FIGURE 2.1 represents an incomplete cross section between points 4 and 5 on the orthophoto map - English General - NSC Geography - Question 2 - 2017 - Paper 2

The position of spot height 203 should be marked at the appropriate elevation of 203 m on the cross section, clearly indicated within the grid lines.

There is no intervisibility between points 4 and 5 likely due to the presence of elevation or terrain features that obstruct sightlines; for example, a hill or dip that would block the view.

To find vertical exaggeration (VE), we use the formula:

$VE = \frac{VS}{HS}$
Where the vertical scale (VS) is given as 1:1,000 (derived from the height of 5 mm), and the horizontal scale (HS) is 1:10,000. Thus:

$VE = \frac{1 \text{ (vs)}}{10} = 10$
This means the vertical exaggeration is 10 times.

The vertical scale in a cross section is often exaggerated to make features appear more pronounced, enabling better visualization of the terrain's relief and slope, which may not be discernible at a smaller scale.

First, calculate the vertical interval (VI) as:

$VI = 726.5 m - 526.2 m = 200.3 m$
Next, the horizontal equivalent (HE) should be determined, which is given as the distance between the two points, such as 3,750 m. Thus the gradient (G) is calculated as:

$G = \frac{200.3 m}{3750 m} = 0.0534 \text{ (rounded to 1:18.7) }$

No, the gradient calculated is not a true reflection of the actual landscape due to variations in terrain such as steep slopes or flat areas that are not accounted for. The average gradient may overlook sudden changes in elevation, especially when traversing between significantly different landforms.

The area can be calculated using the formula:

$Area = \text{length} \times \text{breadth}$
Given the length is 9.3 cm and breadth is 7.4 cm, we can calculate:

$Area = 9.3 \text{ cm} \times 7.4 \text{ cm} = 68.82 \text{ cm}^2$
Converting cm² to km² gives:

$Area = \frac{68.82 \text{ cm}^2}{10^4} = 0.006882 \text{ km}^2$

The apparent size difference can be attributed to the scale factors. The topographic map has a scale of 1:10,000, whereas the orthophoto has a larger scale of 1:5,000, resulting in the orthophoto appearing disproportionately larger. Thus, when represented on a smaller-scale map, it appears reduced in area.