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A photo is enlarged on a photocopier to 150% size. What changes — and what stays the same? Think about the angles, the shape, and the lengths of sides.
Two figures are similar (symbol $\sim$) when they have the same shape but not necessarily the same size. All corresponding angles are equal, and all corresponding sides are in the same ratio — the scale factor $k$. Congruent figures are a special case of similarity where $k = 1$.
The area ratio is $k^2$, not $k$. If a shape is enlarged by scale factor 3, its area increases by a factor of $3^2 = 9$, not 3. Similarly, perimeter scales by $k$ (not $k^2$) — perimeter is a length, area is a length squared. Do not confuse these two.
For two figures to be similar:
The scale factor $k$ is calculated as:
$$k = \frac{\text{image side length}}{\text{corresponding object side length}}$$$\triangle ABC \sim \triangle PQR$. Given $AB = 4$ cm, $PQ = 6$ cm, and $AC = 5$ cm. Find the scale factor $k$ and the length $PR$.
Step 1: Find $k$ using a known pair of corresponding sides.
$$k = \frac{PQ}{AB} = \frac{6}{4} = 1.5$$Step 2: Use $k$ to find the unknown side. $AC$ corresponds to $PR$, so:
$$PR = AC \times k = 5 \times 1.5 = 7.5 \text{ cm}$$Check: All sides of $\triangle PQR$ should be $1.5\times$ the corresponding sides of $\triangle ABC$. ✓
When two shapes are similar with linear scale factor $k$:
$$\frac{\text{Area of image}}{\text{Area of object}} = k^2$$Why? Area is length × length — so it scales by $k \times k = k^2$.
Worked example: Two similar rectangles. The small rectangle has length 8 cm and area 24 cm². The large rectangle has length 12 cm.
$$k = \frac{12}{8} = 1.5$$ $$\text{Area of large} = 24 \times k^2 = 24 \times 1.5^2 = 24 \times 2.25 = 54 \text{ cm}^2$$Perimeter scales by $k$: if the small perimeter is $P$, the large perimeter is $1.5P$.
Similar figures ($\sim$): corresponding angles equal; corresponding sides in ratio $k$
Scale factor: $k = \dfrac{\text{image side}}{\text{object side}}$
Missing side: object side $\times\, k$ = image side
Area ratio: $\dfrac{\text{image area}}{\text{object area}} = k^2$
Perimeter ratio: $\dfrac{\text{image perimeter}}{\text{object perimeter}} = k$
$\triangle ABC \sim \triangle PQR$. $AB = 6$ cm and $PQ = 9$ cm. What is the scale factor $k$?
Two similar triangles have corresponding sides $4$ cm and $10$ cm. Another side of the smaller triangle is $7$ cm. What is the corresponding side of the larger triangle?
Two similar shapes have a linear scale factor of $k = 4$. What is the ratio of their areas (larger : smaller)?
Two similar figures have scale factor $k = 3$. The smaller figure has perimeter 20 cm. What is the perimeter of the larger figure?
A shape has a side of 10 cm. Its similar image has the corresponding side of 15 cm. What is the scale factor?
Q6. $\triangle ABC \sim \triangle XYZ$. $AB = 5$ cm, $BC = 8$ cm, $AC = 6$ cm, and $XY = 7.5$ cm. Find the scale factor $k$, and then find the lengths $YZ$ and $XZ$.
Q7. Two similar triangles have a scale factor of $k = 2.5$. The smaller triangle has area $16$ cm². Find the area of the larger triangle, showing your use of the area ratio rule.
Q8. A model car uses a scale of $1:20$. The real car is $4.2$ m long. (a) What is the length of the model in cm? (b) The real windscreen has area $0.8$ m². What is the area of the model windscreen in cm²?
$k = \dfrac{XY}{AB} = \dfrac{7.5}{5} = 1.5$
$YZ = BC \times k = 8 \times 1.5 = 12$ cm
$XZ = AC \times k = 6 \times 1.5 = 9$ cm
Area ratio $= k^2 = 2.5^2 = 6.25$
Area of larger triangle $= 16 \times 6.25 = 100$ cm²
(a) Scale $1:20$ means model length $= \dfrac{4.2 \text{ m}}{20} = 0.21$ m $= \mathbf{21}$ cm
(b) Linear scale factor (model to real) $k = 20$, so real-to-model is $k = \dfrac{1}{20}$.
Model windscreen area $= 0.8 \text{ m}^2 \times \left(\dfrac{1}{20}\right)^2 = 0.8 \times \dfrac{1}{400} = 0.002$ m²
Convert: $0.002 \text{ m}^2 = 0.002 \times 10{,}000 \text{ cm}^2 = \mathbf{20}$ cm²
Two similar cones have radii in ratio $2:5$. (a) What is the ratio of their curved surface areas? (b) What is the ratio of their volumes? (Recall: surface area scales as $k^2$ and volume scales as $k^3$.)