an open metallic bucket in the shape of frustrum of a come. if the diameter of the two circular ends of the bucket are 45cm and 25cm and the vertical height of the bucket is 24cm,find the area of the metallic sheet used to make the bucket. also find the volume of water it can hold.
Answers
Answer:
Metallic sheet used = 3351.07 cm²
Volume of water = 23728 cm³
Step-by-step explanation:
Please see the attached image for a frustum.
Here it is given that R = 45/2 = 22.5
r = 25/2 = 12.5
h = 24.
The slant height of the bucket is given by following equation
= √(h² + (R - r)²)
= √(24² + (22.5 - 12.5)²)
= 26 cm.
The metal sheet required is equal to surface area of frustum and area of the base circle.
Metal sheet needed = Curved surface area of frustum + Base circle area
= π(R + r)l + πr²
= (22*(22.5 + 12.5)*26* + 22*12.5*12.5)/7
= (22 * 35 * 26 + 22 * 12.5 * 12.5) / 7
= 3351.07 cm²
The volume of water = Volume of frustum
= πh(R² + r² = Rr)/3
= 22 * 24 (22.5² + 12.5² + 22.5*12.5)/(7 * 3)
= 23728 cm³
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