Question

The total surface area of a hollow metal cylinder, open at both ends, of external radius 8 cm and height 10 cm is 338π cm². Taking r cm to be inner radius, write down an equation in r and use it to find the thickness of metal in the cylinder.​

Answer 1

☆ External radius of hollow cylinder :

• Radius = 8 cm
• Height = 10 cm

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❆ Total Surface Area = [2π Rh + 2πrh + 2π (R² - r²)] cm²

➝ 2π [h(R + r) + (R² - r²)] cm²

➝ 2π (R + r) + (h + R - r) cm²

➝ 2π (8 + r) + (10 + 8 - r) cm²

➝ 2π (8 + r) + (18 - r) cm²

★ According to given question :

• 2π(8 + r) (18 - r) = 338π

➸ (8 + r) (18 - r) = 169

➸ 144 + 10r - r² = 169

➸ r² - 10r + 25 = 0

➸ (r - 5)² = 0

➸ r = 5

∴ Internal radius = 5 cm

∴ Thickness of metal in the cylinder = (8 - 5) cm = 3cm.

▣ Therefore,

• Hence, the thickness of the metal in the cylinder is 3 cm.

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Answer 2

Given :

• External radius, R = 8 cm
• Height, h = 10 cm
• Total surface area = 338π cm²

To Find :

• Thickness of metal = ?

Solution :

We know that :

[Total surface area = 2πRh + 2πrh + 2π(R² - r²)]

We have total surface area = 338π

Hence,

2πRh + 2πrh + 2π(R² - r²) = 338π

We have :

• R = 8 cm
• h = 10 cm

So, by substituting values :

=> 2π × 8 × 10 + 2πr × 10 + 2π((8)² - r²) = 338π

=> 2π(80 + 10r + (64 - r²)) = 2π(169)

=> 80 + 10r + 64 - r² = 169

=> - r² + 10r + 144 = 169

=> - r² + 10r + 144 - 169 = 0

=> - r² + 10r - 25 = 0

=> - (r² - 10r + 25) = 0

=> r² - 10r + 25 = 0

=> r² - 5r - 5r + 25 = 0

=> r(r - 5) - 5(r - 5) = 0

=> (r - 5) (r - 5) = 0

=> (r - 5)² = 0

=> r - 5 = 0

=> r = 5

Hence, inner radius, r = 5 cm.

Now,

Thickness of metal = outer radius - inner radius

=> Thickness = 8 cm - 5 cm = 3 cm

Hence, Thickness of given cylinder is 3 cm.