
From a solid cylinder whose height is 2.4cm and diameter 1.4cm, a conical cavity of the same height and the same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $ {\text{c}}{{\text{m}}^2} $ .
Answer
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Hint: In this question remember that Total surface area of the remaining solid will be equal to Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone and to find the surface area of remaining solid use the formula Curved surface area of the cylinder = $ 2\pi rh $ , Area of the base of the cylinder $ \pi {r^2} $ and Curved surface area of the cone $ \pi rl $ to approach the solution.
Complete step-by-step answer:
According to the given information height of 2.4 cm solid cylinder whose diameter is 1.4 cm where same conical cavity is hollowed out of same height and diameter
So, the dimensions we know are
Height of solid cylinder (h) = 2.4cm
Diameter of solid cylinder (d) = 1.4cm
Therefore, radius of solid cylinder (r) = 0.7cm
As we know the formula of total surface area of the remaining solid is equal to Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone
Now, we know that Curved surface area of the cylinder = $ 2\pi rh $ , Area of the base of the cylinder $ \pi {r^2} $ and Curved surface area of the cone $ \pi rl $
Where $ l $ slant height which is given as $ l = \sqrt {{r^2} + {h^2}} $
Therefore, Total surface area of the remaining solid = $ 2\pi rh + \pi {r^2} + \pi r\sqrt {{r^2} + {h^2}} $
Now substituting the values in the above formula, we get
$ 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4 + \dfrac{{22}}{7} \times {(0.7)^2} \times \sqrt {{{(2.4)}^2} + {{(0.7)}^2}} $
$ \Rightarrow $ $ 10.56 + 1.54 + 5.5c{m^2} $
$ \Rightarrow $ $ 17.6c{m^2} $
So, Total surface area of the remaining solid is equal to $ 17.6c{m^2} $ .
Note: Trick behind these types of question is to have knowledge about the shapes like cylinder and cone and also the properties of the shapes like area of shape, curved surface area as in the above solution we came across the three types of shape cylinder, cone and circle where cylinder and cone are three-dimensional shapes whereas circle is a two-dimensional shape.
Complete step-by-step answer:

According to the given information height of 2.4 cm solid cylinder whose diameter is 1.4 cm where same conical cavity is hollowed out of same height and diameter
So, the dimensions we know are
Height of solid cylinder (h) = 2.4cm
Diameter of solid cylinder (d) = 1.4cm
Therefore, radius of solid cylinder (r) = 0.7cm
As we know the formula of total surface area of the remaining solid is equal to Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone
Now, we know that Curved surface area of the cylinder = $ 2\pi rh $ , Area of the base of the cylinder $ \pi {r^2} $ and Curved surface area of the cone $ \pi rl $
Where $ l $ slant height which is given as $ l = \sqrt {{r^2} + {h^2}} $
Therefore, Total surface area of the remaining solid = $ 2\pi rh + \pi {r^2} + \pi r\sqrt {{r^2} + {h^2}} $
Now substituting the values in the above formula, we get
$ 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4 + \dfrac{{22}}{7} \times {(0.7)^2} \times \sqrt {{{(2.4)}^2} + {{(0.7)}^2}} $
$ \Rightarrow $ $ 10.56 + 1.54 + 5.5c{m^2} $
$ \Rightarrow $ $ 17.6c{m^2} $
So, Total surface area of the remaining solid is equal to $ 17.6c{m^2} $ .
Note: Trick behind these types of question is to have knowledge about the shapes like cylinder and cone and also the properties of the shapes like area of shape, curved surface area as in the above solution we came across the three types of shape cylinder, cone and circle where cylinder and cone are three-dimensional shapes whereas circle is a two-dimensional shape.
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