get back to k

Push k to 2. The length bar doubles, the area bar quadruples and the volume bar grows eightfold. This is why a model at 1:10 scale needs a thousandth of the material, not a tenth of it.

03

Working with corresponding parts

Almost every mark lost in this chapter comes from pairing the wrong sides. Before writing any ratio, name the triangles so that corresponding vertices appear in corresponding positions: writing "triangle ABC is similar to triangle PQR" states that A matches P, B matches Q and C matches R.

Once the order is fixed, the ratios follow mechanically: AB/PQ = BC/QR = CA/RP.

Worked example

In triangles ABC and PQR, ∠A = ∠P, ∠B = ∠Q, AB = 6 cm, BC = 9 cm and PQ = 4 cm. Find QR, and the ratio of the areas.

  1. Two pairs of angles are equal, so by AA the triangles are similar with A↔P, B↔Q, C↔R.The third pair of angles is then equal automatically, since angles in a triangle sum to 180°.
  2. Scale factor from ABC to PQR: k = PQ/AB = 4/6 = 2/3.Use the pair of sides you know completely. Keep the direction consistent for the rest of the question.
  3. QR = k × BC = (2/3) × 9 = 6 cm.QR corresponds to BC because Q↔B and R↔C.
  4. Area ratio = k² = (2/3)² = 4/9.Areas scale by the square, so triangle PQR has 4/9 of the area of ABC.

QR = 6 cm; area of PQR : area of ABC = 4 : 9

04

The intercept theorem

A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio. This is really the AA similarity test wearing a disguise: the small triangle at the top and the whole triangle share an angle and have equal corresponding angles at the parallel line, so they are similar.

It is the result behind most "find x" diagrams in this chapter, and it works in both directions — if the two sides are cut in the same ratio, the line must be parallel.

If DE ∥ BC in triangle ABC, then AD/DB = AE/ECand also AD/AB = AE/AC = DE/BCthe first form uses the pieces; the second uses whole sides — mixing them is the usual error

Which ratio does the question want?

The two forms above are both correct but they are not interchangeable. AD/DB compares the two pieces of a side; AD/AB compares a piece with the whole side. Only the second form can be used with DE/BC, because DE corresponds to the whole of BC. Decide which one you need before substituting numbers.

Before you leave this chapter

  1. Similar = same shape, corresponding angles equal and corresponding sides in a fixed ratio k.
  2. For triangles, AA is enough. For any other polygon you must check angles and ratios.
  3. Lengths ×k, areas ×k², volumes ×k³. From an area ratio, square-root to recover k.
  4. Name the triangles in corresponding order before writing a single ratio.
  5. A line parallel to one side of a triangle divides the other two sides in the same ratio.
06

Congruence: the four tests, and the one that is not a test

Similarity asks whether two figures have the same shape. Congruence asks whether they are identical, and it has its own set of tests that the paper expects by name.

For triangles, any one of these four is enough to prove congruence: SSS (three sides), SAS (two sides and the angle between them), ASA (two angles and the side between them, or AAS which reduces to it), and RHS (right angle, hypotenuse and one other side).

What is not a test is AAA. Three matching angles prove similarity, never congruence — any two equilateral triangles have identical angles and can be any size at all. The other near-miss is SSA, two sides and a non-included angle, which can produce two genuinely different triangles from the same data.

Why SAS insists on the included angle

Fix two sticks of length 5 cm and 7 cm hinged at a point. The angle at the hinge decides the third side completely, so SAS pins the triangle down. Give the 5 cm side, the 7 cm side and an angle not between them and the third vertex can often be placed in two different positions — which is exactly why SSA is rejected as a test.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Distinguish between congruent and similar figures.
Model answer

Congruent figures have the same shape and the same size, so corresponding sides are equal. Similar figures have the same shape only; corresponding sides are in a constant ratio k, and congruence is the case k = 1.

Examiner tip. Mentioning that congruence is similarity with k = 1 shows the relationship between the two and usually secures the second mark.

SQ2[2 marks]
Two similar triangles have areas in the ratio 9 : 25. Find the ratio of their corresponding sides.
Model answer

Areas scale by k², so k² = 9/25 and k = 3/5. The sides are in the ratio 3 : 5.

Examiner tip. Square-root an area ratio to get back to lengths. Answering 9 : 25 is the trap this question is built around.

SQ3[2 marks]
Explain why all circles are similar but not all rectangles are.
Model answer

A circle is determined completely by its radius, so any circle is an enlargement of any other. A rectangle needs two independent lengths, so two rectangles are similar only if their length-to-width ratios match — a 2 × 5 and a 3 × 5 rectangle are not similar.

Examiner tip. The general principle is worth stating: a shape family is always similar when a single length determines the whole figure.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
In triangle ABC, DE is parallel to BC with D on AB and E on AC. AD = 4 cm, DB = 6 cm and AE = 5 cm. Find EC and the ratio DE : BC.
Full working
  1. By the intercept theorem AD/DB = AE/EC, so 4/6 = 5/ECpieces compared with pieces[1]
  2. 4 × EC = 30, so EC = 7.5 cm[1]
  3. For the ratio of DE to BC use whole sides: AD/AB = 4/10 = 2/5AB = AD + DB = 10 cm — this is the step most often skipped[1]
  4. DE : BC = 2 : 5accept 0.4[1]

EC = 7.5 cm; DE : BC = 2 : 5

Examiner tip. The switch from pieces (4 : 6) to whole sides (4 : 10) between the two parts is deliberate. Writing DE : BC = 4 : 6 is the standard wrong answer.

N2[4 marks]
A model aeroplane is built to a scale of 1 : 50. The model has a wing area of 120 cm² and a volume of 400 cm³.
  1. Find the wing area of the real aeroplane in cm².
  2. Find the volume of the real aeroplane in cm³.
Full working
  1. Area scale factor = 50² = 2500areas scale by the square of the length ratio[1]
  2. Real wing area = 120 × 2500 = 300 000 cm²accept 30 m²[1]
  3. Volume scale factor = 50³ = 125 000[1]
  4. Real volume = 400 × 125 000 = 5 × 10⁷ cm³accept 50 000 000 cm³ or 50 m³[1]

(a) 300 000 cm² (b) 5 × 10⁷ cm³

Examiner tip. Scale-model questions are testing k² and k³ and nothing else. Write the scale factor down as a number before multiplying anything.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Two similar cylindrical water tanks have heights 1.2 m and 1.8 m.
  1. Find the scale factor from the smaller tank to the larger.
  2. The smaller tank has a curved surface area of 4.8 m². Find the curved surface area of the larger.
  3. The larger tank holds 2 430 litres. Find the capacity of the smaller.
Mark scheme
  1. k = 1.8 / 1.2 = 1.5from smaller to larger, so k > 1[1]
  2. Area factor = k² = 2.25[1]
  3. Larger surface area = 4.8 × 2.25 = 10.8 m²[1]
  4. Volume factor = k³ = 3.375[1]
  5. Going from larger to smaller, divide: 2430 / 3.375direction of the scale factor matters here[1]
  6. = 720 litres[1]

(a) k = 1.5 (b) 10.8 m² (c) 720 litres

Examiner tip. Part (c) reverses the direction, so you divide rather than multiply. Before every step, ask whether the answer should be larger or smaller than what you started with — it catches an inverted scale factor instantly.