Sec 2 Maths: Map Scales (Length & Area Scales), practice questions & worked solutions

(a)

$3$ km $=300\,000$ cm.

$$\begin{aligned}2\text{ cm}&:300\,000\text{ cm}\\ 1&:150\,000\end{aligned}$$

So $n=150\,000$.

(b)

The scale is $1\text{ cm}:1.5\text{ km}$.

$$\begin{aligned}\text{Map perimeter}&=\frac{21}{1.5}\\ &=14\text{ cm}\end{aligned}$$

(c)

$$\begin{aligned}1\text{ cm}&:1.5\text{ km}\\ 1\text{ cm}^2&:1.5^2=2.25\text{ km}^2\\ 2\text{ cm}^2&:2\times 2.25=4.5\text{ km}^2\end{aligned}$$

The actual area is $4.5$ km².

(a)

$$\begin{aligned}1\text{ cm}&:200\,000\text{ cm}\\ &=200\,000\div 100\div 1000\text{ km}\\ &=2\text{ km}\end{aligned}$$

So $n=2$.

(b)

$$\begin{aligned}\text{Map length}&=\frac{11.2}{2}\\ &=5.6\text{ cm}\end{aligned}$$

(c)

$$\begin{aligned}1\text{ cm}&:2\text{ km}\\ 1\text{ cm}^2&:2^2=4\text{ km}^2\\ \text{Actual area}&=1.5\times 4\\ &=6\text{ km}^2\end{aligned}$$

$50\,000$ cm $=0.5$ km, so the scale is $1\text{ cm}:0.5\text{ km}$.

(a)

$$\begin{aligned}\text{Actual distance}&=8.8\times 0.5\\ &=4.4\text{ km}\end{aligned}$$

(b)

$$\begin{aligned}1\text{ cm}^2&:0.5^2=0.25\text{ km}^2\\ \text{Map area}&=\frac{2.25}{0.25}\\ &=9\text{ cm}^2\end{aligned}$$

(a)

$$\begin{aligned}1\text{ cm}&:25\,000\text{ cm}\\ 1\text{ cm}&:250\text{ m}\\ 1\text{ cm}&:0.25\text{ km}\\ 8.2\text{ cm}&:8.2\times 0.25=2.05\text{ km}\end{aligned}$$

The actual distance is $2.05$ km.

(b)

$$\begin{aligned}1\text{ cm}^2&:0.25^2=0.0625\text{ km}^2\\ \text{Map area}&=\frac{2.8}{0.0625}\\ &=44.8\text{ cm}^2\end{aligned}$$

$60\,000$ cm $=0.6$ km, so the scale is $1\text{ cm}:0.6\text{ km}$.

(a)

$$\begin{aligned}\text{Map distance}&=\frac{7.68}{0.6}\\ &=12.8\text{ cm}\end{aligned}$$

(b)

$$\begin{aligned}\text{Actual distance}&=7.5\times 0.6\\ &=4.5\text{ km}\end{aligned}$$

(c)

$$\begin{aligned}1\text{ cm}^2&:0.6^2=0.36\text{ km}^2\\ \text{Actual area}&=15\times 0.36\\ &=5.4\text{ km}^2\end{aligned}$$

(a)

$120\,000$ cm $=1.2$ km.

$$\begin{aligned}1\text{ cm}&:1.2\text{ km}\\ 15\text{ cm}&:15\times 1.2=18\text{ km}\end{aligned}$$

The actual distance is $18$ km.

(b)

$$\begin{aligned}1\text{ cm}^2&:1.2^2=1.44\text{ km}^2\\ \text{Map area}&=\frac{0.17}{1.44}\\ &=0.11805\ldots\\ &=0.118\text{ cm}^2\ \text{(3 s.f.)}\end{aligned}$$

(a)

Find the area scale for $1$ cm² first, then take the square root to get the length scale.

$$\begin{aligned}3\text{ cm}^2&:75\text{ km}^2\\ 1\text{ cm}^2&:25\text{ km}^2\\ 1\text{ cm}&:\sqrt{25}=5\text{ km}\end{aligned}$$

So $x=5$.

(b)

The actual road distance is $8.1\times 5=40.5$ km. From 09 15 to 10 00 is $45$ minutes $=\dfrac34$ h.

$$\begin{aligned}\text{Speed}&=\frac{\text{distance}}{\text{time}}\\ &=40.5\div\frac34\\ &=54\text{ km/h}\end{aligned}$$

(a)

Use any row of the table, for example the Circle Line:

$$\begin{aligned}7.14\text{ cm}&:35.7\text{ km}\\ 1\text{ cm}&:5\text{ km}\\ 1\text{ cm}&:500\,000\text{ cm}\end{aligned}$$

The scale is $1:500\,000$. (Check: $43.0\div 8.6=5$ and $20.0\div 4.0=5$.)

(b)

The actual length of the Downtown Line is $43.0$ km. Map $B$ has scale $1:20\,000$, and $20\,000$ cm $=0.2$ km, so $1\text{ cm}:0.2\text{ km}$.

$$\begin{aligned}\text{Length on Map }B&=\frac{43.0}{0.2}\\ &=215\text{ cm}\end{aligned}$$

(c)

$$\begin{aligned}1\text{ cm}^2&:0.2^2=0.04\text{ km}^2\\ \text{Actual area of depot}&=2.5\times 0.04\\ &=0.1\text{ km}^2\\ \text{Area of 14 football fields}&=14\times 0.007\\ &=0.098\text{ km}^2\end{aligned}$$

Yes. The depot has an area of $0.1$ km², which is more than the $0.098$ km² needed for 14 football fields. (Equivalently, $0.1\div 0.007=14.28\ldots$, so 14 whole fields fit by area.) This agrees with the table, since $100\,000$ m² $=0.1$ km².