Similarity and geometric transformations

Similar shapes have the same shape but not necessarily the same size. Geometric transformations give precise rules for moving, reflecting or resizing every point of a shape.

These ideas support A-level coordinate geometry, vectors, trigonometry and graph transformations. The difficult parts are matching corresponding sides, distinguishing a length scale factor from an area or volume factor, and giving a complete description of a transformation.

Prerequisites

You should be able to:

• simplify ratios and solve proportions;
• calculate lengths, areas and volumes;
• plot and read coordinates in all four quadrants;
• use column vectors;
• use Pythagoras’ theorem and elementary trigonometry.

Review ratio and proportion, coordinate geometry or vectors if needed.

Congruence and similarity

Two shapes are congruent if one can be placed exactly on the other using translations, rotations and reflections. Corresponding lengths and angles are equal. Their orientation and position may differ.

Two shapes are similar if:

• corresponding angles are equal;
• corresponding lengths are in one constant ratio.

Congruent shapes are therefore similar with linear scale factor $1$.

If shape $A$ is transformed into shape $B$ with linear scale factor $k$, then

$$\boxed{k=\frac{\text{length in }B}{\text{corresponding length in }A}}.$$

The order matters. Reversing the transformation gives reciprocal scale factor $1/k$.

Worked example 1: identify corresponding sides

Triangles $ABC$ and $PQR$ are similar. Their equal angles are

$$\angle A=\angle P,\qquad \angle B=\angle Q,\qquad \angle C=\angle R.$$

Suppose $AB=8$, $BC=12$, $PQ=14$ and $PR=17.5$. Find $QR$ and $AC$.

The angle correspondence gives the vertex correspondence

$A\leftrightarrow P,qquad B\leftrightarrow Q,qquad C\leftrightarrow R.$

Hence

$AB\leftrightarrow PQ,qquad BC\leftrightarrow QR,qquad AC\leftrightarrow PR.$

From $ABC$ to $PQR$, the scale factor is

$k=\frac{PQ}{AB}=\frac{14}{8}=\frac74.$

Therefore

$QR=12\times\frac74=21.$

Also,

$17.5=AC\times\frac74,$

so

$AC=17.5\times\frac47=10.$

Do not match sides merely because they occupy similar positions in a sketch. Match the endpoints of equal angles.

Similar triangles

Triangles can be proved similar using any of these tests:

• AA: two corresponding angles are equal;
• SSS: all three pairs of corresponding sides are in the same ratio;
• SAS: two pairs of sides are in the same ratio and their included angles are equal.

For right-angled triangles, one additional equal acute angle is enough for AA similarity.

Worked example 2: use similarity in a diagram

In triangle $ABC$, point $D$ lies on $AB$ and point $E$ lies on $AC$, with $DE$ parallel to $BC$. Suppose

$AD=6,qquad DB=4,qquad AE=7.5.$

Find $EC$.

Because $DE\parallel BC$, corresponding angles are equal. Therefore

$\triangle ADE\sim\triangle ABC.$

First find the whole side:

$AB=AD+DB=6+4=10.$

The scale factor from the small triangle to the large triangle is

$k=\frac{AB}{AD}=\frac{10}{6}=\frac53.$

Thus

$AC=AE\times\frac53=7.5\times\frac53=12.5.$

Finally,

$EC=AC-AE=12.5-7.5=5.$

The common error is to treat $DB$ and $EC$ as corresponding sides. They are leftover segments, not sides of the two similar triangles.

Length, area and volume scale factors

If the linear scale factor is $k$, then every length is multiplied by $k$. Areas contain two independent length dimensions, and volumes contain three:

Quantity	Scale factor
Length or perimeter	$k$
Area	$k^2$
Volume	$k^3$

Thus

$$\frac{A_2}{A_1}=k^2, \qquad \frac{V_2}{V_1}=k^3.$$

These rules apply to all similar two-dimensional shapes and solids, not only rectangles and cubes.

