(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\(km^{2}\). (b) A rectangular...
Question 1 Report
(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\(km^{2}\).
(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeter of the path. Calculate the length of the field.
(c) 
The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\(cm^{2}\). Find, correct to the nearest degree, the value of x. [Take \(\pi = \frac{22}{7}\)].
Answer Details
(a) Area on the map. The linear scale is \(1:20000\), so the area scale is \(1:20000^2=1:4\times10^8\).
Convert the real area: \(1\text{ km}=10^5\text{ cm}\), so \(1\text{ km}^2=10^{10}\text{ cm}^2\), and
\[85\text{ km}^2=85\times10^{10}=8.5\times10^{11}\text{ cm}^2.\]
Map area \[=\frac{8.5\times10^{11}}{4\times10^{8}}=\frac{8.5\times10^{3}}{4}=2125\text{ cm}^2.\]
(b) Length of the field. Let the length be \(L\) m. With a \(6\) m path all round, the outer rectangle is \((L+12)\) m by \((18+12)=30\) m.
Area of the path \(=30(L+12)-18L=12L+360\) (m\(^2\)); perimeter of the path (its outer boundary) \(=2[(L+12)+30]=2L+84\) (m). Equating the field's area with the perimeter of the path,
\[18L=2L+84\Rightarrow 16L=84\Rightarrow L=5.25\text{ m}.\]
(c) Finding x. Radius \(r=\frac{7}{2}=3.5\) cm, so the whole circle has area
\[\pi r^2=\frac{22}{7}\times3.5^2=\frac{22}{7}\times12.25=38.5\text{ cm}^2.\]
In the diagram the shaded part is the major region, i.e. the circle less the sector \(POQ\) of angle \(x°\):
\[38.5-\frac{x}{360}\times38.5=27.5\Rightarrow \frac{x}{360}\times38.5=11\]
\[\frac{x}{360}=\frac{11}{38.5}=\frac{2}{7}\Rightarrow x=\frac{2}{7}\times360=102.86\approx103°.\]
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