Question

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 ₹ 3.50 per m^2 . ​

Answer 1

Answer:

The cost of canvas is ₹ 5365.80.

The cost of canvas is ₹ 5365.80.Step-by-step explanation:

Given:

Diameter of the cylinder = 36 m

Diameter of the cylinder = 36 mRadius of the cylinder , r = 36 / 2 m = 18 m

Diameter of the cylinder = 36 mRadius of the cylinder , r = 36 / 2 m = 18 mThe height of the tent = 77 dm

Diameter of the cylinder = 36 mRadius of the cylinder , r = 36 / 2 m = 18 mThe height of the tent = 77 dmHeight of the cylindrical part ,H = 44 dm

Diameter of the cylinder = 36 mRadius of the cylinder , r = 36 / 2 m = 18 mThe height of the tent = 77 dmHeight of the cylindrical part ,H = 44 dmHeight of the right circular cone , h = (77

– 44) dm = 33 dm = 33/10 = 3.3 m

– 44) dm = 33 dm = 33/10 = 3.3 m[1 dm = 1/10 m]

– 44) dm = 33 dm = 33/10 = 3.3 m[1 dm = 1/10 m]Let the slant height of the cone be (l).

– 44) dm = 33 dm = 33/10 = 3.3 m[1 dm = 1/10 m]Let the slant height of the cone be (l).l² = r² + h

l² = (18)² + (3.3)²

l² = (18)² + (3.3)²l² = 324 + 10.89

l² = (18)² + (3.3)²l² = 324 + 10.89l² = 334.89

l² = (18)² + (3.3)²l² = 324 + 10.89l² = 334.89l = √334.89

l² = (18)² + (3.3)²l² = 324 + 10.89l² = 334.89l = √334.89l = 18.3 m

l² = (18)² + (3.3)²l² = 324 + 10.89l² = 334.89l = √334.89l = 18.3 mslant height of the cone is 18.3 m.

l² = (18)² + (3.3)²l² = 324 + 10.89l² = 334.89l = √334.89l = 18.3 mslant height of the cone is 18.3 m.Curved surface area of the cylinder = 2πrh

= 2π × 18 × 4.4 m² ……………….. (1)

= 2π × 18 × 4.4 m² ……………….. (1)Curved surface area of the cone = πrl

= π × 18 × 18.3 m²…………………….. (2)

= π × 18 × 18.3 m²…………………….. (2)Total curved surface area of the tent = Curved surface area of the cylinder + Curved surface area of the cone

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)= 18 × π (27.1)

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)= 18 × π (27.1)Cost of canvas = 18 × π (27.1) × ₹ 3.5

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)= 18 × π (27.1)Cost of canvas = 18 × π (27.1) × ₹ 3.5 = 18 × 22/7 × 27.1 × 3.5

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)= 18 × π (27.1)Cost of canvas = 18 × π (27.1) × ₹ 3.5 = 18 × 22/7 × 27.1 × 3.5 = ₹ 5365.80

= 2π × 18 × 4.4 m² + π × 18 × 18.3 m²[From eq. 1 & 2]= 18π(2 × 4.4 + 18.3)= 18π (8.8 + 18.3)= 18π(27.1)= 18 × π (27.1)Cost of canvas = 18 × π (27.1) × ₹ 3.5 = 18 × 22/7 × 27.1 × 3.5 = ₹ 5365.80Hence, the cost of canvas is ₹ 5365.80.

#Hope you have satisfied with this answer.