Question

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

Answer 1

An equation in r is r^2 - 10r +25 = 0

The thickness of the metal in the cylinder is 3 cm

Let the external radius of the cylinder = R cm = 8 cm

Let the internal radius of the cylinder = r cm

Let the height of the cylinder = h cm = 10 cm

The thickness of the cylinder = (R-r) cm

TSA of the hollow cylinder = 338π cm^2

⇒ 2πRh + 2πrh + 2(πR^2 - πr^2) = 338π

⇒ 2π (Rh + rh + R^2 - r^2) = 338π

⇒ 8 × 10 + r × 10 + 8^2 - r^2 = 169

⇒ 80 + 10r + 64 - r^2 = 169

⇒ 10r - r^2 = 169 - 80 - 64

⇒ 10r - r^2 = 25

⇒ r^2 - 10r +25 = 0

⇒ (r-5)^2=0

⇒ r - 5 = 0

∴ r = 5.

The thickness of the cylinder = (R-r) = 8 - 5 = 3 cm