Question

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Q3

Answer 1

Solution of Question No. 3 :-

Diameter of the upper end = 20 cm = 2 dm

Radius r1 = 2/2 = 1 dm

Diameter of the lower end = 10 cm or 1 dm

Radius r2 = 1/2 = 0.5 dm

Volume of the tub (frustum of a cone) = 1/3π(r1² + r2² + r1*r2)

⇒ 1/3*22/7(1 + 0.25 + 1*0.5)

⇒ 1/3*22/7*1.75 

⇒ 38.5/21

⇒ 1.83 dm³

So, volume of the tub is 1.83 dm³

Slant height 'l' = √H² + (r1 - r2)²

⇒ √(12)² + (1 - 0.5)²

⇒ √144 + (0.5)²

⇒ √ 144 + 0.25

⇒ √144.25

⇒ 12.01 dm

Slant height is 12.01 dm.

Area of the tin sheet used for making the tub = CSA of the tub + Area of the base

= π(r1 + r2)l + πr2²

⇒ 22/7*(1 + 0.5)*12 + 22/7*0.5*0.5

⇒ (22/7*1.5*12) +  (22/7*0.25)

⇒ 396/7 + 5.5/7

⇒ 56.57 + 0.785

⇒ 57.355 dm²

Cost of tin sheet = 1.20*57.355

= Rs. 68.826

Answer.

Answer 2

Answer:

Diameter of the upper end = 20 cm = 2 dm

Radius r1 = 2/2 = 1 dm

Diameter of the lower end = 10 cm or 1 dm

Radius r2 = 1/2 = 0.5 dm

Volume of the tub (frustum of a cone) = 1/3π(r1² + r2² + r1*r2)

⇒ 1/3*22/7(1 + 0.25 + 1*0.5)

⇒ 1/3*22/7*1.75  

⇒ 38.5/21

⇒ 1.83 dm³

So, volume of the tub is 1.83 dm³

Slant height 'l' = √H² + (r1 - r2)²

⇒ √(12)² + (1 - 0.5)²

⇒ √144 + (0.5)²

⇒ √ 144 + 0.25

⇒ √144.25

⇒ 12.01 dm

Slant height is 12.01 dm.

Area of the tin sheet used for making the tub = CSA of the tub + Area of the base

= π(r1 + r2)l + πr2²

⇒ 22/7*(1 + 0.5)*12 + 22/7*0.5*0.5

⇒ (22/7*1.5*12) +  (22/7*0.25)

⇒ 396/7 + 5.5/7

⇒ 56.57 + 0.785

⇒ 57.355 dm²

Cost of tin sheet = 1.20*57.355

= Rs. 68.826 ⇒ Answer