Q. 84.2( 303 Votes )

# From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^{2}

Answer :

**Given:**

Height (*h*) of the conical part = Height (*h*) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Radius (*r*) of the cylindrical part = 0.7 cm

we know, slant height of cone, l = √(r^{2} + h^{2})

Slant height of the cylindrical part (l) =

=

=

= 2.5

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2πrh + πrl + πr^{2}

=( 2 × × 0.7 × 2.4 )+( × 0.7 × 2.5) +( × 0.7 × 0.7)

= (4.4 × 2.4) +( 2.2 × 2.5) +( 2.2 × 0.7)

= 10.56 + 5.50 + 1.54

= 12.10 + 5.50 cm^{2}

= 17.60 cm^{2} (approx)**Hence, the total surface area of remaining solid is 17.60 cm ^{2}.**

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