If the radius of the circular ends of a truncated conical bucket of height 45 cm are 14 cm and 7 cm, then find the capacity in cubic cm of the bucket. 

16,170

Volume of the bucket = \(\frac{1}{3}\) \(\pi \)H(r₁² + r₂² + r₁r₂)

                               = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 45 (14² + 7² + 14 × 7)

                               = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 45 × 343

                               = 16,170