Give me the best method to solve the surface areas and volumes of ncert class 10 , Find the easiest solution of the exercise 13.1 question number 8 . Also find the simplest method of this 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 cm2.

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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. Q.8

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The diagram for the question is as follows:

From the question we know the following:

The diameter of the cylinder = diameter of conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder

= Ï€rl+(2Ï€rh+Ï€r

^{2})= Ï€r(l+2h+r)

= (22/7)Ã— 0.7(2.5+4.8+0.7)

= 2.2Ã—8 = 17.6 cm

^{2}So, the total surface area of the remaining solid is 17.6 cm

^{2}