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Sakthivel

· started a discussion

· 1 Months ago

pr^2+2prh+prl

Question:

From a solid cylinder of height 4 cm and radius 3 cm, a conical cavity of height 4 cm and of base radius 3 cm is hollowed out. What is the total surface area of the remaining solid?

Options:
A)

15 π square cm

B)

22 π square cm

C)

33 π square cm

D)

48 π square cm

Solution:

Ans: (d)


Total surface area of the remaining solid 

= CSA of cylinder + CSA of the conical cavity + Area of the base of cylinder

CSA Cylinder = 2πrh

= 2π×3×4

= 24π

CSA conical cavity = πrl

= \(π×3×\sqrt[]{9+16}\)

= π×3×5

= 15π

Area of the base of cylinder = πr2

= π×9

= 9π

Total surface area of the remaining solid = 24π+15 π+ 9π = 48 π

Knowledge Expert

· commented

· 1 Months ago

Dear student
We resolved it

Best wishes
Team TR

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