Worked example 3: area from a length factor

Two similar shapes have corresponding sides $9$ cm and $15$ cm. The smaller shape has area $54\text{ cm}^2$. Find the area of the larger shape.

The linear scale factor is

$k=\frac{15}{9}=\frac53.$

The area scale factor is

$k^2=\left(\frac53\right)^2=\frac{25}{9}.$

Therefore

$A_{\text{large}}=54\times\frac{25}{9}=150\text{ cm}^2.$

Multiplying the area by $5/3$ would ignore its second dimension.

Worked example 4: recover a length factor from areas

Two similar triangles have areas $80\text{ cm}^2$ and $125\text{ cm}^2$. A side of the smaller triangle is $12$ cm. Find the corresponding side of the larger triangle.

The area scale factor is

$\frac{125}{80}=\frac{25}{16}.$

Take the positive square root to obtain the linear scale factor:

$k=\sqrt{\frac{25}{16}}=\frac54.$

Hence the required side is

$12\times\frac54=15\text{ cm}.$

Worked example 5: volume and inverse reasoning

Two mathematically similar bottles have capacities $250$ ml and $432$ ml. The smaller bottle is $15$ cm high. Find the height of the larger bottle.

Capacity is volume, so

$k^3=\frac{432}{250}=1.728.$

Therefore

$k=\sqrt[3]{1.728}=1.2.$

The larger height is

$15\times1.2=18\text{ cm}.$

When working backwards from volume, take a cube root. When working backwards from area, take a square root.

What a transformation must specify

A transformation maps every point $P$ to an image point $P'$. The original shape is the object and the transformed shape is the image.

A complete description needs the transformation type and all defining information:

Transformation	Information required
Translation	translation vector
Rotation	centre, angle and direction
Reflection	mirror line
Enlargement	centre and scale factor

Translations, rotations and reflections preserve lengths and angles, so they produce congruent images. An enlargement preserves angles and multiplies every length by $|k|$, so it produces a similar image.

Translations

A translation moves every point by the same vector. The vector

$$\begin{pmatrix}a\\b\end{pmatrix}$$

means $a$ units horizontally and $b$ units vertically. Positive directions are right and up.

The coordinate rule is

$$\boxed{(x,y)\mapsto(x+a,y+b)}.$$

Worked example 6: translate and reverse

Triangle $ABC$ has vertices

$A(-3,2),\qquad B(1,1),\qquad C(0,5).$

Translate it by

$$\begin{pmatrix}4\\-3\end{pmatrix}.$$

Add $4$ to each $x$ coordinate and subtract $3$ from each $y$ coordinate:

$$\begin{aligned} A(-3,2)&\mapsto A'(1,-1),\\ B(1,1)&\mapsto B'(5,-2),\\ C(0,5)&\mapsto C'(4,2). \end{aligned}$$

The inverse translation uses the negative vector:

$$\begin{pmatrix}-4\\3\end{pmatrix}.$$

Reflections

A reflection maps each point to the opposite side of a mirror line. The mirror line is the perpendicular bisector of $PP'$.

Useful coordinate rules are:

Mirror line	Coordinate rule
$x$ axis, $y=0$	$(x,y)\mapsto(x,-y)$
$y$ axis, $x=0$	$(x,y)\mapsto(-x,y)$
$y=x$	$(x,y)\mapsto(y,x)$
$y=-x$	$(x,y)\mapsto(-y,-x)$
vertical line $x=a$	$(x,y)\mapsto(2a-x,y)$
horizontal line $y=b$	$(x,y)\mapsto(x,2b-y)$

Points on the mirror line do not move. They are invariant points.

Worked example 7: reflect in a non-axis line

Reflect $P(-2,5)$ in the line $x=3$.

