A reservoir is in the shape of a frustum of a right circular cone. The radii of its circular ends are 4 m and 8 m and its depth is 7 m. How many kiloliter of water (correct up to one decimal place) can it hold? (Take $π =\frac{22}{7}$)

821.3

We know that,

Volume of a frustum of a cone = \(\frac{1}{3}\) πh (r² + R² + Rr)

The radii of its circular ends = 4m and 8m

Depth = 7m  = h

The volume of reservoir =

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × (8² + 4² + 8 × 4) × 7

= \(\frac{1}{3}\) × 22 × (64 + 16 + 32)

= \(\frac{1}{3}\) × 22 × 112 = 821.3 m³ or 821.3 litres