A circus tent is in the shape of cylinder surmounted by a conical top of same diameter. If their common diameter is 56 m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent.
Solution
Total height of the tent above the ground = 27 m
Height of the cylinderical part,
\[h_1\]= 6 m
Height of the conical part,
\[h_2\] = 21 m
Diameter = 56 m
Radius = 28 m
Curved surface area of the cylinder, CSA1 = \[2\pi r h_1 = 2\pi \times 28 \times 6 = 336\pi\]
Curved surface area of the cylinder, CSA2 will be
\[\pi rl = \pi r\left( \sqrt{h^2 + r^2} \right) = \pi \times 28 \times \left( \sqrt{{21}^2 + {28}^2} \right) = 28\pi\left( \sqrt{441 + 784} \right)\]
\[ = 28\pi \times 35\]
\[ = 980\pi\]
Total curved surface area = CSA of cylinder + CSA of cone
= CSA1 + CSA2
\[= 336\pi + 980\pi\]
\[ = 1316\pi\]
\[ = 4136 m^2\]
Thus, the area of the canvas used in making the tent is 4136 m2.
