This unit facilitates you in,defining cylinder, cone and sphere.explaining the properties of cylinder, coneand sphere and hemisphere.deriving the formula to find the area andvolume of cylinder, cone, sphere andhemisphere.calculating the LSA and TSA of cylinder,cone, sphere and hemisphere.calculating the volume of cylinder, cone,sphere and hemisphere.solving application problems based on areaand volume of cylinder, cone and sphere.explaining the meaning of frustum of a cone.calculating the area and volume of frustumof a cone.identifying objects made up of two or moresolids.finding the area and volume of combinedsolid objects.constructing scale drawing for irregularshaped figures.calcualting area of irregular shaped figures.

Mensuration16

Give me a place to stand,and I will move the earth.

- Archimedes of Syracuse

Meaning of cylinder, cone,frustum of a cone, sphere andhemisphere.

Surface area and volume ofcylinder, cone, frustum of acone, sphere and hemisphere.

Combined figures, their areaand volume.

Scale drawing.

Archimedes(287BC-212BC)

Archimedes of syracuse, Sicily, isremembered as the greatest Greekmathematician of the ancient era.He contributed significantly ingeometry regarding the areas ofplane figures and the areas andvolumes of solid objects.

He proved that the volume of asphere is equal to two-thirds thevolume of a circumscribed cylinder.

378 UNIT-16

The study of measurement of objects is very essential because of its applications inevery day life, industry and engineering. We have already learnt to find the surface areaand volume of solid objects such as cube, cuboid, prism and pyramid.

Now let us consider some examples where measurement of objects areapplied.

• A village panchayat has built an over head water storage tank whichis cylindrical in shape. How to determine the size of the tank? Whatis the cost incurred in painting the tank?

• How do you calculate the amount of dieselthat can be stored in a cylindrical shapedfuel tank of a truck?

• How many basin laddoos can be made from given weight of mixtureof basin laddoo?

The solution for all the above mentioned problems depends on finding the surfacearea and volume of solid objects like cylinder, cone and sphere.

How to find the surface area and volume of these objects? Which are the formulaeused?

In this unit, let us learn about the surface area and volume of cylinder, cone andsphere.

CylinderWe have studied that when a number of rectangular sheets of paper are stacked up,

we get a cuboid or rectangular based prism.What do we get when a number of congruent

circular sheets of paper are stacked up?Observe the figures given.The solid object obtained is called Cylinder. Cylinders are sort of prisms with circular bases.

Activity: Take a thick paper and cut a rectangle ABCD. Paste a long thick stick along oneside, say AB. Hold the stick with both your palms and rotate it fast (you can also paste along thick string, hold the string with hands, on either side and rotate it fast).Do this activity in groups. Discuss and sharethe ideas.What shape can you recognise? From the aboveactivity, we can conclude that

If a rectangle revolves about one of its sidesand completes a full rotation, the solid formedis called a right circular cylinder.

• What is the quantity ofcolour paper required toprepare a birthday cap?

A

BC

D

Mensuration 379

Top circularbase

Bottom circularbase

Curved orlateral surface

In our daily life, we haveseen many objects which arecylindrical in shape. Some of themare given below.

The cylinders can be eithersolid or hollow. Observe that thecylinder has two circular bases,bottom base and top base which are congruent to each other.

Activity: Place a cylinder on the table and hold a piece of cardboardparallel to the top of the table and touching the top of the cylinder.

If we find the perpendicular distance between the top of the table(i.e bottom of the cylinder) to the cardboard (i.e top of the cylinder), it isthe height of the cylinder.Now let us consider a cylinder having the following properties. • It has two congruent and parallel circular bases. • It has a curved surface joining the edges of the two bases. • The line segment joining the centres of the two bases is perpendicular to the base.

It is the height (h) of the cylinder and also called axis of the cylinder.Such a cylinder is called right circular cylinder.It is called so because, the cylinder has a circular base and its axis is atright angle to the base.In this unit, we will be dealing with only right circular cylinders. So unless stated

otherwise, the word cylinder would mean a right circular cylinder.Surface area of a cylinderWe know that a cylinder can be either a hollow cylinder or a solid cylinder.

