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# 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.

Radius of the cylinder $$= \ \frac{1.4}{2} \ = \ 7$$cm

Height of the cylinder $$= \ 2.4$$ cm

Radius of the cone $$= \ 0.7$$ cm

Height of the cone $$= \ 2.4$$ cm

Slant height of the cone $$= \ \sqrt{(0.7)^2 \ + \ (2.4)^2} \ = \ \sqrt{0.49 \ + \ 5.76}$$

$$= \ \sqrt{6.25} \ = \ 2.5$$ cm

Surface area of the remaining solid = Curved surface area of cylinder + Curved surface of the cone + Area of upper circular base of cylinder

$$= \ 2 \pi rh \ + \ \pi rl \ + \ \pi r^2 \ = \ \pi r(2h \ + \ l \ + \ r)$$

$$= \ \frac{22}{7} \ × \ 0.7 \ × \ (2 \ × \ 2.4 \ + \ 2.5 \ + \ 0.7)$$

$$= \ 22 \ × \ 0.1 \ × \ (4.8 \ + \ 2.5 \ + \ 0.7)$$

$$= \ 2.2 \ × \ 8.0 \ = \ 17.6 \ = \ 18$$ cm2