The total surface area of hollow cylinder, which is open from both sides, is 3575 cm 2; area of the base ring is 357.5 cm 2 and height is 14 cm. Find the thickness of the cylinder.
Answers
Given parameters
Total surface area of cylinder = 3575 cm2
Area of the base ring = 357.5 cm2
Height = 14 cm
Let us consider
The inner radius = r
Outer radius = R
Area of the base ring
π(R2−r2) = 357.5 cm2
R2−r2 = 357.5/π
Value of π = 22/7 = 3.143
R2−r2 = 357.5/3.143
R2−r2 = 113.74 sq.cm
(R + r)(R – r) = 113.74 sq.cm…………………….(1)
Total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)
3575 = 2πRh + 2πrh + 2π (R2−r2)
3575 = 2πRh+2πrh+ 2 × 357.5
3575 = 2πh(R+r) + 715
2πh(R+r) = 3575 – 715
2πh(R+r) = 2860
(R + r) = 2860/(2 × 3.143 × 14)
(R + r) = 2860/88.004
(R + r) = 32.498…………………………….(2)
Substitute the value of equation (2) in equation (1) we get
(32.498) × (R – r) = 113.74
(R – r) = 3.49 cm
The thickness of the cylinder is 3.49 cm.
Given parameters
Total surface area of cylinder = 3575 cm2
Area of the base ring = 357.5 cm2
Height = 14 cm
Let us consider
The inner radius = r
Outer radius = R
Area of the base ring
π(R2−r2) = 357.5 cm2
R2−r2 = 357.5/π
Value of π = 22/7 = 3.143
R2−r2 = 357.5/3.143
R2−r2 = 113.74 sq.cm
(R + r)(R – r) = 113.74 sq.cm…………………….(1)
Total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)
3575 = 2πRh + 2πrh + 2π (R2−r2)
3575 = 2πRh+2πrh+ 2 × 357.5
3575 = 2πh(R+r) + 715
2πh(R+r) = 3575 – 715
2πh(R+r) = 2860
(R + r) = 2860/(2 × 3.143 × 14)
(R + r) = 2860/88.004
(R + r) = 32.498…………………………….(2)
Substitute the value of equation (2) in equation (1) we get
(32.498) × (R – r) = 113.74
(R – r) = 3.49 cm
The thickness of the cylinder is 3.49 cm.