## Enlargement

Subject: Optional Mathematics

#### Overview

An enlargement is a transformation which changes the size of an object without changing its shape. If the shape of an object increase, we call it as enlargement and if the shape of an object decrease, we call it as reduction. Enlargement and reduction are commonly known as dilation.
##### Enlargement

An enlargement is a transformation which changes the size of an object without changing its shape. If the shape of object increases, we call it as enlargement and if the shape of an object decreases, we call it a reduction. Enlargement and reduction are commonly known as dilation. It is obtained by allowing the length to vary in such a way that the angles remain constant. If all sides of an object figure are enlarged proportionally, then the resulting figure is called the image of the object under an enlargement. Similarly, if all sides of an object figure are reduced proportionally, then the resulting figure are reduced is called the image of the object under a reduction. In both cases, object and image are similar but not congruent. So, enlargement and reduction are non-isometric translations.

In the figure alongside, ABCDEF is a regular hexagon and O is the centre.

Join OA and produce it to A' such that OA' = 2OA.

Join OB and produce it to B' such that OB' = 2OB.

Similarly, locate C', D', E' and F' taking OC' = 2OC, OD' = 2OD, OE' = 2OE AND OF' = 2OF respectively. Join A', B', C',D', E' and F' in order.

In $\triangle$OA'B', A and B are the middle points of OA' and OB' respectively.

Then, A'B' = 2AB and AB // A'B'.

Similarly, B'C' = 2BC, C'D' = 2CD, D'E' = 2DE, E'F' = 2EF and F'A' = 2FA, BC // B'C', CD // C'D', DE // D'E', EF // E'F' and FA // F'A'. Here, each side of the hexagon A'B'C'D'E'F'are twice the length of the corresponding sides of the hexagon. ABCDEF and these two hexagons are similar to each other.

Here, hexagon A'B'C'D'E'F' is the image of hexagon ABCDEF under an enlargement. The fixed point O is called the centre of enlargement and $\frac{OA'}{OA}$ = $\frac{OB'}{OB}$ = .... = 2 is called the scale factor.

Now we can define enlargement in terms of centre of enlargement and scale factor as follows:

Let O be the given point and k be the given positive real number. An enlargement is a transformation that maps each point P of the plane to an image point P' such that:

1. The points O, P and P' are collinear
2. OP' = K. OP

The fixed point O is called the centre andnon- zero real number k is called the magnitude of enlargement or the scale factor.

When K> 1, the transformation is an enlargement.

When K< 1, the transformation is a reduction.

When K = 1, the transformation is identity transformation.

This scale factor is known as linear scale factor because the centre of enlargement, object point and image point lie in the straight line.

The centre of enlargement may be:

1. Inside the object figure
2. Outside the object figure
3. At the boundary of the object figure

When the centre of enlargement is inside the object figure:

Let O be the centre, k be the scale factor and $\triangle$ABC be the object figure. Join OA, OB, OC and locate the points A', B', C' such that OA' = k. OA, OB and OC' = k. OC.

In figure (a), $\frac {OA'}{OA}$ = $\frac {OB'}{OB}$ = $\frac {OC'}{OC}$ = k > 1. Here, $\triangle$A'B'C' is image of $\triangle$ABC under enlargement with centre O and scale factor k. Image $\triangle$A'B'C' is outside the object $\triangle$ABC. This is an example of enlargement.>

In figure (b), $\frac {OA'}{OA}$ = $\frac {OB'}{OB}$ = $\frac {OC'}{OC}$ = k = 1. Here, $\triangle$A'B'C' is image of $\triangle$ABC under enlargement with centre O and scale factor k. Image $\triangle$A'B'C' and object $\triangle$ABC are coincident. This is an example of identity transformation.

In figure (c), $\frac {OA'}{OA}$ = $\frac {OB'}{OB}$ = $\frac {OC'}{OC}$ = k< 1.Here, $\triangle$A'B'C' is image of $\triangle$ABC under reduction with centre O and scale factor k. Image $\triangle$A'B'C' is inside the object $\triangle$ABC. This is an example of reduction.

When the centre of enlargement is outside the object figure:

Let $\triangle$ABC be the object figure, k be the scalar factor and O be the centre of enlargement outside the object. Join OA, OB, OC and locate the point A', B', C' such that OA' = k. OA, OB' = k. OB and OC' = k. OC.

Case (a):When k> 1, the object $\triangle$ABC and the image $\triangle$A'B'C' lie on the same side of the centre O. $\triangle$ABC is enlarged to $\triangle$A'B'C' and $\triangle$ABC is between centre O and image $\triangle$A'B'C'.

Case (b):When 0< k < 1, the object $\triangle$ABC and image $\triangle$A'B'C' lie on the same side of the centre O. $\triangle$ABC is reduced to $\triangle$A'B'C' and $\triangle$A'B'C' is between centre O and object $\triangle$ABC.

