This question is the combination of cone and cylinder structure to find answer of this question use cylinder and cone surface area and volume formulas. Question number 22 from RS Aggarwal book exercise 17A.
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From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid.
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Volume of the solid left = Volume of cylinder – Volume of cone
= πr²h – ( [1/3]πr²h )
= [22/7] × 6² × 8
= 603. 428 cm³
The slant length of the cone,
Using Pythagoras Theorem,
l = √[r²+ h²]
= √36 +64
l = 10cm
Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone
= πr²+2πrh +πrl
= πr (r+2h +l)
= [22/7] × 6 × (6 +16 +10)
= 603. 42 cm²
∴ Total surface area of the remaining solid is 603. 42 cm²