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]
Show Hint
The shaded region is one equilateral triangle of side 300 m; find its area, multiply by thickness and by bulk density.
Step 1: Place the bore holes on a grid.
Put BH-4 at the origin $(0,0)$ and BH-5 at $(300,0)$ on the bottom row, spaced 300 m apart. Since the pattern is a row of equilateral triangles, the top row sits at height $h=300\sin60^{\circ}=259.8$ m, with BH-1 at $(150,259.8)$ and BH-2 at $(450,259.8)$.
Step 2: Use the shoelace formula for the shaded triangle.
The shaded triangle has corners BH-1 $(150,259.8)$, BH-2 $(450,259.8)$ and BH-5 $(300,0)$. The shoelace area formula for a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is
\[ A=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \]
Step 3: Plug in the coordinates.
\[ A=\frac{1}{2}\left|150(259.8-0)+450(0-259.8)+300(259.8-259.8)\right| \]
\[ A=\frac{1}{2}\left|38970-116910+0\right|=\frac{1}{2}(77940)=38970\ \text{m}^2 \]
This matches the standard formula result of about $38971$ m$^2$; the tiny gap is just rounding in the height value.
Step 4: Turn area into ore volume.
With an average ore thickness of $30$ m,
\[ V=38971\times30=1169130\ \text{m}^3 \]
Step 5: Turn volume into tonnage.
Using the given bulk density of $4400$ kg/m$^3$,
\[ M=1169130\times4400=5144172000\ \text{kg} \]
Divide by $1000$ to get metric tons, then by another $10^6$ to get million tons:
\[ M=\frac{5144172000}{1000\times10^{6}}=5.14\ \text{million tons} \]
\[ \boxed{5.14\ \text{million tons}} \]
