Intermediate level

Similarity and the intercept theorem

Two similar figures have the same shape and a different size, and everything that ties them together fits into a single number — the scale factor. The intercept theorem turns parallelism into a proportion, and the three powers of the scale explain why a model twice as large weighs eight times as much.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

  • The height of a tree from the length of a shadow
    A 1.5 m pole casts a 2 m shadow while the tree beside it casts an 18 m one. The triangles are similar, so the tree is 1.5 · 18 / 2 = 13.5 m tall. The whole measurement comes down to one proportion and a tape measure.
  • Map and ground
    On a 1 : 25,000 map an 8 cm segment corresponds to 8 · 25,000 = 200,000 cm, that is 2 km on the ground. Careful with areas: a plot covering 4 cm² on the map is really 4 · 25,000² cm² = 250,000 m², not 4 · 25,000.
  • An architectural model
    A 1 : 50 model of a building has a volume of 24 dm³. The building is 50 times larger in every dimension, so its volume is 24 · 50³ = 3,000,000 dm³, that is 3000 m³. Multiplying by fifty alone would understate the result 2500 times.
  • A cake twice the size
    A 30 cm cake has twice the diameter of a 15 cm one, but at the same layer thickness its area is 2² = 4 times larger — so four times as much filling and icing has to be prepared, not twice as much.
  • Projection and image size
    A projector 2 m from the screen throws an image 1.2 m wide. Moved back to 5 m, the light cone is similar with a scale of 5 / 2 = 2.5, so the image is 1.2 · 2.5 = 3 m wide — and covers more than six times the area, which is why it looks noticeably dimmer.

All formulas

  • Scale factor

    k=aak = \frac{a'}{a}

    the image side divided by the original side

  • The intercept theorem

    SASC=ABCD\frac{|SA|}{|SC|} = \frac{|AB|}{|CD|}

    segments on the arms of an angle cut by parallels

  • Perimeter of a similar figure

    L=kLL' = k \cdot L

    every length grows k times

  • Area of a similar figure

    P=k2PP' = k^2 \cdot P

    an area depends on two dimensions

  • Volume of a similar solid

    V=k3VV' = k^3 \cdot V

    a volume depends on three dimensions

Two figures are similar when they have the same shape and differ only in size. That whole difference fits into a single number — the scale factor kk, the ratio of corresponding lengths:

k=aak = \frac{a'}{a}

A factor above one is an enlargement, below one a reduction, and k=1k = 1 is congruence. It is the same idea that sits behind a map scale, carried over from a drawing to any figure at all.

When triangles are similar

As with congruence, not everything has to be checked:

CriterionWhat must agree
angle-angletwo angles of one triangle equal two angles of the other
side-side-sideall three pairs of sides proportional
side-angle-sidetwo pairs of sides proportional and the angle between them equal

The first is the one used in practice: since the angles of a triangle add up to 180180^\circ, agreeing on two forces agreement on the third.

The intercept theorem

Cut the arms of an angle with two parallel lines. Two triangles appear that share a vertex and have equal angles — hence they are similar. Written as proportions, that observation is the intercept theorem (Thales's theorem):

ACBD|SA||AB||SC||CD|S
The lines AC and BD are parallel. The segments cut off on one arm are in the same ratio as those on the other.
SASC=ABCD\frac{|SA|}{|SC|} = \frac{|AB|}{|CD|}

The theorem works both ways: if the corresponding segments are proportional, the cutting lines are parallel. It is that second version that lets parallelism be checked with a tape measure alone, without a protractor.

On one arm of an angle |SA| = 4 cm and |AB| = 6 cm, on the other |SC| = 10 cm. Find |CD|, given that AC and BD are parallel.

A line parallel to a side of a triangle

The same drawing, with the arms ending at the outer parallel, is the commonest school situation: a line parallel to a side of a triangle cuts off a triangle similar to the original. All three sides of the smaller triangle are then scaled by the same number.

DEAB43C
The segment DE is parallel to the side AB, so triangle CDE is similar to triangle CAB with a scale of 4 : 7.

What has to be watched here is where the segment is measured from. The proportion CDCA\tfrac{|CD|}{|CA|} uses segments measured from the vertex, while CDDA\tfrac{|CD|}{|DA|} uses adjacent segments. Both are correct, they give different numbers, and mixing them up is the commonest mistake in intercept-theorem exercises.

Similar figures and solids: k, k², k³

The scale acts on lengths. On areas and volumes it is its power that acts, and that is the one thing in this lesson that surprises on first meeting.

L=kL,P=k2P,V=k3VL' = k \cdot L, \qquad P' = k^2 \cdot P, \qquad V' = k^3 \cdot V

The reason is simple: an area is made by multiplying two lengths, a volume by multiplying three. If every length grows kk times, a product of two grows k2k^2 times and a product of three grows k3k^3 times.

2a2b2c
A cuboid with every dimension doubled. It holds eight copies of the original, even though it is 'only twice as big'.
Scale kkPerimeterAreaVolume
222×2\times4×4\times8×8\times
333×3\times9×9\times27×27\times
12\tfrac{1}{2}0.5×0.5\times0.25×0.25\times0.125×0.125\times
Two prisms are similar with k = 3. The smaller one has a surface area of 20 cm² and a volume of 12 cm³. Find both for the larger one.

Read backwards, the same relation recovers the scale from areas or volumes: if P:P=16P' : P = 16 then k=16=4k = \sqrt{16} = 4, and if V:V=64V' : V = 64 then k=643=4k = \sqrt[3]{64} = 4.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0
scale factor: a = 6 cm, a′ = 18 cm

Common mistakes

  • Multiplying an area by the scale instead of its square — at k=3k = 3 the area grows ninefold, not threefold.
  • Multiplying a volume by the square of the scale — a volume depends on three dimensions, so it grows k3k^3 times.
  • Mixing "from the vertex" segments with "between the parallels" ones in the intercept theorem — pick one convention and keep it on both sides of the proportion.
  • Calling figures with equal angles congruent — equal angles give similarity, not equal sides.
  • Forgetting the converse of the intercept theorem — proportional segments are a sufficient condition for parallelism, not merely a consequence of it.
  • Computing the scale from sides that do not correspond — a side must be divided by the side matching it, not by any other.

Formula card

Topic: Similarity and the intercept theorem

  • Scale factor

    k=aak = \frac{a'}{a}

    the image side divided by the original side

  • The intercept theorem

    SASC=ABCD\frac{|SA|}{|SC|} = \frac{|AB|}{|CD|}

    segments on the arms of an angle cut by parallels

  • Perimeter of a similar figure

    L=kLL' = k \cdot L

    every length grows k times

  • Area of a similar figure

    P=k2PP' = k^2 \cdot P

    an area depends on two dimensions

  • Volume of a similar solid

    V=k3VV' = k^3 \cdot V

    a volume depends on three dimensions

ACBD|SA||AB||SC||CD|S
The arms of an angle cut by two parallel lines. The segments on one arm are proportional to the corresponding segments on the other.
2a2b2c
The three powers of the scale: at k = 2 the perimeter doubles, the area quadruples and the volume grows eightfold.

Frequently asked questions

Related articles