Question

the total surface area of a hollow metal cylinder, opened at both ends,of external radius of 10cm and height 12 cm is 512\pi cm(square) ......taking x as the inner radius ,write down an equation in x and solve it to find the thickness of the metal in the cylinder......

Answer 1

Given TSA of cylinder with both open ends = 338π sq cm
External radius, R = 8 cm and height, h = 10 cm
That is 2πrh +  2πRh = 338π
⇒ (2π × r × 10) + (2π × 8 × 10) = 338π
⇒ (20πr) = 338π – 160π = 158π  
∴ r = 7.8 cm
Thickness = R – r = 8 – 7.8 = 0.2 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