The point is $5$ units to the left of $x=3$, so its image is $5$ units to the right. Therefore $x'=8$ and the $y$ coordinate is unchanged:

$P(-2,5)\mapsto P'(8,5).$

Using the coordinate rule confirms this:

$x'=2(3)-(-2)=8.$

To recover an unknown mirror line from a point and its image, find the perpendicular bisector of the segment joining them.

Rotations

A rotation turns every point through the same angle about a fixed centre. Distances from the centre are preserved.

About the origin, these rules are useful:

Rotation	Coordinate rule
$90^\circ$ anticlockwise	$(x,y)\mapsto(-y,x)$
$90^\circ$ clockwise	$(x,y)\mapsto(y,-x)$
$180^\circ$	$(x,y)\mapsto(-x,-y)$

For a centre other than the origin, think in three stages: translate the centre to the origin, rotate, then translate back.

Worked example 8: rotate about a general centre

Rotate $P(5,1)$ by $90^\circ$ anticlockwise about $C(2,-1)$.

First write the displacement from the centre to the point:

$P-C=(5-2,1-(-1))=(3,2).$

A $90^\circ$ anticlockwise rotation maps

$(3,2)\mapsto(-2,3).$

Add the centre back:

$P'=C+(-2,3)=(2-2,-1+3)=(0,2).$

Hence

$\boxed{P(5,1)\mapsto P'(0,2)}.$

The centre, angle and direction are all needed when describing this transformation. For $180^\circ$, clockwise and anticlockwise give the same result.

Enlargements

An enlargement with centre $C$ and scale factor $k$ obeys the vector equation

$$\boxed{\overrightarrow{CP'}=k\overrightarrow{CP}}.$$

Equivalently, if $C=(a,b)$ and $P=(x,y)$, then

$$P'=C+k(P-C).$$

This formula contains the complete geometry:

• if $k>1$, the image is farther from $C$;
• if $0<k<1$, the image lies between $C$ and $P$;
• if $k<0$, the image lies on the opposite side of $C$;
• if $k=1$, every point is unchanged;
• if $k=0$, every point maps to the centre, so the image collapses.

The centre is invariant for every enlargement.

Worked example 9: positive fractional enlargement

Enlarge $P(8,1)$ by scale factor $1/3$ about $C(2,4)$.

Find the displacement from the centre:

$P-C=(6,-3).$

Multiply it by $1/3$:

$\frac13(6,-3)=(2,-1).$

Add the centre back:

$P'=C+(2,-1)=(4,3).$

Because $0<1/3<1$, $P'$ lies between $C$ and $P$, one third of the way from $C$.

Worked example 10: negative scale factor

Enlarge $P(4,5)$ by scale factor $-2$ about $C(1,1)$.

$P-C=(3,4).$

Multiply by $-2$:

$-2(3,4)=(-6,-8).$

Therefore

$P'=C+(-6,-8)=(-5,-7).$

The negative sign places $P'$ on the opposite side of the centre. The magnitude $|-2|=2$ makes the distance twice as large:

$CP'=2CP.$

A negative enlargement is not a reflection in a line. For $k=-1$, it has the same effect as a $180^\circ$ rotation about the centre.

Worked example 11: find the centre of enlargement

A shape is enlarged to an image. One pair of corresponding points is

$A(2,1)\mapsto A'(5,4),$

and another is

$B(4,1)\mapsto B'(9,4).$

The image side $A'B'$ has length $4$ while $AB$ has length $2$, so $k=2$.

For $A'=C+2(A-C)=2A-C$, rearrange to obtain

$C=2A-A'.$

Thus

$C=2(2,1)-(5,4)=(-1,-2).$

Check with the second pair:

$2B-B'=2(4,1)-(9,4)=(-1,-2).$

Geometrically, the centre lies where the straight lines $AA'$ and $BB'$ meet. Extending those lines is often necessary.

Combined transformations and order

Two transformations performed in sequence form a combined transformation. Order usually matters.

