Question

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is $85$ m and height of the cylindrical part is $50$m. If the diameter of the base is $168$m, find the quantity of canvas required to make the tent. Allow $20$ extra for fold and far stitching. Give your answer to the nearest $m^2$.

• A. $504053.012$
• B. $50553.702$
• C. $40423.712$
• D. $50423.712$

Answer 1

The correct option is D $50423.712$

The total canvas required will be equal to sum of lateral surface area of cone and cylinder.

Given: Diameter of base of cylinder and cone $168m\Rightarrow r=84m$

Total height of tent$85m$

Height of cyliner $50m$

Height of cone $35m$

slant height (l) of cone$=\sqrt{{35}^2+{84}^2}=\sqrt{h^2+r^2}$

$l=\sqrt{8281}=91m$

L.S.A of cylinder $2\pi rhcylinder$

=$2\times \pi \times 84\times 50$

LSA.cylinder=$263809.378m^2$

L.S.A of cone=$\pi rl$

$=\pi \times 84\times 91$

LSA cone $=24014.334m^2$

It is given that add $20m^2$ extra for fold and for a stitching

$\therefore$Total canvas required $=LSAcylinder+LSAcona+20m^2$

$26389.378+24014.334+20$

$50423.712m^2$