Case (c):When k< 0, the object $\triangle$ABC and image $\triangle$A'B'C' lie to the opposite side of the centre O. If the numerical value of k is greater than unity i.e. if |k| > 1, $\triangle$ABC is enlarged to $\triangle$A'B'C' and if |k| < 1, $\triangle$ABC is reduced to $\triangle$A'B'C'. In this case inverted image is formed.

Case (d):When k = 1, the enlargement maps the object $\triangle$ABC onto itself. It means object $\triangle$ABC and image $\triangle$A'B'C' are coincide. So, k = 1 produces an identity trsnsformation.

When k = -1, it will not produce an identity transformation. In this case the enlargement maps $\triangle$ABC onto $\triangle$A'B'C' in such a way that $\triangle$ABC is congruent to $\triangle$A'B'C'. When k = -1, the enlargement is equivalent to the half turn about the centre of enlargement O.

Properties of Enlargement

1. Object figure and image figure are similar.
2. If the scale factor k is more than unity i.e. k> 1, the object is said to be enlarged. In this case, object and image lie on the same side of the centre of enlargement.
3. If 0< k< 1, the object is said to be reduced. Object and image lie on the same side of the centre of enlargement.
4. If k < 0, the object and image figure lie to the opposite side of the centre of enlargement. The image is inverted.
5. If k = 1, object and image figures are coincident.
6. If k = -1, object an image are congruent and the enlargement is equivalent to the half turn about the centre of enlargement.
7. Centre of enlargement is invariant point.
8. Under enlargement, sides of the object change proportionately but the angle between any two sides remain unchanged.
9. Corresponding sides of object and image are parallel.
10. The object point A, image point An' and the centre of enlargement lie in a line.

#### Use of coordinates in Enlargement

When centre of enlargement is origin and scalar factor k
Let E (0, k) be the enlargement with centre at the origin and scalar factor k.
Let P (x, y) be an object point. Draw perpendicular PM from P to the axis. Then:
OM = x and MP = y.Let P' (x', y') be the image point of P (x, y) under the enlargement E (0, k).
Draw perpendicular PN from P' to the X- axis. Then:
ON = x' and NP' = y
Now by the definition of enlargement,
$\frac{OP'}{OP}$ = k or, OP' = k × OP.
Here, $\triangle$POM and $\triangle$P'ON are similar.
$\frac{OP'}{OP}$ =$\frac{ON'}{OM}$ =$\frac{NP'}{MP}$ = k
$\therefore$ ON = k, OM snd NP' = k× MP
Now, x' = ON = k× OM = kx snd y' + NP' = k× MP = ky
Hence,E{0,k} : P(x, y)→P (kx, ky)

When the centre of enlargement is the point (a, b) and the scale factor k.
Let E{(a,b), k} be the enlargement with centre P (a, b) and scale factor k.
The position vector of P is $\overrightarrow{OP'}$ =$\begin{vmatrix} a \\ b \end{vmatrix}$
Let A (x, y) be an object point. Then position vector of A is$\overrightarrow{OA}$ =$\begin{vmatrix} x \\ y \end{vmatrix}$
Let A' (x', y') be the image of A () under the enlargementE[(a,b), k].
Then, position vector of A' is $\overrightarrow{OA'}$ =$\begin{vmatrix} x' \\ y' \end{vmatrix}$
Now, $\overrightarrow{PA}$ =$\overrightarrow{OA}$ -$\overrightarrow{OP}$ =$\begin{vmatrix} x \\ y \end{vmatrix}$ -$\begin{vmatrix} a \\ b \end{vmatrix}$ =$\begin{vmatrix} x - a\\ y - b\\ \end{vmatrix}$
$\overrightarrow{PA'}$ =$\overrightarrow{OA'}$ -$\overrightarrow{OP}$ =$\begin{vmatrix} x '\\ y' \end{vmatrix}$ -$\begin{vmatrix} a \\ b \end{vmatrix}$ =$\begin{vmatrix} x' - a\\ y' - b\\ \end{vmatrix}$
By definition of enlargement,
$\overrightarrow{PA'}$ = k×$\overrightarrow{PA'}$
$\therefore$ $\begin{vmatrix} x' - a\\ y' - b\\ \end{vmatrix}$ = k$\begin{vmatrix} x - a\\ y - b\\ \end{vmatrix}$ = $\begin{vmatrix}kx -ka\\ky - kb\\ \end{vmatrix}$
Equation corresponding components we get
x' - a = kx - ka
x' = kx - ka + a
y' - b = ky - kb
y'= ky - kb + b.
\9\therefore\) A' (kx - ka + a, ky - kb +b) is the image point od A(x,y) under the enlargement E{(a,b), k}.
Hence, E{(a, b), k} : A(x,y)→A'(kx - ka + a, ky - kb +b)

