Music, asked by MASTERblasterrr, 7 months ago

From a solid cylinder of height 12 centimetre and diameter 10 cm a conical cavity of same height and same diameter is hollowed out find the total surface area of remaining solid​

Answers

Answered by KrishnaKumar01
5

Answer:

Volume of remaining solid = Volume of cylinder - Volume of Cone

In the given case,

r = 5cm

h = 12cm

Volume of Cylinder = pi * r * r * h = 3.14 * 5 * 5 * 12 = 942 cm^3

Volume of Cone = (1/3) * pi * r * r * h = (1/3) * 3.14 * 5 * 5 * 12 = 314 cm^3

Remaining Volume = 942 - 314 = 628cm^3

Total Surface of Remaining Solid = Conical surface area of Cone + Area of Circle + Cylindrical area of Cylinder

Conical surface area of Cone = pi * r * ((r^2 + h^2)^(1/2)) = 282.74cm^2

Area of Circle = pi * r * r = 78.54cm^2

Cylindrical area of Cylinder = 2 * pi * r * h = 376.8cm^2

Total Surface Area = 738.08cm^2

Explanation:

Answered by Anonymous
10

Volume of remaining solid = volume of cylinder - volume of cone.

r=5 cm (given)

h=12 cm

Volume of cylinder = πr

2

h

=3.14×5

2

×12

=942cm

2

Volume of cone =

3

1

πr

2

h=

3

1

×3.14×5

2

×12

=314cm

3

Remaining volume = 942-314

=628cm

3

Total surface of remaining solid = conical sutface area of cone + Area of circle + Cylindrical area of cylinder.

Conical sutface area of cone =πr

r

2

+h

2

=204.203 cm²

Conical of cirle = πr² =78.54cm²

Cylindrical area of cylinder = 2πrh

=376.8cm²

Total surface area 659.543cm²

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