Question

The interior of a building is in the form of a right circular cylinder of a diameter 4.2m and height 4m, surmounted by a cone. The vertical height of the cone is 2.1m. Find the outer surface area and volume of the building. (Take $\pi = \dfrac{{22}}{7}$ )
A. $Outer\,surface\,area = 72.4{m^2};Volume = 68.142{m^3}$
B. $Outer\,surface\,area = 72.4{m^2};Volume = 65.142{m^3}$
C. $Outer\,surface\,area = 75.4{m^2};Volume = 68.142{m^3}$
D. $Outer\,surface\,area = 75.4{m^2};Volume = 65.142{m^3}$

Answer 1

Hint: The outer surface of the building will be the sum of the curved surface area of the cylindrical part and the curved surface of the conical part and the volume of the building will be the sum of volume of the cylindrical part and the conical part.

Complete step-by-step answer:

Given, the diameter (d) of the right circular cylinder is 4.2 m and height is 4m. The vertical height of the cone is 2.1 m.
Outer surface area of the building = Curved surface area of cylindrical part + Curved surface area of the conical part.
First, let us find the curved surface area of cylinder i.e.., the formula to find the curved surface are of a cylinder with radius “r” and height “h” $ = 2\pi rh$
Radius of the cylindrical part$ = \dfrac{{Diameter}}{2} = \dfrac{{4.2}}{2} = 2.1m$.
So, the radius is 2.1 m and the height of the cylinder is 4 m.
Therefore, the curved surface area of the cylinder is $2\pi rh = 2 \times \dfrac{{22}}{7} \times 2.1 \times 4 = 52.75{\text{ }}{{\text{m}}^2} \to (1)$.
Now, coming to the curved Surface area of a cone, we know the formula of curved surface area of a cone is $\pi rl$ where r is the radius of the cone and l is the slant height.
For a cone, $l = \sqrt {{h^2} + {r^2}} = \sqrt {{{2.1}^2} + {{2.1}^2}} \left( {h = 2.1,r = 2.1} \right)$
$l = 2.1\sqrt 2 m$.
Hence curved surface area of cone is $\pi rl = \dfrac{{22}}{7} \times 2.1 \times 2.1\sqrt 2 = 19.58{\text{ }}{{\text{m}}^2} \to (2)$.
Therefore, from the values of equation (1) and (2),
Outer surface area of the building = Curved surface area of cylindrical part + Curved surface area of the conical part.
Outer surface area of the building=$52.75 + 19.58 = 72.4{\text{ }}{{\text{m}}^2}$.
Now, coming to volume of the building, it can be written as
Volume of building= Volume of the cylindrical part+ Volume of conical part
We are going to apply the formula of,
Volume of cylinder = $\pi {r^2}h = \dfrac{{22}}{7} \times {2.1^2} \times 4 = 55.44{\text{ }}{{\text{m}}^3}\left( {\because h = 4,r = 2.1} \right)$
Volume of cone = $\dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3} \times \dfrac{{22}}{7} \times 2.1 \times 2.1 \times 2.1 = 9.702{\text{ }}{{\text{m}}^3}\left( {\because h = 2.1,r = 2.1} \right)$
Therefore,
Volume of building = $55.44 + 9.702 = 65.142{m^2}$

Note: In these types of questions, we need to find volume and surface area with the help of formulae. So, before solving the question, make sure to know the formulae of the volume and surface area of given shapes. Make sure to note the difference of height (h) and slant height (l) of the cone.