# 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. (Use $\pi = \dfrac{{22}}{7}$)

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Hint- Here, we will be using the formulas for curved surface areas of cylinder and cone and the area of cylindrical base.

Given, height of the solid cylinder $H = 2.4{\text{ cm}}$

Diameter of the solid cylinder $D = 1.4{\text{ cm}}$

Radius of the solid cylinder $R = \dfrac{D}{2} = \dfrac{{1.4}}{2} = 0.7{\text{ cm}}$

Also, given the conical cavity is of the same height and same diameter as that of the solid cylinder so the radius of the conical cavity will also be equal.

Height of the conical cavity, $h = H = 2.4{\text{ cm}}$

Radius of the conical cavity, $r = R = 0.7{\text{ cm}}$

Clearly from the figure we can see that the total surface area of the remaining solid consists of the curved surface area of the cylindrical part, curved surface area of the conical part and area of one of the cylindrical base.

As we know that the curved surface area of any cylinder with radius as $R$ and height as $H$ is given by $2\pi RH$.

Also we know that the curved surface of any cone with radius of the base as $r$, height as $h$ and slant height as $l$ is given by $\pi rl$ where $l = \sqrt {{r^2} + {h^2}} $

Also, area of the base of cylinder with radius as $R$ and height as $H$ is given by $\pi {R^2}$.

i.e., Total surface area of the remaining solid$ = $Curved surface area of the cylindrical part$ + $Curved surface area of the conical part$ + $Area of one of the cylindrical base

Total surface area of the remaining solid\[ = 2\pi RH + \pi rl + \pi {R^2} = 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4 + \dfrac{{22}}{7} \times 0.7 \times \left( {\sqrt {{{0.7}^2} + {{2.4}^2}} } \right) + \dfrac{{22}}{7} \times {0.7^2} = 10.56 + 5.5 + 1.54 = 17.6{\text{ c}}{{\text{m}}^2}\]

Hence, the total surface area of the remaining solid is 17.6${\text{ c}}{{\text{m}}^2}$.

Note- In these type of problems, we have to observe the shape of the solid which is left after cutting the cavity. In this particular problem when a conical cavity is hollowed out of a solid cylinder, the surfaces which remain are the curved surface area of the cylinder, one cylindrical base and the curved surface of the cone.

Given, height of the solid cylinder $H = 2.4{\text{ cm}}$

Diameter of the solid cylinder $D = 1.4{\text{ cm}}$

Radius of the solid cylinder $R = \dfrac{D}{2} = \dfrac{{1.4}}{2} = 0.7{\text{ cm}}$

Also, given the conical cavity is of the same height and same diameter as that of the solid cylinder so the radius of the conical cavity will also be equal.

Height of the conical cavity, $h = H = 2.4{\text{ cm}}$

Radius of the conical cavity, $r = R = 0.7{\text{ cm}}$

Clearly from the figure we can see that the total surface area of the remaining solid consists of the curved surface area of the cylindrical part, curved surface area of the conical part and area of one of the cylindrical base.

As we know that the curved surface area of any cylinder with radius as $R$ and height as $H$ is given by $2\pi RH$.

Also we know that the curved surface of any cone with radius of the base as $r$, height as $h$ and slant height as $l$ is given by $\pi rl$ where $l = \sqrt {{r^2} + {h^2}} $

Also, area of the base of cylinder with radius as $R$ and height as $H$ is given by $\pi {R^2}$.

i.e., Total surface area of the remaining solid$ = $Curved surface area of the cylindrical part$ + $Curved surface area of the conical part$ + $Area of one of the cylindrical base

Total surface area of the remaining solid\[ = 2\pi RH + \pi rl + \pi {R^2} = 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4 + \dfrac{{22}}{7} \times 0.7 \times \left( {\sqrt {{{0.7}^2} + {{2.4}^2}} } \right) + \dfrac{{22}}{7} \times {0.7^2} = 10.56 + 5.5 + 1.54 = 17.6{\text{ c}}{{\text{m}}^2}\]

Hence, the total surface area of the remaining solid is 17.6${\text{ c}}{{\text{m}}^2}$.

Note- In these type of problems, we have to observe the shape of the solid which is left after cutting the cavity. In this particular problem when a conical cavity is hollowed out of a solid cylinder, the surfaces which remain are the curved surface area of the cylinder, one cylindrical base and the curved surface of the cone.

Last updated date: 20th Sep 2023

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