From a solid cylinder whose height is 2.4 cm and diameter of 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 cm².

The figure of the solid can be created as per the given information as shown below.

As the conical cavity is of the same height and diameter has been hollowed out, it can be seen that one of the bases of the cylinder is not included in the total surface area of the solid.

TSA of the remaining solid = CSA of the cylindrical part + CSA of conical part + Area of one cylindrical base

Let us find the area of the remaining solid by using formulae;

 CSA of cylinder= 2πrh

Area of the base of the cylinder = πr², where r and h are radius and height of the cylinder respectively.

CSA of the cone = πrl

Slant height of the cone, l = √[r² + h²]

where r, h and l are radius, height and slant height of the cone respectively.

Height of the cylinder = Height of the cone = h = 2.4 cm

Diameter of the cylinder = diameter of the cone = d = 1.4 cm

Radius of the cylinder = radius of the cone = r = d / 2 = 1.4 / 2 cm = 0.7 cm

Slant height of the cone, l = √[r² + h²]

l = √[(0.7 cm)² + (2.4 cm)²]

= √[0.49 cm² + 5.76 cm²]

= √[6.25 cm²]

= 2.5 cm

TSA of the remaining solid = CSA of the cylindrical part + CSA of conical part + Area of one cylindrical base

= 2πrh + πrl + πr²

= πr (2h + l + r)

= 22/7 × 0.7 cm × (2 × 2.4 cm + 2.5 cm + 0.7 cm)

= 2.2 cm × 8 cm

= 17.6 cm²

Hence, the total surface area of the remaining solid to the nearest cm² is 18 cm².