Aggregate reserve: sandstone 300 m, 227 km2

Principle: reserve (volume) of a bed = area $\times$ true thickness, where the true thickness is obtained from the vertical thickness by $t = t_v\cos\delta$.

Given

Vertical thickness $t_v = 300$ m, attitude S72°W/43° (dip $\delta = 43^\circ$), area $A = 227\ \text{km}^2$.

Calculation

$$\begin{aligned} t &= 300\cos 43^\circ = 300 \times 0.7314 = 219.41\ \text{m} = 0.2194\ \text{km} \\ V &= A \times t = 227 \times 0.2194 = 49.81\ \text{km}^3 \end{aligned}$$ $$V = 49.81\ \text{km}^3 = 4.98 \times 10^{10}\ \text{m}^3$$

Answer: reserve of sandstone aggregate = 49.8 km³ (4.98 × 10¹⁰ m³).

(If the 227 km² is the plan area on the map, the volume under it is $A\times t_v = 227 \times 0.3 = 68.1\ \text{km}^3$.)