Aggregate reserve: quartzite 300 m, 437 km2

Principle: reserve = area $\times$ true thickness, with $t = t_v\cos\delta$.

Given

$t_v = 300$ m, attitude S72°W/43° (dip $43^\circ$), area $A = 437\ \text{km}^2$.

Calculation

$$\begin{aligned} t &= 300\cos 43^\circ = 219.41\ \text{m} = 0.2194\ \text{km} \\ V &= 437 \times 0.2194 = 95.88\ \text{km}^3 = 9.59 \times 10^{10}\ \text{m}^3 \end{aligned}$$

Answer: reserve of quartzite aggregate = 95.9 km³ (9.59 × 10¹⁰ m³).

(If 437 km² is the plan area, the volume under it is $437 \times 0.3 = 131.1\ \text{km}^3$.)