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The total surface area of a hollow cylinder which is open from both sides is 4620 sq.cm, area of base ring is 115.5 sq.cm and height 7 cm. Find the thickness of the cylinder. |
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Answer» 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. Step-by-step explanation: hope you like it !!! |
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