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.


Answer :




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

NCERT solutions of related questions for Surface Areas and Volumes

NCERT solutions of related chapters class 10 maths

NCERT solutions of related chapters class 10 science