Question

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 sq. cm

Answer 1

Given that,
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm
Diameter of the cylindrical part = 1.4 cm
Therefore, radius (r) of the cylindrical part = 0.7 cm

Total surface area of the remaining solid will be
= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

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

Answer 2

Given height h = 2.4cm, diameter d= 1.4cm then radius will be d/2 = 1.4/2 = 0.7

We need to find the total surface area of the remaining solid.

We know that Total surface of remaining of the remaining solid = Curved area of cylinder + Area of cylinder + Curved surface area of the cone.

1) Curved area of the cylinder:

   2pirh

= 2 * 22/7 * 0.7 * 2.4

= 10.56cm^2.


2)  Area of cylinder = pir^2

                                = 22/7 * (0.7)^2

                               = 1.54cm^2

Height of cone l^2 = root h^2 + r^2

                                = root (2.4)^2 + (0.7)^2

                                = 5.76+0.49

                                = 6.25

                                = 2.5cm

3) curved surface area of cone = pirl

                                                    = 22/7 * 0.7 * 2.5

                                                    = 5.5cm^2


Total surface area of remaining solid = 10.56 + 1.54 + 5.5

                                                              = 17.6 cm^2

                                                              = ~18 cm^2.


Hope this helps!