# 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 - Mathematics

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

#### Solution

Let the inner radii of hollow cylinder rcm
Outer radii of hollow cylinder Rcm
Then,

2πrh + 2πRh +2πR^2  - 2πr^2 = 4620 → (1)

πR^2 =πr^2 =115.5→(2)

⇒ 2πh(R+r)+2(πR^2-πr^2)=4620 and πR^2-m^2 = 115.5

⇒2πh(R+r)+231=4620 and π(R^2-r^2)=115.5

⇒2π× 7(r+R)=4389 and π(R^2-r^2)=115.5

⇒π (R+r)=313.5 and π (R+r) (R-r)=115.5

⇒(π(R+r)(R-r))/(π(R+r))=115.5/313.5

⇒ R-r=7/19cm.

Concept: Surface Area of Cylinder
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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 19 Surface Areas and Volume of a Circular Cylinder
Exercise 19.1 | Q 5 | Page 8

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