question_answer

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameter of the base and the top of the frustum are 20 m and 6 m respectively and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.

A)  \[924.71\text{ }{{m}^{2}}\]          

B)  \[1402.23\text{ }{{m}^{2}}\]

C)  \[1124.56\,{{m}^{2}}\]   

D)         \[1068.57\,{{m}^{2}}\]

Correct Answer: D

Solution :

Height of the cone part \[({{h}_{1}})=(28-24)=4\,m.\]. Radius of base of the cone \[({{r}_{1}})=\frac{6}{2}=3m.\] Radius of base of frustum \[({{r}_{2}})=\frac{20}{2}=10m\] Height of the frustum \[({{h}_{2}})=24\,m\] Let \[{{l}_{1}}\] and \[{{l}_{2}}\] be the slant height of the cone and frustum respectively. So,                        \[\Rightarrow \]  \[{{l}_{2}}^{2}={{({{r}_{2}}-{{r}_{1}})}^{2}}+{{({{h}_{2}})}^{2}}\] \[\Rightarrow \] \[{{l}_{2}}^{2}={{(10-3)}^{2}}+{{24}^{2}}\] \[\Rightarrow \] \[{{l}_{2}}^{2}=49+576=625\] \[\Rightarrow \] \[{{l}_{2}}=25\,m\] And \[{{l}_{1}}^{2}={{r}_{1}}^{2}+{{h}_{1}}^{2}\] \[\Rightarrow \] \[{{l}_{1}}^{2}={{3}^{2}}+{{4}^{2}}\Rightarrow {{l}_{1}}=5\,m\] Quantity of canvas required = curved surface area of frustum + curved surface area of cone \[=\pi \,({{r}_{1}}+{{r}_{2}}){{l}_{2}}+\pi \,{{r}_{1}}{{l}_{1}}=\pi [(10+3)25+3\times 5]\] \[=\pi [325+15]=1068.57{{m}^{2}}\]