A hollow cylinder is formed by the curved surface only. Pipe, straw, tubes, flute, areexamples for hollow cylinders.

A solid cylinder is the object bounded by two circular plane surfaces and also thecurved surface. A cylindrical pillar, the wheels of road roller, a garden roller, pencil, filledcylindrical tanks and filled cans are examples of solid cylinder.

The surface area of a cylinder refers to the area of the external surface of thecylinder.

A hollow cylinder has only the curved surface, which is also called the lateral surface.The area of this surface is the curved surface area (CSA) or thelateral surface area (LSA) of the hollow cylinder.

A solid cylinder has two circular surfaces and also the curvedsurface. Hence, the area of the curved surface is the curved orlateral surface area of the solid cylinder.

The total surface area (TSA) of the solid cylinder refers tothe sum of areas of the two circular regions and the curved surfacearea.

h

Milk can Pipe Gas cylinder Pillars

380 UNIT-16

Now, let us derive the formulae used to find the curved surface area and total surfacearea of cylinders.

i. Curved surface area of a cylinder: Take a rectangular sheet of paper whose lengthis just enough to go round the given cylinder and whose breadth is equal to theheight of the cylinder. This activity can also be done by taking a cylinder made ofcardboard. Cut open the lateral surface along a line and unfold it, to get a rectangle.

b b

l

h

Fold

Cut

2 r

Observe that,

Area of the rectangular sheet = Curved surface area of the cylinder

Length of the rectangle = Circumference of the base of the cylinder l = 2r

Breadth of the rectangle = Height of the cylinder b = h

We know that, Area of rectangle = l × b = 2r × h = 2rh

Curved surface area of the cylinder = 2rh sq. unitsii. Total surface area of a solid cylinder:

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

Volume of a cylinderYou have already learnt that volume of a prism is the product of area of itsbase and height. Since, cylinder is also a type of prism with circular base, itsvolume should be calculated in the same manner.

Volume of a solid cylinder = Area of base × height = r2 × h = r2h Volume of a cylinder = r2h cubic units

ILLUSTRATIVE EXAMPLES

Example 1 : Find the CSA of a right circular cylinder whose height and radius of baseare 30 cm and 3.5 cm respectively.

Sol. CSA of cylinder = 2rh = 22

27

3.50.5

cm×30cm = 660cm2

CSA of the cylinder = 660cm2

Total surface areaof the cylinder

Curved surface area of the cylinder

Area of thecircular bases = + 2 ×

= 2rh + 2 × r2

= 2rh + 2r2 = 2r (h + r)

h

r

A = r

2

Mensuration 381

Example 2 : The CSA of a right circular cylinder of height 14 cm is 88 cm2. Find theradius of the base of the cylinder.

Sol. CSA of a cylinder = 2rh 88 cm2= 22

2 14cm7

r

r = 1 2 4 88 2cm 7

2 22 14 2cm = 1 cm

radius of the base of the cylinder = 1 cmExample 3 : The radii of two right circular cylinders are in the ratio 2 : 3 and theirheights are in the ratio 5 : 4. Calculate the ratio of their curved surface areas.Sol. Let the radii of 2 cylinders be 2r and 3r respectively and their heights be 5h and 4h

respectively.Let S1 and S2 be the curved surface areas of two cylinders.S1 CSA of cylinder of radius 2r and height 5h = S1 = 2 × × 2r × 5h = 20rhS2 CSA of cylinder with radius 3r and height 4h= S2= 2 × × 3r × 4h = 24rh

1

2

SS =

5 206 24

rh 56rh

S1 : S2 = 5 : 6Example 4 : The diameter of a roller 120cm long is 84 cm. If it takes 500 completerevolutions to level a play ground, determine the cost of levelling it at the rate of0.50 paise per square metre.Sol. h = 120 cm, r = 42 cm.

