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 the 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 folds and for stitching. Give your answer to the nearest $m^2$ .

Answer 1

Radius of the tent, r =$\frac{168}{2}$ m = 84 m.

Height of the tent = 85 m.

Height of the cylindrical part, H = 50 m.

Height of the conical part, h = (85 - 50) m = 35 m.

Slant height of the conical part, l = $\sqrt{h^2+r^2}=\sqrt{(35{)}^2+(84{)}^2}=\sqrt{8281}m$ = 91 m

Quantity of canvas required = Curved surface area of the tent

= Curved surface area of the cylindrical part + Curved surface area of the conical part

= 2$\pi$rH + $\pi$rl = $\pi$r(2H + I)
= $\frac{22}{7}\times 84(2\times 50+91)m^2=(22\times 12\times 191)m^2=50424m^2$.

Area of canvas required for folds and stitching = (20% of 50424) $m^2$

= ($\frac{20}{100}$) × 50424 $m^2$

= 10084.80 $m^2$.

Total quantity of canvas required to make the tent

= (50424 + 10084.80) $m^2=60508.80m^2=60509m^2$ (to the nearest $m^2$)