Similar & Congruent Figures

Congruent figures are identical: they have exactly the same shape and the same size, so all corresponding sides and angles are equal. (Same shape but different size describes similar figures.)

Similar figures have the same shape: corresponding angles are equal and corresponding sides are in the same ratio (the scale factor), but the sizes may differ.

SSS (side-side-side) proves congruence when all three pairs of corresponding sides are equal. (AAA compares only angles and does not prove congruence.)

Capacity (volume) scales with the cube of the length scale factor. With a scale factor of $2$: $250 \times 2^3 = 250 \times 8 = 2000$ mL. ($500$ mL used the scale factor $2$ instead of $2^3$.)

A scale factor of 2 multiplies every side by 2: $3 \to 6$, $4 \to 8$, $5 \to 10$. (Scale factor 3 would give $9, 12, 15$.)

The scale factor is the ratio of corresponding sides: $\dfrac{10}{4} = 2.5$. (0.4 is the ratio the other way, large to small.)

Convert to the same units first: $1.2$ m $= 1200$ mm. The scale factor is $\dfrac{1200}{24} = 50$. ($0.02$ is the ratio the other way; using $\dfrac{1.2}{24}$ forgets to convert the units.)

Paint covers the surface, and surface area scales with the square of the length scale factor: $3^2 = 9$. So the larger statue needs $2 \times 9 = 18$ litres. ($6$ litres used the scale factor $3$; $54$ litres used the volume factor $3^3$.)

Area scales with the square of the scale factor: $3^2 = 9$. (27 would be the volume ratio, which scales with the cube.)

Area scales with the square of the scale factor: $4^2 = 16$. So the enlarged photo has 16 times the area.

Since area ratio is the square of the side ratio, take square roots: $\sqrt{25} : \sqrt{4} = 5 : 2$.

The mirror creates two similar triangles, so $\dfrac{\text{tree}}{20} = \dfrac{1.6}{2.5}$. The tree $= \dfrac{1.6}{2.5} \times 20 = 12.8$ m. ($32$ m multiplied by $2.5$ instead of dividing.)

The triangles are similar, so $\dfrac{\text{tree}}{12} = \dfrac{2}{3}$, giving tree $= \dfrac{2}{3} \times 12 = 8$ m.

Real distance $= 4 \times 50\,000 = 200\,000$ cm $= 2000$ m $= 2$ km. (Convert the map length using the scale, then to sensible units.)

Area scales with the square of the scale factor: $3^2 = 9$, so the larger area is $12 \times 9 = 108$ cm². (36 used the scale factor 3 instead of $3^2$.)

AAA (three equal angles) only guarantees the triangles are the same shape, i.e. similar. They could be different sizes, so AAA does not prove congruence. The other four are valid congruence tests.

Each real length is $1000$ times the map length: $5$ cm $\to 5000$ cm $= 50$ m and $3$ cm $\to 3000$ cm $= 30$ m. The real area is $50 \times 30 = 1500$ m². (Area scales by $1000^2$, not $1000$.)

The model is $\dfrac{1}{20}$ of the real length: $\dfrac{4}{20} = 0.2$ m $= 20$ cm.

Volume scales with the cube of the scale factor: $1^3 : 2^3 = 1 : 8$. (Area would scale as $1 : 4$.)

A line parallel to one side creates a similar triangle. With a scale factor of $\tfrac{1}{2}$, the base is $14 \times \tfrac{1}{2} = 7$ cm.

RHS (right angle, hypotenuse, side) is a valid congruence test for right-angled triangles. SSA/ASS and AAA are not sufficient to prove congruence.

Multiply by the scale factor: $DE = 4 \times \dfrac{3}{2} = 6$.

The area ratio is $\dfrac{200}{50} = 4$. The scale factor is the square root of the area ratio: $\sqrt{4} = 2$.

Because $DE \parallel BC$, triangle $ADE$ is similar to triangle $ABC$. The scale factor is $\dfrac{AB}{AD} = \dfrac{4 + 6}{4} = \dfrac{10}{4} = 2.5$, so $BC = 8 \times 2.5 = 20$. ($12$ used $DB$ instead of the full side $AB$.)

The length scale factor is $4 : 6 = 2 : 3$. Surface area scales with the square: $2^2 : 3^2 = 4 : 9$. (Volume would scale as $8 : 27$.)