if the radii of the circular ends of a conical bucket, which is 16 cm high are 20 cm and 8 cm respectively.find the capacity of bucket and T.S.A of bucket
Answers
Since a bucket have a shape of frustum cone.
TSA of frustum cone = π(r+R)×{√(R-r)^2 + h^2} + πr^2 + πR^2
Therefore, TSA of the bucket = π(8+20)×{√(20-8)^2 + 16^2} + π8^2 + π20^2
= π × 28 × {20} + π × 64 + π × 400
= 1759.29 + 201.06 + 1256.63
= 3216.98 cm^2
Capacity of the bucket = 1/3 × πh (R^2 + rR + r^2)
= 1/3 × π×16 (20^2 + 8×20 + 8^2)
= 1/3 × 50.265 (400 + 160 + 64)
= 1/3 × 31365.36
= 10455.12 cm^3
Hope you got it!
Answer:
10459.43 cm³, 1961.14 cm²
Step-by-step explanation:
Given, h = 16 cm, r₁ = 20 cm, r₂ = 8 cm.
The shape of a conical bucket is like a frustum of cone.
∴ Volume of bucket = Volume of frustum of cone.
= 1/3 * πh(r₁² + r₂² + r₁r₂)
= 1/3 * 22/7 * 16(20² + 8² + 20 * 8)
= 22/21 * 16(400 + 64 + 160)
= 219648/21
= 10459.43 cm³.
∴ Slant height of the bucket(l) = √h² + r²
= √16² + (20 - 8)²
= √256 + 144
= 20 cm.
∴ Total surface area of the bucket = π(r₁ + r₂)l + πr₂²
= (22/7)[(20 + 8) * 20 + (8)²]
= 22/7[28 * 20 + 64]
= 22/7[560 + 64]
= 22/7[624]
= 13728/7
= 1961.14 cm².
Therefore:
⇒ Capacity of bucket = 10459.43 cm³
⇒ T.S.A of bucket = 1961.14 cm²
Hope it helps!