Double the Sides of a Triangle and Its Area Becomes Four Times as Big

If every side of a triangle is doubled, the base doubles and the height doubles. Since area is $\frac{1}{2}\times \text{base}\times \text{height}$, **the area becomes $2\times 2=4$ times as large.**

In general, if lengths scale by $k$:

$$\text{lengths}\times k\text{areas}\times k^2\text{volumes}\times k^3$$

This part covers the area theorem, ratios from medians, altitudes and perimeters, maps and models, and enlargement and reduction.

**If $△ABC\sim △PQR$, then the ratio of their areas equals the square of the ratio of any pair of corresponding sides.**

$$\frac{\text{ar}(△ABC)}{\text{ar}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$$

Worked example 1. $AB:PQ=3:5$ and $\text{ar}(△ABC)=54{\text{ cm}}^2$.

$$\frac{54}{\text{ar}(△PQR)}=\frac{9}{25}\text{ar}(△PQR)=\frac{54\times 25}{9}=150{\text{ cm}}^2$$

Worked example 2. Two similar triangles have areas $64{\text{ cm}}^2$ and $121{\text{ cm}}^2$. A side of the smaller is $12$ cm.

$$\frac{12}{x}=\sqrt{\frac{64}{121}}=\frac{8}{11}x=16.5\text{ cm}$$

Worked example 3. In $△ABC$, $DE\parallel BC$ and $AD:DB=2:3$. Then $AD:AB=2:5$, so

$$\text{ar}(△ADE):\text{ar}(△ABC)=4:25\text{ar}(△ADE):\text{ar}(DECB)=4:21$$

An everyday example. A kite-maker who cuts a triangular sail with every side twice as long needs four times as much paper.

The trap. **Use $AD:AB$, not $AD:DB$**, when comparing the small triangle with the whole.

In similar triangles, corresponding medians, altitudes and perimeters are all in the same ratio as the sides, so the ratio of areas is the square of any of these ratios.

Worked example 1 — altitudes. Corresponding altitudes are $6$ cm and $9$ cm.

$$\text{area ratio}={\left(\frac{6}{9}\right)}^2=\frac{4}{9}$$

Worked example 2 — perimeters. Perimeters are $30$ cm and $45$ cm, so the area ratio is again ${\left(\frac{2}{3}\right)}^2=\frac{4}{9}$.

Worked example 3 — medians. Corresponding medians are in the ratio $5:7$ and the larger triangle has area $196{\text{ cm}}^2$.

$$\text{smaller area}=196\times \frac{25}{49}=100{\text{ cm}}^2$$

Worked example 4 — reverse. Areas are in the ratio $16:81$. Then the perimeters are in the ratio $\sqrt{16}:\sqrt{81}=4:9$.

An everyday example. Two similar triangular flower beds in a park: fencing depends on perimeter, which scales by $k$, while soil depends on area, which scales by $k^2$.

The substance. Perimeters do not get squared — only areas do.

**For a scale of $1:n$, multiply model lengths by $n$, model areas by $n^2$ and model volumes by $n^3$ to get the real values, then convert units.

Worked example 1 — map.** Scale $1:50000$.

$$4\text{ cm on the map}=4\times 50000=200000\text{ cm}=2\text{ km}$$
$$3{\text{ cm}}^2\text{ on the map}=3\times 50000^2=7.5\times 10^9{\text{ cm}}^2=0.75{\text{ km}}^2$$

since $1{\text{ km}}^2=10^{10}{\text{ cm}}^2$.

Worked example 2 — model ship. Scale $1:200$.

$$\text{length }60\text{ cm}\to 60\times 200=12000\text{ cm}=120\text{ m}$$
$$\text{deck area }150{\text{ cm}}^2\to 150\times 40000=6000000{\text{ cm}}^2=600{\text{ m}}^2$$
$$\text{tank }2\text{ L}\to 2\times 200^3=16000000\text{ L}=16000{\text{ m}}^3$$

An everyday example. An architect's model of a housing society shows the real building shrunk by a fixed scale, so its floor areas shrink by the square of that scale.

The substance. Convert units only at the end, and remember $1{\text{ m}}^2=10^4{\text{ cm}}^2$ and $1{\text{ m}}^3=1000$ L.

**Multiply lengths by $k$, areas by $k^2$ and volumes by $k^3$; $k>1$ gives an enlargement and $0<k<1$ gives a reduction.

Worked example 1 — enlargement.** A triangle of area $12{\text{ cm}}^2$ is enlarged with $k=3$.

$$\text{new area}=12\times 3^2=108{\text{ cm}}^2$$

Worked example 2 — reduction. A triangle with a $10$ cm side and area $48{\text{ cm}}^2$ is reduced with $k=\frac{1}{2}$: the side becomes $5$ cm and the area $48\times \frac{1}{4}=12{\text{ cm}}^2$.

Worked example 3 — solid. A cuboid $2\times 3\times 4$ cm is enlarged with $k=2$.

$$\text{volume }24\to 24\times 8=192{\text{ cm}}^3\text{surface area }52\to 52\times 4=208{\text{ cm}}^2$$

Worked example 4 — finding k. A figure's area changes from $20{\text{ cm}}^2$ to $45{\text{ cm}}^2$.

$$k^2=\frac{45}{20}=\frac{9}{4}k=\frac{3}{2}$$

An everyday example. **A photocopier's $141\%$ setting enlarges an A4 page to A3.** That is $k\approx \sqrt{2}$, so the area doubles.

The substance. **To find $k$ from areas, take the square root**; from volumes, take the cube root.