Question

The total surface area of ahollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.

Answer 1

The thickness of the hollow cylinder is 0.358 cm.

• Given,

Total surface area of the hollow cylinder (T.S.A) = 4620 sq. cm

Area of the base ring = 115.5 sq. cm

Height of the cylinder (h) = 7 cm

• Total surface area of a hollow cylinder  (T.S.A.) = Outer curved surface area of the cylinder + Inner curved surface area of the cylinder + Area of the base ring

• Let the radius of the outer cylinder be R cm, and that of the inner cylinder be r cm.

• Now,

Outer curved surface area of the cylinder = 2πRh

Inner curved surface area of the cylinder = 2πrh

Area of the base ring = Area of outer cylinder - Area of inner cylinder

=> Area of base ring = πR² - πr²

=> 115.5 sq. cm = π (R² - r²)

=> 115.5 sq. cm = π (R - r) (R + r)

[ Applying a² - b² = (a - b) (a + b) ]

• Therefore,

T.S.A. = 2πRh + 2πrh + π (R - r) (R + r)

Or, 4620 sq. cm = 2πh (R + r) + 115.5 sq. cm

Or, 2πh (R + r) = 4620 sq. cm - 115.5 sq. cm

Or, 2 × (22 / 7) × 7 cm × (R + r) = 4504.5 sq. cm

Or, 2 × 22 cm × (R + r) = 4504.5 sq. cm

Or, 44 cm × (R + r) = 4504.5 sq. cm

Or, R + r = 4504.5 sq. cm / 44 cm

Or, R + r = 102.375 cm

• Thickness of a hollow cylinder = Outer Radius - Inner Radius

=> Thickness of the hollow cylinder = R - r

• Putting the value of (R + r) in eq. (i), we get,

115.5 sq. cm = π (R - r) × 102.375 cm

Or, π (R - r) = 115.5 sq. cm / 102.375 cm

Or, (22 / 7) (R - r) = 1.128 cm

Or, R - r = (1.128 cm × 7) / 22

Or, R - r = 7.817 cm / 22

Or, R - r = 0.358 cm

∴   Thickness of the hollow cylinder = 0.358 cm.

Answer 2

Answer:

Step-by-step explanation:• Given,

Total surface area of the hollow cylinder (T.S.A) = 4620 sq. cm

Area of the base ring = 115.5 sq. cm

Height of the cylinder (h) = 7 cm

• Total surface area of a hollow cylinder  (T.S.A.) = Outer curved surface area of the cylinder + Inner curved surface area of the cylinder + Area of the base ring

• Let the radius of the outer cylinder be R cm, and that of the inner cylinder be r cm.

• Now,

Outer curved surface area of the cylinder = 2πRh

Inner curved surface area of the cylinder = 2πrh

Area of the base ring = Area of outer cylinder - Area of inner cylinder

=> Area of base ring = πR² - πr²

=> 115.5 sq. cm = π (R² - r²)

=> 115.5 sq. cm = π (R - r) (R + r)

[ Applying a² - b² = (a - b) (a + b) ]

• Therefore,

T.S.A. = 2πRh + 2πrh + π (R - r) (R + r)

Or, 4620 sq. cm = 2πh (R + r) + 115.5 sq. cm

Or, 2πh (R + r) = 4620 sq. cm - 115.5 sq. cm

Or, 2 × (22 / 7) × 7 cm × (R + r) = 4504.5 sq. cm

Or, 2 × 22 cm × (R + r) = 4504.5 sq. cm

Or, 44 cm × (R + r) = 4504.5 sq. cm

Or, R + r = 4504.5 sq. cm / 44 cm

Or, R + r = 102.375 cm

• Thickness of a hollow cylinder = Outer Radius - Inner Radius

=> Thickness of the hollow cylinder = R - r

• Putting the value of (R + r) in eq. (i), we get,

115.5 sq. cm = π (R - r) × 102.375 cm

Or, π (R - r) = 115.5 sq. cm / 102.375 cm

Or, (22 / 7) (R - r) = 1.128 cm

Or, R - r = (1.128 cm × 7) / 22

Or, R - r = 7.817 cm / 22

Or, R - r = 0.358 cm

∴   Thickness of the hollow cylinder = 0.358 cm