Answer it fast!!
The total surface area of a hollow cylinder which is open from both the sides is 4620sq cm, area of base ring is 115.5sq cm and height 7cm. Find the thickness of the cylinder.
Correct answer will be mark as brainliest.
Answers
Given Area of base ring pi(R^2 - r^2) = 115.5cm^2. ----- (1)
Given TSA of a hollow cylinder = 4620cm^2.
NOTE: We know that TSA of a hollow cylinder = Outer curved surface area + Inner curved surface area + 2(Area of circular base). --- (2)
We know that Outer curved surface area = 2piRh --- (3)
We know that inner curved surface area = 2pirh --- (4)
We know that Area of circular base = pi(R^2 - r^2). --- (5)
Substitute (3),(4),(5) in (2), we get
4620 = 2piRh + 2pirh + 2pi(R^2 - r^2).
4620 = 2pih(R + r) + 2 * 115.5
4620 = 2 * 22/7 * 7(R + r) + 231
4620 - 231 = 2 * 22 * (R + r)
4389 = 44 * (R + r)
4389/44 = (R + r)
99.75 = (R + r) ---- (1)
Now,
Area of the base ring = 115.5cm^2.
pi(R^2 - r^2) = 115.5
pi(R + r)(R - r) = 115.5
22/7 * 99.75(R - r) = 115.5
22 * 14.25(R - r) = 115.5
313.5 (R - r) = 115.5
R - r = 115.5/313.5
R - r = 0.368
R - r = ~0.37.
Therefore the thickness of the cylinder = 0.37cm.
Hope this helps!
Answer:
Let the radii of outer and inner surfaces be R and r.
(I) TSA of hollow cylinder :
TSA = Outer CSA + Inner CSA + 2(Area of circular base)
➳ 4620 = 2πRh + 2πrh + 2π(R² - r²)
➳ 4620 = 2πh(R + r) + 2 × 115.5
➳ 4620 = 2πh(R + r) + 231
➳ 4620 - 231 = 2πh(R + r)
➳ 4389 = 2πh(R + r)
➳ 4389 = 2 × 22/7 × 7 × (R + r)
➳ 4389 = 44 × (R + r)
➳ 4389/44 = (R + r)
➳ 399/4 = (R + r) ...........[Equation (i)]
_____________________
(II) Area of base ring :
Area of base ring = π(R² - r²)
➳ 115.5 = 22/7(R² - r²)
➳ 115.5 × 7 = 22(R² - r²)
➳ 808.5/22 = R² - r²
➳ 8085/22 = R² - r²
➳ 147/4 = (R + r) (R - r).......[Equation (ii)]
____________________
Now, Substituting equation (I) in equation (II) we get,
➳ 147/4 = (R + r) (R - r)
➳ 147/4 = (399/4) (R - r)
➳ (R - r) = 399/147
➳ (R - r) = 7/19
➳ (R - r) = 0.36842 cm
Therefore, the thickness of the cylinder is 0.36842 cm.