Two conditions, both required: every pair of corresponding angles is equal, and every pair of corresponding sides is in the same ratio — the linear scale factor $k = \dfrac{\text{new length}}{\text{old length}}$. Congruent shapes are just the special case $k=1$. Watch for rotated or reflected shapes: corresponding sides are always the ones opposite equal angles, not necessarily the ones that "look" matched.

Scaling every linear dimension by $k$ scales area by $k^2$ and volume by $k^3$ — this follows directly from multiplying two or three scaled lengths together (e.g. a rectangle's area $L\times W$ becomes $(kL)(kW)=k^2LW$).

Shape A (area $10\text{ cm}^2$) enlarged by $k=3$ to shape B: area scale factor $=3^2=9$, so area B $=10\times9=90\text{ cm}^2$.

Reverse direction — areas $100\text{ cm}^2$ and $36\text{ cm}^2$, large length $25$ cm, find the small length: area scale factor $=\frac{36}{100}=0.36$, so $k=\sqrt{0.36}=0.6$, giving small length $=25\times0.6=15$ cm.

You can't jump directly between them — always pivot through $k$: area ratio $\to\sqrt{\ }\to k \to k^3\to$ volume ratio, or volume ratio $\to\sqrt[3]{\ }\to k\to k^2\to$ area ratio.

Two similar cones have volumes in ratio $8:27$ — find the ratio of their surface areas: $k=\sqrt[3]{27/8}=\frac32$, so area ratio $=k^2=\frac94$, i.e. $4:9$.

Mass works the same way as volume for two similar objects of the same material (constant density), since $\text{mass}\propto\text{volume}$: $\dfrac{\text{mass}_2}{\text{mass}_1}=k^3$.

Convert the linear scale to the target units before squaring — squaring first and converting the huge resulting number afterwards is far more error-prone.

Map scale $1:50\,000$, park is $4\text{ cm}^2$ on the map — find the real area in km². $1$ cm represents $50\,000$ cm $=0.5$ km, so the area scale factor is $0.5^2=0.25\text{ km}^2/\text{cm}^2$, giving real area $=4\times0.25=1\text{ km}^2$.

A frustum is a large cone (or pyramid) with a smaller, similar cone removed from the top — always subtract: $V_{\text{frustum}} = V_{\text{large}} - V_{\text{small}}$. The fastest route to $V_{\text{small}}$ is via the volume scale factor, not recomputing dimensions from scratch.

Cone height $30$ cm, base radius $10$ cm, cut to leave a top radius of $6$ cm — find the frustum's volume.

1. $V_{\text{large}} = \frac13\pi(10)^2(30) = 1000\pi\text{ cm}^3$.
2. The removed cone is similar with $k=\frac{6}{10}=0.6$, so $\text{VSF}=0.6^3=0.216$.
3. $V_{\text{small}} = 1000\pi \times 0.216 = 216\pi\text{ cm}^3$.
4. $V_{\text{frustum}} = 1000\pi - 216\pi = 784\pi\text{ cm}^3$.

Two similar spheres have surface areas in the ratio $16:25$. The smaller sphere has volume $80\text{ m}^3$. Find the volume of the larger sphere.

1. Area ratio $\Rightarrow k = \sqrt{25/16} = \frac54$.
2. Volume scale factor $= k^3 = \left(\frac54\right)^3 = \frac{125}{64}$.
3. Larger volume $= 80 \times \frac{125}{64} = 156.25\text{ m}^3$.

Always identify what you're given (length/area/volume) and what's being asked before choosing which root or power of $k$ to apply — that's the one decision that determines the whole method.