Question

An exhibition tent is in the form of a cylinder surmounted by a cone. The height
the tent above the ground is 85 m and the height of the cylindrical part is 50 m If
diameter of the base is 168 m, find the quantity of canvas required to make t
Allow 20% extra for folds and stitching. Give your answer to the nearest m2.

please explain me how you do the 20% part...I just need that
plus I'll mark as brainliest​

Answer 1

Step-by-step explanation:

Height of the cone = 35 m

Height of cylinder = 50 m

Curved surface area of the tent = 2πrh + πrl

= 2 × 22/7 × 84 × 50 + 22/7 × 84 × 91 [l2 = (35)2 + (84)2] i.e., l = 91]

44 ×12 × 50 + 22 × 12 × 91

= 26400 + 24024

= 50424 sq. m

Area of the canvas required with (20% extra)

= 120/100 × 50424 = 60508.8 sq. m

= 60509 m2

Answer 2

Total height of the tent = 85 m

Diameter of the base = 168 m

Therefore,

radius (r) = 84 m

Height of the cylindrical part = 50 m

Then height of the conical part = (85 - 50) = 35 m

slant height =

$\sqrt{ {r}^{2} + {h}^{2} } = \sqrt{ {84}^{2} + {35}^{2} }$

Slant height = 91 cm

Total surface area of the tent = 2πrh + πrl = π(2h + l)

$= \frac{22}{7} \times 84(2 \times 50 + 91)$

= 264(100+91)

= 264 × 191

Total surface area of the tent = 50424 m²

Since 20% extra is needed for folds and stitching,

total area of canvas needed = 50424(120/100)

= 60508.8

= 60509 m²