Math, asked by KanishkGangwar, 4 months ago

C and C2 are two cylinders having equal total surface areas. The radius of each
cylinder is equal to the height of the other. The sum of the volumes of both the
cylinders is 250pi cm^2. Find the sum of their curved surface areas.

PLEASE ANSWER IN CORRECT WORDS.​

Answers

Answered by BrainlyAmazinGirl
9

Let A be the surface area of two cylinders (C1 and C2) having heights and radii h₁, r₁ and h₂, r₂ respectively.

The radius of each cylinder is equal to the height of the other.

r₁ = h₂ and r₂=h₁

Volume of C1 = πr₁²h₁ = π h₂²h₁

Volume of C2 = πr₂²h₂ = π h₁²h₂

Sum of volumes = π h₂²h₁+ π h₁²h₂

250π = π(h₂²h₁+ h₁²h₂)

250 = h₁h₂(h₂+h₁) —-(1)

Curved surface area of C1 = 2πr₁h₁ = 2πh₁h₂

Total surface area of C1 =2πh₁h₂+ 2πr₁²= 2πh₁h₂+ 2πh₂²

Curved surface area of C2 = 2πr₂h₂ = 2πh₁h₂

Total surface area of C2 =2πh₁h₂+ 2πr₂²= 2πh₁h₂+ 2πh₁²

Both total surface areas are equal.

2πh₁h₂+ 2πh₂² =2πh₁h₂+ 2πh₁²

h₂² =h₁²

h₁=h₂

From (1), 250 = h₁h₁(2h₁) = 2h₁³

h₁³ =125

h₁=h₂=5

Sum of curved surface area = 2π(5)(5)+2π(5)(5) = 100π

Ans: 100π:

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Answered by 21raghavb9
1

Step-by-step explanation:

Let A be the surface area of two cylinders (C1 and C2) having heights and radii h₁, r₁ and h₂, r₂ respectively.

The radius of each cylinder is equal to the height of the other.

r₁ = h₂ and r₂=h₁

Volume of C1 = πr₁²h₁ = π h₂²h₁

Volume of C2 = πr₂²h₂ = π h₁²h₂

Sum of volumes = π h₂²h₁+ π h₁²h₂

250π = π(h₂²h₁+ h₁²h₂)

250 = h₁h₂(h₂+h₁) —-(1)

Curved surface area of C1 = 2πr₁h₁ = 2πh₁h₂

Total surface area of C1 =2πh₁h₂+ 2πr₁²= 2πh₁h₂+ 2πh₂²

Curved surface area of C2 = 2πr₂h₂ = 2πh₁h₂

Total surface area of C2 =2πh₁h₂+ 2πr₂²= 2πh₁h₂+ 2πh₁²

Both total surface areas are equal.

2πh₁h₂+ 2πh₂² =2πh₁h₂+ 2πh₁²

h₂² =h₁²

h₁=h₂

From (1), 250 = h₁h₁(2h₁) = 2h₁³

h₁³ =125

h₁=h₂=5

Sum of curved surface area = 2π(5)(5)+2π(5)(5) = 100π

Ans: 100π:

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