A frustum of a cone has an upper base radius of 3 meters and a lower base of 6 meters. If the altitude of the frustum is 9 meters, determine the volume and it's surface area.
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sol'n:
let:
R = lower base radius
r = upper base radius
Solving volume:
V = πh/3 * (R² + r² + Rr)
= π(9)/3 * [6² + 3² + 6(3)]
= 189π
= 593.761 m³
Solving surface area:
S = πR² + πr² + π(R + r)L
where:
L = slant length
L² = (R - r)² + h²
L² = (6 - 3)² + 9²
L² = 90
L = √90
L = 9.487 m
S = π(6)² + π(3)² + π(6 + 3)(9.487)
= 36π + 9π + 268.2386
= 409.61 m²
Answer: volume = 593.761 m³ ; sufrace area = 409.61 m².
let:
R = lower base radius
r = upper base radius
Solving volume:
V = πh/3 * (R² + r² + Rr)
= π(9)/3 * [6² + 3² + 6(3)]
= 189π
= 593.761 m³
Solving surface area:
S = πR² + πr² + π(R + r)L
where:
L = slant length
L² = (R - r)² + h²
L² = (6 - 3)² + 9²
L² = 90
L = √90
L = 9.487 m
S = π(6)² + π(3)² + π(6 + 3)(9.487)
= 36π + 9π + 268.2386
= 409.61 m²
Answer: volume = 593.761 m³ ; sufrace area = 409.61 m².