The Total Surface Area of a hollow Cylinder which is open from both sides is 4620 cm^2 and area of base ring is 115.5 cm^2 and Height is 7 cm. Find the thickness of Cylinder. [ Answer :- 7/19 cm].
Please post your answer with proper explanation.
Answers
∴ Area of the base ring = π(R² - r²)
⇒ 115.5 = π(R² - r²)
⇒ (R² - r²) = 115.5 ÷ 22/7
(R² - r²) = (115.5*7)/22
(R + r) (R - r) = (1155*7)/220
(R + r) (R - r) = 147/4 sq cm ............(1)
Total surface area of the cylinder = 4620 sq cm
Now, total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)
= 2πRh + 2πrh + 2π(R² - r²)
⇒ 2πRh + 2πrh + 2π(R² - r²) = 4620
⇒ 2πh (R + r) + (2 × 115.5) = 4620
⇒ 2πh (R + r) + 231 = 4620
⇒ 2πh (R + r) = 4620 - 231
⇒ 2 × 22/7 × 7 × (R + r) = 4389
⇒ (R + r) = 4389/44
⇒ (R + r) = 399/4
Substituting the value of (R + r) = 399/4 in equation (1), we get.
(R + r)(R - r) = 147/4
399/4 (R - r) = 147/4
R - r = 147/4 ÷ 399/4
R - r = (147/4) × (4/399)
R - r = 147/399
R - r = 7/19 cm
So, the thickness of the cylinder is 7/19.
Hope it helped you, Do mark me as the Brainliest....
Answer:
7/19 cm
Step-by-step explanation:
Let the radii of outer and inner surfaces be R and r.
(i) TSA of hollow cylinder:
TSA = Outer curved surface area + Inner C.S.A + 2(Area of circular base)
⇒ 4620 = 2πRh + 2πrh + 2π(R² - r²) [∴ Area of base ring = π(R² - r²)]
⇒ 4620 = 2πh(R + r) + 2 * 115.5
⇒ 4620 = 2πh(R + r) + 231 cm²
⇒ 4620 - 231 = 2πh(R + r)
⇒ 4389 = 2 * 22/7 * 7 * (R + r)
⇒ 4389 = 44 * (R + r)
⇒ 4389/44 = (R + r)
⇒ 399/4 = (R + r)
(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)
Substitute (i) in (ii), we get
⇒ (R + r)(R - r) = 147/4
⇒ (399/4) (R - r) = 147/4
⇒ (R - r) = 147/399
⇒ (R - r) = 7/19 cm.
Therefore, Thickness of the cylinder = 7/19 cm.
Hope it helps!