question_answer

The interior of a building is in the form of cylinder of diameter \[4.3\text{ }m\]and height 3.8 m, surmounted by a cone whose vertical angle is a right angle. Find the volume and curved surface area of the building respectively. (Take\[\pi =3.14\]).

A)   \[65.56\text{ }{{m}^{3}},\text{ }71.83\text{ }{{m}^{2}}\]

B)  \[70.24\text{ }{{m}^{3}},\text{ }62.24\text{ }{{m}^{2}}\]

C)  \[62.26\text{ }{{m}^{3}},\text{ }75.56\text{ }{{m}^{2}}\]

D)  \[72.26\text{ }{{m}^{3}},\text{ }66.46\text{ }{{m}^{2}}\]

Correct Answer: A

Solution :

Radius of cone = Radius of cylinder \[=\frac{4.3}{2}=2.15\,cm\] Also,   \[\angle FAC={{90}^{o}}\] (given)     \[\Rightarrow \]            \[\angle BAC={{45}^{o}}\] Now, in \[\Delta \text{ }ABC,~\] \[\tan {{45}^{o}}=\frac{BC}{AB}\] \[1=\frac{2.15}{AB}\,\,\,\Rightarrow \,\,AB=2.15\] \[\therefore \]  Height of cone\[=2.15\text{ }m\].                 Volume of the building = Volume of cylinder    + Volume of cone \[=\pi {{(2.15)}^{2}}\times 3.8+\frac{1}{3}\pi \,{{(2.15)}^{2}}\times (2.15)\] \[=\pi [17.5655+3.3127]=3.14\times 20.8782=65.56{{m}^{3}}\]           Now, slant height of the cone \[l=\sqrt{{{(2.15)}^{2}}+{{(2.15)}^{2}}}=2.15\sqrt{2}=3.04\,m\] Surface area of the building = Surface Area of cylinder + Surface area of cone \[=2\times \pi \times 2.15\times 3.8+\pi \times 2.15\times 3.04\] \[=\pi [16.34+6.536]=3.14\times 22.876=71.83{{m}^{2}}\]