228

Mensuration is a branch of mathematics which deals with the measurements oflengths of lines, areas of surfaces and volumes of solids.

Mensuration may be divided into two parts.

1. Plane mensuration

2. Solid mensuration

Plane mensuration deals with perimeter, length of sides and areas of two dimensionalfigures and shapes.

Solid mensuration deals with areas and volumes of solid objects.

After learning this chapter you will be able to* Recognise a cylinder, a cone and a sphere.

* Understand the properties of cylinder, cone and sphere.

* Distinguish between the structure of cylinder and cone.

* Derive the formula to find the surface area and volume of cylinder, cone and sphere.

* Solve simple problems pertaining to the surface area and volume of cylinder, coneand sphere.

CYLINDER

Observe the following figures :

Road roller Circular based storage Tank Cylinder

Wheels of a road roller, a circular based storage tank etc will suggest you, theconcept of a right circular cylinder.

9 MENSURATION

229

1. The right circular cylinder

A right circular cylinder is a soliddescribed by revolution of a rectangleabout one of its sides which remainsfixed.

AP = Radius of the circular plane

AB = Axis of the cylinder

PQ = Height of the cylinder

Features of a right circular cylinder1) A right circular cylinder has two plane surfaces, circular in shape.

2) The curved surface joining the plane surfaces is the lateral surface of the cylinder.

3) The two circular planes are parallel to each other and also congruent.

4) The line joining the centers of the circular planes is the axis of the cylinder.

5) All the points on the lateral surface of the right circular cylinder are equidistant fromthe axis.

6) Radius of circular plane is the radius of the cylinder.

Two types of cylinders :1. Hollow cylinder and

2. Solid cylinder

A hollow cylinder is formed by the lateral surface only.

Example : A pipe

230

A solid cylinder is the region bounded by two circular plane surfaces and alsothe lateral surface.

Example : A garden roller

2. Surface area of a right circular cylinderA. Lateral Surface area :

Activity :1. Take a strip of paper having width equal to the height of the cylinder.

2. Wrap the strip around the lateral surface of the cylinder and cut the overlappingstrip along the vertical line. (say PQ)

3. You will get a rectangular paper cutting which exactly covers the lateral surface.

4. Area of the rectangle is equal to the area of the curved surface of the cylinder.

Expression for the lateral surface area :

(i) Length of the rectangle is equal to the circumference, l = 2 rπ(ii) Breadth of the rectangle b is the height of the cylinder = h

Area of the rectangle A = l x b

Lateral surface area of the Cylinder A = 2 rπ h

A = 2 rπ h sq. units

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Observe :

Surface area of a thin hollow cylinder having circumference P andheight is ‘h’ = Ph or = 2 rπ h (∴ P = 2 rπ )

B. Total surface area of a cylinder :

The total surface area Area of two Lateral surfaceof the cylinder = circular bases + area of cylinder

= rh2rr 22 π+π+π

= rh2r2 2 π+π= r2π (r + h) sq. units

Lateral surface area of a cylinder = rh�2 sq. units.

Total surface area of a cylinder = r�2 (r+h) sq. units.

Remember :

Area is always expressed in square units

Worked examples :Example 1 : Find the lateral surface area of a cylinder whose circumference is

44 cm and height 10 cm.

Given : circumference = r2π = 44 cms

and height = h = 10 cms

Solution : Lateral surface area of cylinder = rh2π= 44 x 10

= 440 sq. cm.

2rπ

rh2π

2rπ

[ [ [ [

232

Example 2 : Find the total surface area of the cylinder, given that the diameter is10 cm and height is 12.5 cm.

Given : Diameter = d = 10 cms

∴ r = 2

d = 5 cms

and height = h = 12.5 cms.

Solution : Total surface area of a cylinder = r2π (r+h)

= x5x7

22x2 (5 + 12.5)

= x5x7

22x2 17.5

= 550 sq.cm.

Example 3 : The lateral surface area of a thin circular bottomed tin is 1760 sq. cmand radius is 10 cm. What is the height of the tin?

