plz solve this friend
the ANS for h is 16
Answers
Answer:
2745.6
Step-by-step explanation:
Given, lower end of the frustum (r₁) = 8 cm.
Given, upper end of the frustum (r₂) = 20 cm.
Let the frustum be 'h' cm.
(i)
We know that volume of frustum(v₁) = (1/3)π(r₁² + r₂² + r₁r₂)h
⇒ (1/3)(22/7)[8² + 20² + 8(20)]h
⇒ (22/21)[64 + 400 + 160]h
⇒ (22/21)[624]
⇒ ~653.4h
(ii)
Given volume of milk container(v₂)= 10459 (3/7) cm³
= 73216/7
= 10459.4
On solving both volumes, we get
⇒ 653.4h = 10459.4
⇒ h = 10459.4/653.4
⇒ h = 16 cm
(iii)
We know that Slant height(l) = √(r₁ - r₂)² + h²
⇒ √(20 - 8)² + 16²
⇒ √12² + 16²
⇒ 20 cm.
We know that Total surface area of the container:
⇒ π(r₁ + r₂) * l + πr₁²
⇒ (22/7) * (8 + 20) * 20 + (22/7) * (8)²
⇒ (22/7)[28 * 20 + 64]
⇒ (22/7)[624]
⇒ 1961.14 cm².
Given that cost of 1 cm² metal sheet = 1.40 per cm².
So, cost of 1961.14 cm² required sheet:
⇒ 1.40 * 1961.14
⇒ 2745.6 .
Therefore, cost of required sheet = 2745.6.
Hope it helps!
Answer:
2745.6
Step-by-step explanation:
Given, lower end of the frustum (r₁) = 8 cm.
Given, upper end of the frustum (r₂) = 20 cm.
Let the frustum be 'h' cm.
(i)
We know that volume of frustum(v₁) = (1/3)π(r₁² + r₂² + r₁r₂)h
⇒ (1/3)(22/7)[8² + 20² + 8(20)]h
⇒ (22/21)[64 + 400 + 160]h
⇒ (22/21)[624]
⇒ ~653.4h
(ii)
Given volume of milk container(v₂)= 10459 (3/7) cm³
= 73216/7
= 10459.4
On solving both volumes, we get
⇒ 653.4h = 10459.4
⇒ h = 10459.4/653.4
⇒ h = 16 cm
(iii)
We know that Slant height(l) = √(r₁ - r₂)² + h²
⇒ √(20 - 8)² + 16²
⇒ √12² + 16²
⇒ 20 cm.
We know that Total surface area of the container:
⇒ π(r₁ + r₂) * l + πr₁²
⇒ (22/7) * (8 + 20) * 20 + (22/7) * (8)²
⇒ (22/7)[28 * 20 + 64]
⇒ (22/7)[624]
⇒ 1961.14 cm².
Given that cost of 1 cm² metal sheet = 1.40 per cm².
So, cost of 1961.14 cm² required sheet:
⇒ 1.40 * 1961.14
⇒ 2745.6 .
Therefore, cost of required sheet = 2745.6.