Worked example 12: order changes the result

Start with $P(1,2)$. Let $T$ be translation by

$$\begin{pmatrix}3\\0\end{pmatrix},$$

and let $R$ be reflection in the $y$ axis.

Translate first, then reflect:

$$(1,2)\xrightarrow{T}(4,2)\xrightarrow{R}(-4,2).$$

Reflect first, then translate:

$$(1,2)\xrightarrow{R}(-1,2)\xrightarrow{T}(2,2).$$

The outputs differ. Always apply transformations in the order stated and record an intermediate image.

How to recognise an unknown transformation

Compare an object with its image systematically:

1. If size and orientation are unchanged and all point displacements are equal, it is a translation.
2. If size is unchanged and corresponding points are equally distant from one fixed point, test a rotation.
3. If corresponding segments are cut perpendicularly in half by one line, it is a reflection.
4. If size changes, it is an enlargement. Join corresponding vertices to locate the centre, then calculate the scale factor.

One point pair is rarely enough to identify a unique transformation. Use at least two corresponding pairs and check a third where possible.

Common misconceptions

Using the linear factor for area or volume

If lengths double, areas multiply by $2^2=4$ and volumes by $2^3=8$. Decide what kind of quantity is being scaled before calculating.

Giving an incomplete description

“A rotation” is not complete. State, for example, “rotation $90^\circ$ clockwise about $(2,-1)$”. Likewise, a reflection needs its mirror line and an enlargement needs its centre and scale factor.

Enlarging coordinates from the origin automatically

The rule $(x,y)\mapsto(kx,ky)$ works only when the centre is $(0,0)$. For any other centre, use

$P'=C+k(P-C).$

Treating a negative scale factor as a negative size

Lengths remain non-negative and are multiplied by $|k|$. The sign determines which side of the centre contains the image.

Self-check

1. Similar lengths

Two similar quadrilaterals have corresponding sides $7$ cm and $11.2$ cm. Another side of the smaller quadrilateral is $12.5$ cm. Find its corresponding larger side.

Answer

$k=\frac{11.2}{7}=1.6,$

so the required length is

$12.5\times1.6=20\text{ cm}.$

2. Area and volume factors

Similar solids have linear scale factor $3/2$ from smaller to larger. State the area and volume scale factors.

Answer

$$\text{area factor}=\left(\frac32\right)^2=\frac94,$$ $$\text{volume factor}=\left(\frac32\right)^3=\frac{27}{8}.$$

3. Reflection

Reflect $(-4,7)$ in the line $y=x$.

Answer

Reflection in $y=x$ swaps the coordinates:

$(-4,7)\mapsto(7,-4).$

4. Rotation

Rotate $(6,-2)$ by $90^\circ$ clockwise about the origin.

Answer

Use $(x,y)\mapsto(y,-x)$:

$(6,-2)\mapsto(-2,-6).$

5. Enlargement

Enlarge $P(-1,5)$ by scale factor $-1/2$ about $C(3,1)$.

Answer

$P-C=(-4,4).$

Multiply by $-1/2$:

$-\frac12(-4,4)=(2,-2).$

Add the centre:

$P'=C+(2,-2)=(5,-1).$

6. Reverse area scaling

Two similar shapes have areas $45\text{ cm}^2$ and $245\text{ cm}^2$. A corresponding length in the smaller shape is $9$ cm. Find the length in the larger shape.

Answer

$$k=\sqrt{\frac{245}{45}} =\sqrt{\frac{49}{9}} =\frac73.$$

Therefore the length is

$9\times\frac73=21\text{ cm}.$

Where this leads

You should now be able to prove or use similarity, convert between length, area and volume factors, and describe the four standard transformations completely.

Next, connect these ideas to graph transformations, where equations transform whole graphs. Coordinate geometry develops distance and midpoint methods, while vectors gives a more powerful language for displacement and geometric proof. Similar triangles also recur throughout Pythagoras and trigonometry and circle geometry.