##### Things to remember
• An enlargement is a transformation which changes the size of an object without changing its shape.
• Enlargement and reduction are non-isometric translations.
• Object figure and image figure are similar in enlargement.
• Under enlargement, sides of the object change proportionately but the angle between any two sides remain unchanged.
• Corresponding sides of object and image are parallel.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Enlargement  ##### Maths Made Easy! Transformations #4: Enlargement ##### Transformations - Enlargements of Shapes

Soln:

Consttruction procedure:

To enlarge the triangle ABC with centre at O and scale factor 2 i. e. join O and A and produce it upto A' where OA' = 2OA. Also join O and B and O and C. Then produce OB and OC upto B' and C' where OB' = 2OB and OC' = 2OC. Then, if we join A', B' and C' then the triangle A'B'C' is the image of triangle ABC under the enlargemant with centre O and scale factor 2.

Soln:

Construction procedure:

Here, at first B, C and D are joined with O and then OB, OC and OD are produced upto B', C' and D' such that OB' = 3OB, OC' = 3OC and OD' = 3OD. Now, B', C' and D' are joined so that the triangle A'B'C'D' is the image of triangle BCD under the enlargement with centre at O and scale factor 3.

Soln:

Construction procedure:

Here, at first CO, DO and Eo are produced upto C', D' and E' respectively such that OC' =$\frac{1}{2}$OC, OD' =$\frac{1}{2}$OD and OE' =$\frac{1}{2}$OE.

Then C", D', E' and O are joined so that the parallelogram OC'D'E' is the image of parallelogram OECD under the enlargement with centre at O and scale factor $-\frac{1}{2}$

Soln:

Construction procedure:

Here, OF, OG, OH, OI and OJ are produced upto F', G', H', I' and J' as in the figure so that OF' = $1\frac{1}{2}$OF, OG' =$1\frac{1}{2}$OG, OH' =$1\frac{1}{2}$OH, OI' =$1\frac{1}{2}$ and OJ' =$1\frac{1}{2}$OJ. Then join O, F, G, I and J so that the hexagon OF'G'H'I'J' is the image of hexagon OFGHIJ under the enlargement with the centre at O and scale factor $1\frac{1}{2}$.

Soln:

Construction procedure:

Here, as in the figure, QO, RO and PO are produced upto Q', R' and P' respectively so that OQ' = 2OQ, OR' = 2OR and OP = 2OP'. Now O, Q', R' and P' are joined so that the figure OP'Q'R' is the image of the figure OPQR under the enlargement with centre at O and scale factor -2.

Soln:

Construction procedure:

Here, at first points B, C, D, E and F are joined with O. Then the points A', B', C', E', F' and G' are marked on the line OA, OB, OC, OD, OE, OF and OG respectively where OA' =$\frac{3}{4}$OA, OB' =$\frac{3}{4}$OB, OC' =$\frac{3}{4}$OC, OD' =$\frac{3}{4}$OD, OE' =$\frac{3}{4}$OE, OF' =$\frac{3}{4}$OF and OG' =$\frac{3}{4}$OG. Then the points A', B', C' D', E', F' and G' are joined so that the shaded figure A'B'C'E'F'G'F'G' is the image of the given figure under [0,$\frac{3}{4}$].

Soln:

Construction procedure:

Here, as in figure, points A, B, C, D and E are joined with O. Then points A', B', C', D' and E' are marked on the line OA, OB, OC, OD and OE respectively where OA' =$\frac{1}{2}$OA, OB' =$\frac{1}{2}$OB, OC' =$\frac{1}{2}$OC, OD' =$\frac{1}{2}$OD and OE' =$\frac{1}{2}$OE. Now, the points A', B', C', D' and E' are joined so that the pentagon A'B'C'D'E' is the image of the pentagon ABCDE under the enlargement with centre at O and scale factor$\frac{1}{2}$.

Soln:

Procedure: Here lines joining AA', BB; and CC; intersect at the point O. So, by the actual measurement,

OA = 4.2 cm

OA' = 7.2 cm

So, $\frac{OA'}{OA}$ = $\frac{7.2}{4.2}$ =$\frac{12}{7}$ = 1.7

∴ Scale factor = 1. 7

In symbol [0,$\frac{12}{7}$

Soln:

Procudure: Here, the lines AA', BB', CC' and DD' intersect at the point O. So that O is the centre of enlargement. Also from actual measurement,

OA' = 4 cm

OA = 1.5 cm

$\frac{OA'}{OA}$ =$\frac{4}{1.5}$ =$\frac{8}{3}$

∴ Scale factor =$\frac{8}{3}$

Soln: Here, let the centre be P (2,1) and scale factor k = 2 and A (2,0) be a point. so,

