The total surface area of a hollow metal cylinder, open at both ends of external re
8 cm and height 10 cm is 338 p cm. Taking r to be inner radius, obtain an equation
and use it to obtain the thickness of the metal in the cylinder.
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Answer:
external radius= 8cm
height= 10cm
inner radius= r
total surface area
2πrh+2πrh+2π (R^-r^)
2πh(R+r) + 2π(R^-r^)
2π * 10 (8+r) + 2π (8^-r^)
2π[80+10r+64-r^] cm^
therefore = 2π[80+10r+64-r^] = 338π
288+20r-2r^=338
2r^-20r+50=0
r^-10r+25=0
(r-5)^=0
r=5
therefore,the thickness of the cylinder = R-r = 8-5=3cm
it is help u
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