A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides as axis.

**Surface Area**

Surface area is the measurement of all exposed area of a solid
object.

Observe the given figures in Fig.7.2 and identify the shape.

These objects resemble the shape of a cylinder.

A right circular cylinder is a solid generated by the revolution
of a rectangle about one of its sides as axis.

If the axis is perpendicular to the radius then the cylinder is
called a right
circular cylinder. In the Fig.7.3, AB = *h* represent the height and* _{
}AD *=

A solid cylinder is an object bounded by two circular plane
surfaces and a curved surface. The area between the two circular bases is
called its ‘Lateral Surface Area’ (L.S.A.) or ‘Curved Surface Area’ (C.S.A.).

**Formation of a Right Circular Cylinder – Demonstration**

(i) Take a rectangle sheet of a paper of length *l* and
breadth *b*.

(ii) Revolve the paper about one of its sides, say *b* to
complete a full rotation (without overlapping).

(iii) The shape thus formed will be a right
circular cylinder whose circumference of the base is *l* and the
height is *b.*

Curved surface area (C.S.A.) of a right circular cylinder

= Area of the corresponding
rectangle

* = l *×*b*

= 2*πr* ×*h*
(since, *l* is the circumference of the base, *b* is the height)[see
Fig. 7.5]

= 2*πrh*

C.S.A. of a right circular cylinder = 2πrh sq. units.

Total surface area refers to the sum of areas of the curved
surface area and the two circular regions at the top and bottom.

That is, total surface area (T.S.A.) of right circular cylinder

= C.S.A + Area of top circular region +Area of
bottom circular region.

= 2*πrh* + *πr* ^{2}
+ *πr* ^{2} (Refer Fig.7.4)

= 2*πrh* + 2*πr* ^{2}

= 2*πr* (*h* + *r* )

T.S.A. of a right circular cylinder = 2π*r* (*h* + *r*)
sq. units

We always consider** π **=** **22/7** **, unless
otherwise stated.

The term ‘surface area’ refers to ‘total surface area’.

A cylindrical drum has a height of** **20** **cm and base radius of** **14** **cm. Find its** **curved surface area and
the total surface area.

Given that, height of the cylinder* **h*** **= 20

Now, C.S.A. of the cylinder = 2*πrh* sq. units

C.S.A. of the cylinder = 2 × (22/7) × 14 ×20 = 2 ×22 × 2 ×20
= 1760 cm^{2}

T.S.A. of the cylinder = 2*πr (h*+* r) *sq. units

= 2 × (22/7) × 14 ×(20 +
14) = 2 × 22/7 × 14 ×34

= 2992 cm^{2}

Therefore, C.S.A. = 1760 cm^{2} and T.S.A. = 2992 cm^{2}

The curved surface area of a right circular cylinder of height** **14** **cm is** **88 cm^{2} . Find
the diameter of the cylinder.

Given that, C.S.A. of the cylinder** **=88

2πrh =88

2 × (22/7) × *r* ×14 =88 (given *h*=14 cm)

2r = (88×7) / (22×14) = 2

Therefore, diameter = 2 cm

**Example 7.3**

A garden roller whose length is** **3** **m long and whose diameter is** **2.8** **m is** **rolled to level a
garden. How much area will it cover in 8 revolutions?

*Solution*

Given that, diameter* **d*** **= 2.8

radius *r* = 1.4 m

Area covered in one revolution = curved surface area of the
cylinder

= 2πrh sq. units

= 2 × (22/7) × 1.4 ×3 = 26.4

Area covered in 1 revolution = 26.4 m^{2}

Area covered in 8 revolutions = 8 ×26.4 = 211.2

Therefore, area covered is 211.2 m^{2}

**Thinking Corner**

1. When ‘*h*’ coins each of radius ‘*r*’ units and
thickness 1 unit is stacked one upon the other, what would be the solid object
you get? Also find its C.S.A.

2. When the radius of a cylinder is double its height, find the
relation between its C.S.A. and base area.

3. Two circular cylinders are formed by rolling two rectangular
aluminum sheets each of dimensions 12 m length and 5 m breadth, one by rolling
along its length and the other along its width. Find the ratio of their curved
surface areas.

Tags : Definition, Formula, Solved Example Problems | Mensuration | Mathematics , 10th Mathematics : UNIT 7 : Mensuration

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10th Mathematics : UNIT 7 : Mensuration : Surface Area: Right Circular Cylinder | Definition, Formula, Solved Example Problems | Mensuration | Mathematics

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