Question

The bore hole distribution of a massive sulphide deposit represented in the form of an equilateral triangular pattern is given below. If the average thickness of the ore body is 30 m, the tonnage in the shaded area is million tons (rounded off to two decimal places).

[Use: Bulk density of ore body = 4400 kg/m³]

Hint:

The shaded region is one equilateral triangle of side 300 m; find its area, multiply by thickness and by bulk density.

Answer 1

Correct Answer: 5.14

Step 1: Identify the shaded triangle.
The seven bore holes form two rows of equilateral triangles, each side 300 m. The shaded region is the single inverted triangle formed by BH-1, BH-2 and BH-5, so it is itself an equilateral triangle of side 300 m.

Step 2: Recall the area formula for an equilateral triangle.
For a side length \(a\), the area is
\[ A = \frac{\sqrt{3}}{4}a^2 \]
This comes from splitting the triangle into two right triangles and using the height \(h = a\sin 60^{\circ}\).

Step 3: Substitute the side length.
\[ A = \frac{\sqrt{3}}{4}(300)^2 = \frac{\sqrt{3}}{4}\times 90000 \]
\[ A = 0.4330 \times 90000 = 38971.1 \text{ m}^2 \]

Step 4: Find the volume of ore in the shaded block.
The average thickness of the ore body is 30 m, so
\[ V = A \times t = 38971.1 \times 30 = 1169134.3 \text{ m}^3 \]

Step 5: Convert volume to mass using bulk density.
The bulk density is given as 4400 kg/m\(^3\).
\[ M = V \times \rho = 1169134.3 \times 4400 = 5144190898 \text{ kg} \]

Step 6: Convert mass to tons and then to million tons.
One metric ton equals 1000 kg, so
\[ M = \frac{5144190898}{1000} = 5144190.9 \text{ tons} = 5.14 \text{ million tons} \]

Final Answer:
The tonnage of ore in the shaded triangular block is
\[ \boxed{5.14 \text{ million tons}} \]