∴ $\overrightarrow{PA}$=$\begin{pmatrix} 2&-2 \\ o&-1 \\ \end{pmatrix}$= $\begin{pmatrix} 0 \\ -1 \\ \end{pmatrix}$

∴ A (2,0)→ A' (2×0+2,2×-1+1)=A.(2,-1)

Also, Relation of P (2,1) and B (3,5)

$\overrightarrow{PB}$= $\begin{pmatrix} 3&-2 \\ 5&-1 \\ \end{pmatrix}$=$\begin{pmatrix} 1 \\ 4 \\ \end{pmatrix}$

∴ B (3,5)→ B' (2×1+2,2×4+1) = B'(4,9)

Relation of P (2,1) and C (4,-7) is

$\overrightarrow{PC}$=$\begin{pmatrix} 4&-2 \\ -7&-1 \\ \end{pmatrix}$=$\begin{pmatrix} 2 \\ -8 \\ \end{pmatrix}$

∴ C= (4,-7)→C'(2×2+2,2×-8+1)=C'(6,-15)

Relation of P (2,1) and D (-3,8) is

$\overrightarrow{PD}$=$\begin{pmatrix} -3&-2 \\ 8&-1 \\ \end{pmatrix}$=$\begin{pmatrix} -5 \\ 7 \\ \end{pmatrix}$

∴ D(-3,8)→D'(2×-5+2,2×7+1)=D'(-8,15)

Relation of P (2,1) and F (-5,-7) is

$\overrightarrow{PF}$= $\begin{pmatrix} -5 &-2 \\ -7&-8 \\ \end{pmatrix}$=$\begin{pmatrix} -7 \\ -8 \\ \end{pmatrix}$

∴ F(-5,-7)→F'(2×-7+2,2×-8+1)=F'(-12,-15)

Relation of P (2,1) and G (0,-11) is

$\overrightarrow{PG}$= $\begin{pmatrix} 0&-2 \\ -11&-1 \\ \end{pmatrix}$=$\begin{pmatrix}-2 \\ -1 \\ \end{pmatrix}$

∴ G(0,-11)→G'(2×(-2)+2,2×(-12)+1)=G'(-2,-23)

The graph of the given points and their images under the enlargement with centre at P(2,1) and scale factor 2 is given below:

Soln: Here given, image of point A(1,p) is A' (q,8) and centre of enlargement (o,o) and scale factor k=2

We know that enlargement of P(x,y) under E[o,k] is P'(kx,ky)

Sp, A (1,p)→A'(2.1)2p)=A'(2,2p)

but, image of A is A'(q,8)

So, (2,2P)=(q,8)

or, 2=q amd 2p=8

∴ p=4 and q=2. Ans.

Soln: Here, the enlargement of rectangle ABCD under centre at origin and scale factor2 that is [O,2] by using P(x,y)→P'(kx,ky) is as follow:

A(-2,1)→A'(-2×2,1×2)=A'(-4,2)

B(-2,4)→B'(-2×2,4×2)=B'(-4,8)

C(4,4)→C'(4×2,4×2)-C'(8,8)

D(4,1)→D'(4×2,1×2)=D'(8,2)

Here, rectangle A'B'C'D' is the image of the rectangle under the enlargement [0,2], which is shown in the following graph paper.

Soln:

Construction procudure: Here as shown in the figure, R is joined with O, Here scale factor is 3, So the lines OP,OQ and OR are produced up to P', Q' and R' respectively so that OP'=3OP, OQ'=3OQ and OR' =3OR.Now, P',Q' and R', are joined so that the $\triangle$ P',Q',R' of the images of $\triangle$PQR is prepared with centre at O and scale factor 3.

Soln:

Enlargement of $\triangle$ ABC under E[0,2]:

A(2,4)→A'(2×2,4×2)=A'(4,8) ∴p(x,y)→P'(kx,ky)

B(-3,5)→B'(-3×2,5×2)=B'(-6,10)

C(-2,-3)→C'(-2×2,-3×2)=C'(-4,-6)

$\triangle$A'B'C' is the image of $\triangle$ABC under E[0,2].Which is shown in the following graph.

Soln:

Enlargement of $\triangle$ ABC under E[0,-2]:

A(2,4)→A'(2×-2,4×-2)=A'(-4,-8) ∴p(x,y)→P'(kx,ky)

B(-3,5)→B'(-3×-2,5×-2)=B'(6,-10)

C(-2,-3)→C'(-2×-2,-3×-2)=C'(4,6)

$\triangle$A'B'C' is the image of $\triangle$ABC under E[0,-2].Which is shown in the following graph.