WAEC 2010 · Paper 2 · Q6✱✱

(a)
The scale of a map is $1 : 20\,000$ . Calculate the area, in square centimetres, on the map of a forest reserve which covers $85\text{ km}^2$ .
(b)
A rectangular playing field is $18\text{ m}$ wide. It is surrounded by a path $6\text{ m}$ wide such that its area is equal to the area of the path. Calculate the length of the field.
(c)
In the diagram, $O$ is the centre of a circle of radius $3.5\text{ cm}$ and $\angle POQ = x$ . The area of the shaded major sector is $27.5\text{ cm}^2$ . Find $x$ , correct to the nearest degree. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

On the map, $1\text{ cm}$ stands for $20\,000\text{ cm} = 200\text{ m}$
$= 0.2\text{ km}$ .
Square it for areas: $1\text{ cm}^2$ on the map stands for $0.2 \times 0.2 = 0.04\text{ km}^2$ .
Divide: $85 \div 0.04 = 2125$ .
The forest covers $2125\text{ cm}^2$ on the map.

Let the length of the field be $a\text{ m}$ .
With the path, the outer rectangle is $(a + 12)\text{ m}$ by $18 + 12 = 30\text{ m}$ .
Area of the path = outer area − field area $= 30(a + 12) - 18a = 12a + 360$ .
The path and the field have equal areas: $12a + 360 = 18a$ .
So $6a = 360$ and $a = 60$ .
The field is $60\text{ m}$ long.

The shaded sector has angle $360^\circ - x$ .
Its area is $\dfrac{360 - x}{360} \times \dfrac{22}{7} \times 3.5^2 = 27.5$ .
$\frac{22}{7} \times 3.5^2 = 38.5$ , so $\dfrac{360 - x}{360} = \dfrac{27.5}{38.5}$
$= \dfrac57$ .
Then $360 - x = \frac57 \times 360 = 257.14$ .
So $x = 102.86^\circ$ , which is $103^\circ$ to the nearest degree.