Question

The total surface area of a hollow cylinder open at both ends of external radius 8cm and height 10cm is 338pie cm^2. Taking r
to be inner radius, find thickness of the metal in the cylinder.

Answer 1

Let the outer radius be R 
and inner radius be r
total surface area = inner surface area + outer surface area +2(area of                                            thickness)
                             = 2πrh + 2πRh + 2π(R² -r²)
⇒338π = 20πr + 160π + 128π - 2πr²
⇒2πr² - 20πr + 50π =0
⇒r² -10r + 25 =0
⇒(r -5)²=0
⇒r =5 cm
so thickness of the metal in cylinder = R- r = 8- 5 = 3 cm

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