Question 1 Report
The Table X above shows the number of farm units in three villages of Oju District in the year 2010. Use the data in Table X to answer the questions that follow.
(a) Using the map of Oju District provided on page 3 and radius of 1 cm to represent 3 farm units, Construct proportional circles to represent the data. (b) State two disadvantages of proportional circles.
Data read from Table X
| Village | Number of farm units |
|---|---|
| A | 81 |
| B | 100 |
| C | 64 |
(a) Constructing the proportional circles
In a proportional circle the area of the circle (not the radius) is made proportional to the quantity it represents. Since the area of a circle is \( A = \pi r^{2} \), the area is proportional to \( r^{2} \). Therefore, to make the area proportional to the number of farm units \( N \), the radius must be proportional to the square root of \( N \):
\[ A \propto N \quad\Rightarrow\quad \pi r^{2} \propto N \quad\Rightarrow\quad r \propto \sqrt{N} \]
Applying the given scale. The scale states that a radius of \(1\ \text{cm}\) represents \(3\) farm units. Taking the reference circle of radius \(1\ \text{cm}\) to stand for \(3\) farm units by its area, the area per farm unit is fixed, so:
\[ \pi r^{2} = \frac{N}{3}\,\pi \quad\Rightarrow\quad r = \sqrt{\frac{N}{3}} \ \text{cm} \]
Working for each village:
Notice that because the given values are perfect squares, the radii keep the neat ratio \( \sqrt{81}:\sqrt{100}:\sqrt{64} = 9:10:8 \), so village B has the largest circle, village A the middle one and village C the smallest.
Table of radii for drawing:
| Village | Farm units (N) | \( \sqrt{N} \) | Radius \( r=\sqrt{N/3} \) (cm) |
|---|---|---|---|
| A | 81 | 9 | 5.2 |
| B | 100 | 10 | 5.8 |
| C | 64 | 8 | 4.6 |
Steps to construct on the map of Oju District:
(b) Two disadvantages of proportional circles
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