Given : The lateral surface area of a cylinder = rh2π = 1760 sq. cm.

radius = r = 10 cms

Solution : Lateral surface area of a cylinder = rh2π

1760 = hx10x7

22x2

h = 10x22x2

7x1760

h = 28 cms

Exercise : 9.11) The radius of the circular base of a cylinder is 14 cm and height is 10 cm. Calculate

the curved surface area of the cylinder.

2) The circumference of a thin hollow cylindrical pipe is 44 cm and length is 20 mts.Find the surface area of the pipe.

3) A cylinder has a diameter 20 cm and height 18 cm. Calculate the total surfacearea of the cylinder.

233

4) Lateral surface area of a cylinder is 1056 sq. cm and radius is 14 cm. Find theheight of the cylinder.

5) A mansion has twelve cylindrical pillars each having the circumference 50 cm andheight 3.5 mts. Find the cost of painting the lateral surface of the pillars atRs. 25 per squaremeter.

6) The diameter of a thin cylindrical vessel open at one end is 3.5 cm and heightis 5 cm. Calculate the surface area of the vessel.

7) A closed cylindrical tank is made up of a sheet metal. The height of the tank is1.3 meters and radius is 70 cm. How many square meters of sheet metal is requiredto make?

8) A roller having radius 35 cms and length 1 meter takes 200 complete revolutionsto move once over a play ground. What is the area of the playground?

3. Volume of a right circular cylinder :

Activity :

1) A coin is placed on a horizontal plane.

2) Pile up the coins of same size one upon the other such that they form a right circularcylinder of height ‘h’.

Volume of a cylinder = Bh

The area of the circular base B = πr2 [B is the circular base of radius r]

Height of the cylinder = h

∴ Volume of the cylinder = hr� 2 cubic units

Volume of the right circular cylinder of radius ‘r’ and height

‘h’ = hr� 2 cubic units. Volume is expressed in cubic units.

234

Worked examples :Example 1 : Find the volume of the cylinder whose radius is 7 cm and height is

12 cm.

Given : Radius of the cylinder = r = 7 cm

Height of the cylinder = h = 12 cm

Solution : Volume of the cylinder V = Bh

= hr2π

= 12x7x7x7

22

= 12x49x7

22

= 1848 cubic cms.

∴ Volume of the cylinder = 1848 cc

Example 2 : Volume of the cylinder is 462 cc and its diameter is 7 cm. Find theheight of the cylinder.

Given : Volume = V = 462 cc

Diameter = d = 7 cm ∴ r = 3.5 cm.

Solution : Volume of a cylinder V = hr2π

462 = hx)5.3(x7

22 2

h = 2)5.3(x22

7x462

h = 12 cm

∴ Height of the cylinder h = 12 cms

Exercise : 9.21) Area of the base of a right circular cylinder is 154 sq. cm and height is 10 cm.

Calculate the volume of the cylinder.

2) Find the volume of the cylinder whose radius is 5 cm and height is 28 cm.

235

3) The circumference of the base of a cylinder is 88 cm and its height is 10 cm.Calculate the volume of the cylinder.

4) Volume of a cylinder is 3080 cc and height is 20 cm. Calculate the radius of thecylinder.

5) A cylindrical vessel of height 35 cm contains 11 litres of juice. Find the diameterof the vessel (one litre = 1000 cc.)

6) Volume of a cylinder is 4400 cc and diameter is 20 cm. Find the height of thecylinder.

7) The height of water level in a circular well is 7 mts and its diameter is 10 mts.Calculate the volume of water stored in the well.

8) A thin cylindrical tin can hold only one litre of paint. What is the height of thetin if the diameter of the tin is 14 cm? (one litre = 1000 cc)

THE RIGHT CIRCULAR CONE

Observe the following figures :

Heap of sand Ice cream cone cone

A heap of sand, an Ice cream cone suggests you the concept of a right circularcone.

