Therefore, triangle \(ABC \) and triangle \(EFG\) are similar.

Enlargement definition

Enlargement is a transformation in which all the points of an object move from a fixed point with a constant ratio. The fixed point is known as the centre of enlargement and the constant ratio is known as the scale factor.

The scale factor \(k\), of an enlargement can be determined as follows.

\[\begin{aligned} k&=\frac{\text{D.O.P of image from $P$}}{\text{D.O.P. of object from $P$}}\\ \\&\text{where}\\ &\text{D.O.P. is distance of point.}\end{aligned}\] or \[\begin{aligned} k=\frac{\text{length of side of image}}{\text{length of side of object}} \end{aligned}\]

Example 3

The diagram below shows three objects and their images under the transformation of enlargement. Describe the enlargement by determining the scale factor and the centre of enlargement for the following:

\(R'S'T'U'\) is the image of \(RSTU\) under an enlargement at centre \((13,10)\) with a scale factor of \(\dfrac{1}{3}\).

c) \[\begin{aligned} &\text{Scale factor}\\&=\frac{\text{distance of $L^\prime$ from }P}{\text{distance of $L$ from }P}\\ &=-\frac{6}{3}\\ &=-2.\end{aligned}\]Note: \(P\) is centre of enlargement.

\(L'M'N'\) is the image of \(LMN\) under an enlargement at centre \((2,-5)\) with a scale factor of \(-2\).

Area of the image of an enlargement formula

\(\text{Area of an image} \\= k^2\times\text{Area of the object}\)

where \(k\) is the scale factor.

Example 4

Given an area of object and image are \(\text{5 cm}^2\) and \(\text{45 cm}^2\) respectively. Calculate the scale factor, \(k\).

Solution:

\[\begin{aligned} k&=\sqrt{\frac{\text{Area of the image}}{\text{Area of the object}}}\\ &=\sqrt{\frac{45}{5}}\\ &=\sqrt{9}\\ &=3. \end{aligned}\]

Enlargement

5.2

Enlargement

What is similarity?

Two objects are similar if they have the same shape, or one has the same shape as the mirror image of the other.

For example, all circles are similar to each other, all squares are similar to each other, and all equilateral triangles are similar to each other.

Note: If all the corresponding sides of a pair of triangles are proportional, then all the corresponding angles are equal and vice versa.

Note: If two angles of a triangle have measures equal to the measures of two angles of another triangle, then the triangles are similar.

Example 2

Determine whether the pair of the following geometric objects are similar.

Therefore, triangle \(ABC \) and triangle \(EFG\) are similar.

Enlargement definition

Enlargement is a transformation in which all the points of an object move from a fixed point with a constant ratio. The fixed point is known as the centre of enlargement and the constant ratio is known as the scale factor.

The scale factor \(k\), of an enlargement can be determined as follows.

\[\begin{aligned} k&=\frac{\text{D.O.P of image from $P$}}{\text{D.O.P. of object from $P$}}\\ \\&\text{where}\\ &\text{D.O.P. is distance of point.}\end{aligned}\] or \[\begin{aligned} k=\frac{\text{length of side of image}}{\text{length of side of object}} \end{aligned}\]

Example 3

The diagram below shows three objects and their images under the transformation of enlargement. Describe the enlargement by determining the scale factor and the centre of enlargement for the following:

\(R'S'T'U'\) is the image of \(RSTU\) under an enlargement at centre \((13,10)\) with a scale factor of \(\dfrac{1}{3}\).

c) \[\begin{aligned} &\text{Scale factor}\\&=\frac{\text{distance of $L^\prime$ from }P}{\text{distance of $L$ from }P}\\ &=-\frac{6}{3}\\ &=-2.\end{aligned}\]Note: \(P\) is centre of enlargement.

\(L'M'N'\) is the image of \(LMN\) under an enlargement at centre \((2,-5)\) with a scale factor of \(-2\).

Area of the image of an enlargement formula

\(\text{Area of an image} \\= k^2\times\text{Area of the object}\)

where \(k\) is the scale factor.

Example 4

Given an area of object and image are \(\text{5 cm}^2\) and \(\text{45 cm}^2\) respectively. Calculate the scale factor, \(k\).

Solution:

\[\begin{aligned} k&=\sqrt{\frac{\text{Area of the image}}{\text{Area of the object}}}\\ &=\sqrt{\frac{45}{5}}\\ &=\sqrt{9}\\ &=3. \end{aligned}\]

Chapter : Congruence, Enlargement and Combined Transformation