From a solid cylinder whose height is 2.4cm and diameter 1.4cm, 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 cm². (Use π=22/7)

Given:
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4cm
Diameter of the cylindrical part = 1.4cm
Radius = Diameter/2
Radius (r) of the cylindrical part = 0.7cm
Slant height (l) of conical part = √r² + h² = √0.7² + 2.4² = √0.49 + 5.76 = √6.25 = 2.5
Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
= 2πrh + πrl + πr²
= 2 × 22/7 × 0.7 × 2.4 + 22/7 × 0.7 × 2.5 + 22/7 × 0.7 × 0.7
= 4.4 × 2.4 + 2.2 × 2.5 + 2.2 × 0.7
= 10.56 + 5.50 + 1.54
= 17.60 cm²
The total surface area of the remaining solid to the nearest cm² is 18 cm²