1. Surface area of a right circular cone :

A right circular cone is a solid generated bythe revolution of a right angle triangle aboutone of the sides containing the right angle.

l

O B

h

A

236

Properties of a right circular cone :1) A cone has a circular plane as its base.

2) The point of intersection of the axis of the cone and slant height is the vertex ofthe cone (A).

3) The curved surface which connects the vertex and circular edge of the base is thelateral surface of the cone.

4) The line joining the vertex and the center of the circular base is the height of thecone (AO = h)

Remember :

The distance between the vertex and any point on the circumference of thebase is the slant height.

A. The curved surface area :

Activity :1) Take a right circular cone.

2) Wrap the curved surface with a piece of paper.

3) Cut the paper along the length of slant height say AB

4) Take out the paper which exactly covers the curved surface.

5) Spread the paper on a plane surface.

Observe :

Radius of the circular section is equal to slant height of the cone = l

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This circular section can be divided into small triangles as shown in the figure say T1,

T2, T

3 ... T

n

From the figure lateral surface area of the circular section = sum of the areasof each triangle.

= T1 + T

2 + ..... T

n

Area of the Triangle = 2

1 x base x height

i.e. Area of T1 =

2

1 x B

1 x l

T2 =

2

1 x B

2 x l

T3

= 2

1 x B

3 x l

Area of Tn =

2

1 x B

n x l

Area of circular section = 2

1 B

1l +

2

1 B

2l + ............... +

2

1 B

nl

= 2

1l (B

1 + B

2 + ...... + B

n)

= 2

1l ( r2π ) [B

1 + B

2 + .... + B

n = r2π ]

= 2

1l x r2π

∴ Area of the curved surface of a cone = � rl sq. units

238

B. Total Surface area of a cone :

Total surface area of a cone= Area of circular base + Area of the curved surface

= r�r� 2 + l

= rπ (r + l)

Total surface area of a cone = rπ [r + l] sq. units

Worked Examples :Example 1 : Calculate the curved surface area of cone, given radius of the base as

10 cms. and slant height 28 cms.

Given : Radius of the base r = 10 cm

Slant height l = 28 cms

Solution : Curved surface area of a cone = r� l

= 28x10x7

22

= 22 x 10 x 4

= 880 sq.cm

Example 2 : Find the total surface area of the cone whose radius is 3.5 cm and slantheight is 10 cm.

Given : Radius = r = 3.5 cm

Slant height = l = 10 cm

Solution : Total surface area of cone = rπ (r + l)

= 7

22 x 3.5 (3.5 + 10)

= 7

22 x 3.5 x 13.5

= 148.5 sq. cm

2r�r� l

239

Example 3 : The diameter of a cone is 10 cm and height is 12 cm. Calculate thetotal surface area of the cone.

Given : diameter = d = 10 cm

height = h = 12 cm

Solution : d = 10 cm

∴ r = 5 cm

and l2 = r2 + h2

22 125 +±=l

169±=l

13±=l

1 = 13 cm

Total surface area of cone = r� (r + l)

= 7

22x 5 (5+13)

= 7

22x 5 x 18

= 282.85 sq.cm

Exercise : 9.31) Find the curved surface area of a cone whose circumference of base is 66 cm

and slant height is 12 cm.2) The curved surface area of a cone 440 sq.cm. and slant height is 10 cm. Find

the radius of the cone.3) The diameter of the cone is 14 cm and slant height is 9 cm. Find the total surface

area of the cone.4) Find the total surface area of conical tomb when the slant height is 8 meters and

diameter is 12 meters.5) The slant height and the diameter of a conical tent are 25 mtrs and 14 mtrs

respectively. Calculate the cost of the canvas used at Rs. 15 per sq. meter.6) The curved surface area of a conical tomb is 528 sq. meters and radius is 8 meters.

Find the height of the tomb.7) The height of the cone is 5.6 cms and diameter of the base is 8.4 cms. Find the

area of the curved surface.8) The height of a conical tent is 28 mts and the diameter of the base is 42 meters.

Find the cost of the canvas used at Rs. 20/- per sq. mts.

