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 cm2.
The diameter of the cylinder = diameter of conical cavity = 1.4 cm
Radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7
Height of the cylinder = height of the conical cavity = 2.4 cm
∴ Slant height of the conical (l) = \(\sqrt{h^2 + r^2}\)
= \(\sqrt{(2.4)^2 + (0.7)^2}\)
= \(\sqrt{5.76 + 0.49}\)
= \(\sqrt{6.25}\)
= 2.5 cm
TSA of remaining solid = surface area of conical cavity + TSA of the cylinder
= πrl + (2πrh + πr2)
= πr(l + 2h + r)
= (22/7) × 0.7(2.5 + 4.8 + 0.7)
= 2.2×8 = 17.6 cm2
So, the total surface area of the remaining solid is 17.6 cm2