Question

S. 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 cm?
​

Answer 1

AnSwer :

• Solid cylinder of height 2.4 cm.
• And the diameter of the solid cylinder is 1.4 cm.
• Hence,the radius of the solid cylinder is (1.4/2) = 0.7 cm.
• Height of the cone is 2.4 cm
• Radius of the cone is (1.4/2) = 0.7 cm.

We have to find the total surface area of the

remaining solid to the nearest cm.

◖ Finding slant height of the cone :

⇏Slant Height = √(Height)² + (Radius)²

⇏Slant Height = √(2.4)² + (0.7)²

⇏Slant Height = √5.76 + 0.49

⇏Slant Height = √6.25

⇏Slant Height = 2.5 cm

◖ According to the Question Now :

⇏TSA of remaining solid = CSA of cylinder + Area of cylinder + CSA of cone

⇏TSA of remaining solid = 2πrh + πr² + πrl

⇏TSA of remaining solid = 2 × 22/7 × 0.7 × 2.4 + 22/7 × (0.7)² + 22/7 × 0.7 × 2.5

⇏TSA of remaining solid = 22/7 × 0.7(2 × 2.4 + 0.7 + 2.5)

⇏TSA of remaining solid = 22 × 0.1 (4.8 + 0.7 + 2.5)

⇏TSA of remaining solid = 2.2 × 8

⇏TSA of remaining solid = 17.6 cm²

⛬ The total surface area of the remaining solid is 17.6 cm².