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2. Volume of a right circular cone

Suggested Activity :1) Take a conical cup and a cylindrical vessel of the same radius and height.

2) Fill the conical cup with water up to its brim.

3) Pour the water into cylindrical vessel.

4) Count how many cups of water is required to fill the cylindrical vessel upto itsbrim.

Observe that exactly three cups of water is required to fill the vessel.

Volume of a cylinder = 3 x volume of a cone having same base and height.

∴ Volume of a cone = 3

1 of the volume of a cylinder having same base and height.

= 3

1 x Bh [

∴ Volume of a cylinder = Bh]

= hr3

1 2π [∴

B = 2rπ ]

Volume of a cone of radius r and height h = hr�31 2 cubic units.

Worked Examples :Example 1 : Find the volume of the cone whose base area is 154 sq. cm and height

is 12 cm.

Given : Base area = B = 154 sq. cm.

and height of the cone = h = 12 cm

Solution : Volume of a cone = 3

1Bh

241

= hr3

1 2π [B = 2rπ ]

= 3

1 x 154 x 12

Volume of the cone = 616 cc

Example 2 : Find the volume of the cone having radius 7 cm and height 18 cm.

Given : Radius = r = 7 cm

Height h = 18 cm

Solution : Volume of a cone = 3

1 Bh

= hr3

1 2π [∴ B = 2rπ ]

= 18x7x7

22x

3

1 2

Volume of the cone = 924 cc

Exercise : 9.41) Area of the base of a cone is 300 sq. cm and height is 15 cm. Find the volume

of the cone.

2) Volume of a cone is 550 cc and diameter is 10 cm. Find the height of the cone.

3) The radius of the base is 10 cms and the height is 21 cms. Find the volume ofthe cone.

4) The circumference of the brim of a conical cup is 22 cm and height is 6 cm. Howmuch water does it hold?

5) The volume of a cone is 3080 cc and height is 15 cm. Find the radius of thecone.

6) The volume and the height of a cone are 2200 cc and 21 cm respectively. Findthe diameter of the base.

7) A meter long metal rod (cylindrical in shape) of radius 3.5 cm is melted and recastto form cones of radius 1 cm and height 2.1 cm. Find the number of cones soformed.

8) A right angled triangle of sides 21 cm 28 cm and 35 cm is revolved on the side28 cm. Name the solid formed and find the volume.

242

THE SPHERE

Observe the following figures

Shot put Ball Sphere

A shotput, a ball etc, will suggest you the concept of a sphere.

Properties of a sphere :

A sphere is a solid described by the revolution of asemi circle about a fixed diameter.

1) A sphere has a centre

2) All the points on the surface of the sphere are equidistant from the centre.

3) The distance between the centre and any point on the surface of the sphere is theradius of the sphere.

Remember :

A plane through the centre of the sphere divides it into two equalparts each called a hemisphere.

1. Surface Area of a sphere :

Activity :1) Consider a sphere of radius r.

2) Cut the solid sphere into two equal halves.

3) Fix a pin at the top most point of a hemisphere.

4) Starting from the centre point of the curved

243

surface of the hemisphere, wind a uniformthread so as to cover the whole curved surfaceof the hemisphere.

5) Measure the length of the thread.

6) Similarily, fix a pin at the center of the planecircular surface.

7) Starting from the centre, wind the thread ofsame thickness to cover the whole circularsurface.

8) Unwind and measure the lengths of the threads.

9) Compare the lengths.

What do you observe from both the activities?

It is found that the length of the thread required to cover the curved surface istwice the length required to cover the circular plane surface.

Area of the plane circular surface = 2rπ∴ Curved surface area of a hemisphere = 2r2π

Surface area of the whole sphere = 22 r2r2 π+π

= 2r4π

Surface area of a sphere of radius r = 2r�4 sq. units.

Worked Examples :Example 1 : Calculate the surface area of the sphere whose radius is 14 cm.

Given : Radius of the sphere r = 14 cm.

Solution : Surface area of the sphere = 2r4π

= 2)14(x

7

22x4

= 196x7

22x4

= 2464 sq. cm.

