if the radii of the circular ends of a conical bucket which is 32cm high are 40cm and 16cm find the capacity of total surface area of the bucket?
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》》》Radius of the bigger end of the frustum (bucket) of cone = R = 20 cm
Radius of the smaller end of the frustum (bucket) of the cone = r = 8 cm
Height = 16 cm
Volume = 1/3πh[R² + r² + R*r]
= 1/3*22/7*16[20² + 8² + 20*8]
= 352/21[400 + 64 + 160]
= (352*624)/21
= 219648/21
= 10459.43 cu cm
Now,
Slant height of the cone = l = √(R - r)² + h²
l = √(20 - 8)² + 16²
l = √12² + 16²
l = √144 + 256
l = √400
l = 20 cm
Slant height is 20 cm
Now,
Surface area = π[R² + r² + (R + r)*l]
= 22/7[20² + 8² + (20 + 8)*16]
= 22/7[400 + 64 + 448]
= 22/7*912
= 20064/7
= 2866.29 sq cm
i hope it helps you
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Answer:
Explanation:
Volume of bucket = Volume of the frustum = πh(R^2+Rr+r^2 )/3
Here R = 40 cm
r = 16 cm
h = 32 cm.
Volume of bucket = 22*32*(40*40 + 40*16 + 16*16)/(3*7) = 22 * 32 * 2496/(3*7)
= 83675.42 cubic cm.
The bucket is normally open from the top and closed from bottom. (Add area of bottom circle to lateral surface area)
Hence total surface area = π(R+r)*s+ πr^2
s = Slant height. = √((h^2+(R-r)^2) = √(32^2+(40-16)^2 = 40cm.
Hence Total surface area of the bucket = π(40+16)*40+ π16^2 = 7844.57 Square cm.
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