A tent of height 77 dm is in the form of a right circular cylinder of

1.	A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m2( use π = \(\frac{22}7\)).
Answer» Given: height of the tent, H = 77 dm = 7.7 m Height of the cylindrical portion, h = 44 dm = 4.4 m ∴ Height of the conical portion, h’ = 7.7 – 4.4 = 3.3 m Radius of the cylinder, r = \(\frac{36}2\) = 18 m Curved surface area of the cylindrical portion of the tent = 2πr² h = 2 × π × (18)² × 4.4 = 158.4 π m² Slant height of the conical portion of the tent = \(\sqrt{3.3^2 + 18^2}\) = 18.3 m ∴ Curved surface area of the conical portion = π × 18 × 18.3 = 329.4π m² ∴ Total surface area of the tent = 158.4 π + 329.4 π = 487.8 π = 1533.09 m² ∴ Canvas required to make the tent = 1533.09 m² Total cost of the canvas = 1533.09 × 3.50 = Rs 5365.80

A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m2( use π = \(\frac{22}7\)).

Answer»

Given:

height of the tent, H = 77 dm = 7.7 m

Height of the cylindrical portion, h = 44 dm = 4.4 m

∴ Height of the conical portion, h’ = 7.7 – 4.4 = 3.3 m

Radius of the cylinder, r = \(\frac{36}2\) = 18 m

Curved surface area of the cylindrical portion of the tent

= 2πr² h

= 2 × π × (18)² × 4.4

= 158.4 π m²

Slant height of the conical portion of the tent = \(\sqrt{3.3^2 + 18^2}\)

= 18.3 m

∴ Curved surface area of the conical portion = π × 18 × 18.3

= 329.4π m²

∴ Total surface area of the tent = 158.4 π + 329.4 π

= 487.8 π

= 1533.09 m²

∴ Canvas required to make the tent = 1533.09 m²

Total cost of the canvas = 1533.09 × 3.50

= Rs 5365.80