244

Example 2 : The surface area of the sphere is 616 sq. cm. Find the diameter of thesphere.

Given : Surface area = 2r4π = 616 sq. cm.

d = ?

Solution : Surface area of a sphere = 2r4π

616 = 4 x 7

22 x r2

r2 = 4

616 x

22

7

r2 = 49

r = 49Radius of the sphere = r = 7 cm

d = 2r

= 2 x 7

Diameter of the sphere d = 14 cm

Exercise : 9.51) Find the surface area of a sphere whose radius is 21 cm.

2) The circumference of a globe is 88 cm. Calculate the surface area of the globe.

3) Find the total surface area of a hemisphere of radius 14 cm.

4) The circumference of a hemispherical dome is 44 mts. Calculate the cost of paintingat Rs. 20 per square meter.

5) The surface area of a sphere is 154 sq. cm. Find the diameter of the sphere.

6) A solid sphere of radius 10.5 cm is cut into 2 halves. Find the total surface areaof both the hemispheres.

2. Volume of a sphereObserve the following figures.

r base = B

height = r

245

A solid sphere is made up of miniature cones whose height is equal to the radius ofthe sphere and each having circular base.

Volume of each cone = 3

1 x base x height

Volume of cone 1 = 3

1 x B

1 x r

Volume of cone 2 = 3

1 x B

2 x r etc

Volume of cone n = 3

1 x B

n x r

Volume of the sphere = Sum of the volumes of all the cones

= 3

1 x B

1 x r +

3

1 x B

2 x r + ....

3

1 x B

n x r

= 3

1 r (B

1 + B

2 + ........ + B

n) [

∴T.S.A. of sphere 4 2rπ ]

= 3

1 r (B) [

∴Surface area of sphere]

= 3

1 r x 4 2rπ

= 3r

3

4 π

Volume of the sphere = 3r

3

4 π cubic units

Volume of the hemisphere = 3r�

3

4x

2

1

= 3r�

3

2

∴ Volume of a hemisphere = 3r�

3

2 cubic untis

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Worked Examples :Example 1 : Find the volume of the sphere whose radius is 21 cm.

Given : Radius of the sphere = r = 21 cms

Solution : Volume of a sphere = 3r

3

4 π

= 3)21(x

7

22x

3

4

= 21x21x21x7

22x

3

4

= 4 x 22 x 21 x 21

Volume of the sphere = 38808 cc.

Example 2 : A hemispherical bowl has radius 14 cm. How many litres of water doesit hold?

Given : Radius of the sphere, r = 14 cms

Solution : Volume of a hemisphere = 3r�

3

2

= 314x

7

22x

3

2

= 14x14x14x7

22x

3

2

Capacity of the bowl = 5.75 Litres

[ ∴

1000 cc = 1 litre]

Example 3 : Total volume of 21 steel balls in a bearing is 88 cc. Find the diameterof each ball.

Given : Volume of 21 balls = 88 cc

Volume of each ball = 21

88 cc

Solution :

Volume of a Steel ball = 3r3

4 π

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3rx7

22x

3

4

21

88 =

r3 = 22

7x

4

3x

21

88

r3 = 1

r = 3 1

r = 1 cm

∴ d = 2r

d = 2 cms

∴ Diameter of each ball = 2 cms

Exercise : 9.61) Find the volume of the sphere whose radius is 3 cm.

2) The diameter of a shot put is 9 cm. Calculate the volume of the shot-put.

3) The depth of a hemispherical water tank is 2.1 m at the centre. Find the capacityof the water tank in litres.

4) Twenty one lead marbles of even size are recast to form a big sphere. Find thevolume of the sphere when the radius of each marble is 2 cm.

Remember at a glance :

Solid Curved Surface Total Surface Volume area area

Cylinder rl2π r2π (r+1) hr2π

Cone rlπ rπ (r+1) hr3

1 2π

Sphere 2r4π 2r4π3r

3

4 π

Solid hemisphere 2r2π 2r3π3r

3

2 π

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SCALE DRAWING

After studying this unit you will be able to

* Develop the skill in drawing the polygonal figures to the scale.

