Question

A tent of height 3.3 m is in the form of a right circular cylinder of diameter 12m and height 2.2m, surmounted by a right circular cone of the same diameter. Find the cost of the canvas of the tent at the rate of Rs. 500 per m².

Answer 1

Given: Height of the tent = 3.3 m

The height of the cylindrical portion = 2.2 m

∴ The height of the conical portion

= Height of the tent – Height of the cylindrical portion

= 3.3 – 2.2

= 1.1m

Given Diameter of the cylinder, d = 12m

Radius = 12/2 = 6m

CSA of cylindrical portion = 2πrh

= 2 x 22/7 x 6 x 2.2

= 82.971 m²

Firstly, we have to find the slant height (l) of the conical portion

⇒ l² = h² + r²

⇒ l² = (1.1)² + (6)²

⇒ l² = 1.21 + 36

⇒ l² = 37.21

⇒ l = √37.21

⇒ l = 6.1m

∴ CSA of the conical portion = πrl

= 22/7 x 6 x 6.1

= 115.029 m²

So,

Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion

= 115.029 + 82.971

= 198 m²

∴ Canvas required to make the tent = 198 m²

Cost of 1m² canvas = Rs 500

Cost of 198 m² canvas = Rs 500 × 198

= Rs 99000

Answer 2

Given: Height of the tent =3.3 m
The height of the cylindrical portion
∴ The height of the conical portion
= Height of the tent - Height of the cylindrical portion
=3.3−2.2
=1.1 m
Given Diameter of the cylinder, d=12 m
∴Radius=12/2​=6 m
CSA of cylindrical portion =2πrh
=2×22/7​×6×2.2
=82.971 m2
Firstly, we have to find the slant height t(l) of the conical portion 
⇒l^2=h^2+r^2
⇒l^2=(1.1)^2+(6)^2
⇒l^2=1.21+36
⇒l^2=37.21
⇒l=sq root 37.21​
⇒l=6.1 m
∴ CSA of the conical portion =πrl
=22/7​×6×6.1
=115.029 m2
So, 
Total surface area of the tent = Surface area of conical portion +surface area of cylindrical portion 
=115.029+82.971
=198 m2
∴ Canvas required to make the tent =198 m2
Cost of 1m2 canvas =Rs.500
Cost of 198 m2 canvas =Rs 500×198=Rs.99000