Question

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An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet as shown in the figure. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. (Take π = 22/7)​

Answer 1

Answer:

Area of metallic sheet used=CSA of frustum+CSA of cylinder +CSA of Base

CSA of frustum,

Diameter of bigger circular end=45cm

Radius=45/2=22.5cm

Diameter of smaller circular end=25cm

Radius=25/2=12.5cm

Height of the frustum=Total height of the bucket−Height of the circular base

40−6=34cm

Slant Height= √h²+(r²-r)²

=√34²+(22.5-2.5²)

=√1156+(10)²

=√1256

Slant height=35−44cm

CSA of frustum=π(r 1+r2)l= 22/7 (22.5+12.5)35.44

= 22/7×35×35.44

=3898.4cm ²

Area of circular Base

Base is a circular part with radius 25/2

Area of circular base=πr²=22/7 ×12.5×12=491.07cm

CSA of cylinder=2πrh=2× 22/7×12.5×7=471.428

Area of metallic sheet used=3989.4+491.07+471.428

=4860.898cm².

Answer 2

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Area of sheet used-CSA of frustration+CSA of cylinder+ Base area

= π(r1 r₂)l + π(1₂) + 2πr₂

= 48609 Cm²

Volume of water it can hold,

= π (r² + r ²₂ + r₁ r₂) h

22/7 (25/4² 25.45/4)34

= 33.61