Question

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

Answer 1

Total surface area of external cylinder=2Πr(r+h)
2×(22÷7)×8(8+10)
(352÷7)×18
(6336÷7)cmsq. equation(1)

Total surface area of internal cylinder=2Πr(r+h)
338=2(22÷7)×10rsq.
(338×440÷7) cmsq. equation(2)

adding equation (1)and(2)
(6336÷7)+(338×440÷7)
22150.8571cmsq.
22150.86(approximately)

Answer 2

Answer:

SA of the hollow cylinder = 338π cm²

↠ 2π(8)10 + 2π(r)10 + 2[π(8)² – π(r)²] = 338π

↠ 160 + 20r + 2(64 – r²) = 338

↠ – 2r² + 20r – 50 = 0

↠ r² – 10r + 25 = 0

↠ (r – 5)² = 0

↠ r = 5

∴ Thickness of the metal = 8 – r

= 8 – 5 = 3 cm