* To find the area of triangles and trapeziums.

* Relate the area of polygonal figure to the area of irregular field.

* Represent the irregular shaped field into known geometrical rectilinear figures, takingthe measurements to the scale.

You are familiar with the geometrical figures like triangle, rectangle, trapezium and

the formulae for the area of these figures.

Area of triangle = 21

x base x height = 21

bh

Area of rectangle = length x breadth = lb

Area of trapezium = 21

x height x (sum of two parallel sides) = 21

h (a+b)

Recall :

1) Find the area of triangle when the base is 12 cm and height is 8 cm.

2) Find the area of trapezium given AB = 8 cm, CD = 6 cm, AB || CD and thedistance between AB and CD is 5 cm.

1. Measurement of the area of a land :Observe the following piece of land. How to find the area of the land?

1) Irregular shaped field is divided into known geometrical shaped fragments.

2) Measurements are recorded and a sketch is drawn to the scale.

D

C

B

A

E

4

5Q

R

P

1

2

3

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3) Measurements are recorded in the surveyor’s field book.

4) Total area of the land is the sum of the areas of all the right angled triangles andthe trapezium.

Surveyor’s mode of recording the measurements of a land is given below.

To.D (in mts)

200

140 50 To C

To E 60 120

40 30 to B

From A

Activity :

To measure the area of the land

1) Take suitable scale, say 20m = 1 cm

2) Draw the base line AD

3) Draw the perpendiculars to the base line PB, QE and RC

4) Complete the polygon joining the end points.

5) Find out the area of ∆ ABP, ∆ AQE, ∆ DQE and ∆ DRC and also the are oftrapezium BPRC.

6) Total area of the land is equal to the sum of the all the triangles and trapezium.

D

ED

B

A

RQ

P

60

50

60

100

30

40

250

Total Area of the land = Area of ∆1 + Area of ∆

2 + Area of ∆

3 +

Area of ∆4 + Area of Trapezium.

Remember :

Area of the land is expressed in hectares.

Worked Example :Example 1 : Plan out and find the area of the field from the following notes from

the field book.

Meter to D

150

100 70 to C

To E 80 80

30 40 to B

From A

Solution : Scale 20 mts = 1 cm

Observe that,

AM = 30 mts

AN = 80 mts

MP = (100-30)

= 70 mts

ND = (150-80)

= 70 mts

PD = (150-100)

= 50 mts

1) Area of ∆ ABM = 2

1bh

= 2

1x 30 x 40

= 600 sq.mts

D

P C

N

5070

80E

M B

A

20

50

30

40

251

2) Area of Trapezium MDCP = 2

1h (a+b)

= 2

1 x 70 (70+40)

= 3850 sq.mts

3) Area of ∆ DPC = 2

1 x 50 x 70

= 1750 sq. mts

4) Area of ∆ DEN = 2

1 x 80 x 70

= 2800 sq. mts

5) Area of ∆ NEA = 2

1x 80 x 80

= 3200 sq. mts

Area of the field ABCDE = 600 + 3850 + 1750 + 2800 + 3200

= 12200 sq mts.

= 1.22 hectares

Exercise : 9.71) Draw a plan and calculate the area of a level ground using the information given

below.

Meters to C

220

To D 120 210

120 200 to B

To E 180 80

From A

Observe :1 hectare =10,000 sq.mts.

252

2) Plan out and find the area of the field from the data given from the Surveyor’sfield book

Meters to E

350

To D 100 300 150 to F

To C 75 250

150 100 to G

To B 50 50

From A

3) Sketch a rough plan and calculate the area of the field ABCDEFG from the followingdata.

Meters to D

225

To E 90 175

125 20 to C

To F 60 100

To G 15 80

60 70 to B

From A

4) Calculate the area of the field shown in the diagram below :

[Measurements are in meters]

A

B

30

G30

E

F

D

C

25

35

40 70