Total Surface Area of Cylinder

Concept Explanation

Total Surface Area of Cylinder

A Cylinder of radius r and height h. The total surface area or TSA includes area of circular top and base as well as the curved surface area or CSA.

Length of rectangle= Circumference of base = $large 2pi r$

Breadth of rectangle = h

TSA = Area of base + Area of top + CSA of cylinder

$large =pi r^2 + pi r^2 + 2pi r h$

$=2pi r^2 + 2pi r h$

$=2pi r(r +h)$

Illustration: A solid cylinder has total surface area of 462 $cm^2.$ Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.

Solution: Given: Curved surface area $=frac {1}{3} times Total;Surface;Area$

$Rightarrow$ $2 pi rh = frac {1}{3} times (2 pi rh + 2 pi r^2)$

$2 pi rh = frac {1}{3} times(2pi r) (r+h)$

$h = frac {1}{3} times (r+h)$

$3h =r+h$

$3h -h=r$

$2h=r$

$r=2h$ ...................(i)

Now, given that total surface area = 462

$Rightarrow$ Curved surface area $=frac {1}{3} times 462$

$Rightarrow$ $2pi rh=154$

Using equation (i0, we have

$2pi (2h)h=154$

$4pi h^{2}=154$

$h^{2}=frac{154}{4pi }$

$h^{2}=frac{154}{4}times frac{7}{22}$

$h^{2}=frac{7}{4}times 7$

$h^{2}=frac{49}{4}$

$h=sqrt{frac{49}{4}}$

$h=pm frac{7}{2}$

As $hneq - frac{7}{2}$

So, $h = frac{7}{2}$

Hence height of cylinder = 3.5 cm

Using equation (i), we have

and radius of cylinder = r = 2h

$= 2times frac{7}{2}$

= 7 cm

Sample Questions

(More Questions for each concept available in Login)

Question : 1

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its total curved surface area.
A. 2038.08	B. 3033.04	C. 3506.09	D. 48001.01
Right Option : A
View Explanation

Question : 2

If the curved surface area and the radius of a base of a cylinder are 8.8 cm sq and 0.7 cm respectively, then the height of cylinder is _____________
A. 2 cm	B. 10 cm	C. 0.5 cm	D. 1 cm
Right Option : D
View Explanation

Question : 3

The students of a school were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with the base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The school was to supply the competitors with cardboard. If there were 35 competitors, how much card board was required to be bought for the competition? (in sq cm)
A. 7928	B. 7933	C. 7920	D. 7925
Right Option : C
View Explanation

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Language - English

Chapters

Polynomials II

Direct and Inverse Variation

Linear Equations in One Variable

Introduction to Trigonometry

Introduction to Euclids Geometry

Herons Formula Complete

Polynomial I

Surface Area and Volume II

Surface Area and Volume I

Coordinate Geometry

Constructions

Areas Of Parallelograms And Triangles II

Areas Of Parallelograms And Triangles I

Linear Equations in Two Variables

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