Area covered by the roller in one revolution = CSA of the roller = 2rh

= 22

27

426cm × 120cm = 31,680 cm2

Area covered in 500 revolutions = (31,680 × 500) cm2 = 15,84,000 cm2

Area covered inm2 215840000 m100 100

= 1584 m2

Cost of levelling the playground per m2 = 0.50 ps Cost of levelling the playground= 1584 m2 × 0.50 ps/m2 = ̀ 792.

Example 5 : A metal pipe is 77 cm long. The inner diameter of cross section is 4 cm,the outer diameter being 4.4cm. Find its (a) inner curved surface area (b) Outer curvedsurface area (c) Total surface area.Sol. Outer radius = R = 2.2 cm; Inner radius = r = 2 cm; h = 77cm

(a) Inner CSA = S1 = 2rh = 22

27

2cm 7711

cm = 968 cm2

(b) Outer CSA = S2 = 2Rh = 22

27

2.2cm 7711

cm = 1064.8cm2

(c) TSA = S1 + S2 + area of two bases = S1 + S2 + 2(R2 – r2)

= 968cm2 + 1064.8cm2 +22

27 {(2.2)

2 – 22}

= 44968 1064.87

0.840.12 2cm

= (968 + 1064.8 + 5.28) cm2

TSA = 2038.08 cm2

382 UNIT-16

Example 6 : Find the volume of a right circular cylinder if the radius of its base is 7cmand height is 15 cm.Sol. r = 7cm, h = 15 cm

Volume of a cylinder = r2h = 227

7cm × 7cm × 15cm = 2310 cm3

Example 7 : The circumference of the base of a cylinder is 132cm and its height is 25cm. Find the volume of the cylinder.Sol. Let 'r' be the radius of the cylinder. Circumference = 132 cm

2r = 132 cm

222

7r = 132 cm r =

1326 3

cm 72 22

r = 21 cm

Volume of a cylinder = r2 h = 222

(21) 257

= 227

213

21 25 = 34,650 cm3

Example 8 : A well with 10 m inside diameter is dug 14 m deep. Earth taken out of itis spread all around to a width of 5m to form an embankment. Find the height of theembankment.

Sol. Volume of earth dug out = r2h = 227

25 142 3m

= 22 × 25 × 2m3 = 1100 m3

Area of the embankment (Shaded region) = (R2 – r2)

= 2 2 222 2210 5 75m7 7

Volume of earth dugout = (r2) h = (Area of embankement) × h

height of the embankment = Volume of earth dugout

Area of embankment

= 3

2

1100m22 75m7

= 7 1100 m22 75 = 4.66m

Height of the embankment = 4.66m

Example 9 : The difference between outside and inside surfaces of a cylindrical metallicpipe 14 cm long is 44 cm2. If the pipe is made of 99cm3 of metal, find the outer andinner radii of the pipe.Sol. Let, R be the external radius and r be the inner radius of the metallic pipe. h=14 cm

Outer surface area – inner surface area = 44cm2

2Rh – 2rh = 44 cm2 R – r= 44222 147

= 2 44 7

2 22 14 2

R – r = 12

5m5m

5m

Mensuration 383

Volume of metal used = 99cm3 External volume – internal volume = 99cm3

R2h – r2h = 99 cm3h(R2 – r2) = 99 cm3

227

142(R r)(R r)= 99 cm3 [ a2 – b2 = (a + b)(a – b)] and 12R r

22 2 1(R r)2 = 99 cm

3 R + r = 99

9

22 = 92

R r 92R r 12

1022R

2R = 5

R = 52 = 2.5 cm

External radius = 2.5 cm; Internal radius = 2 cm

Example 10 : The circumference of the base of a cylindrical vessel is 132 cm and itsheight is 25 cm. How many litres of water can it hold?Sol. C= 132 cm, h = 25 cm Let the radius of the base = r

2r = 132cm 222 r7

= 132 cm

r = 132

2227

= 132

6 3

72 22 = 21 r = 21 cm

Volume of the vessel = r2h = 2 322 (21) 25cm7

= 227

213 321 25 cm = 34650cm3

Volume in litres = 346501000

[ 1000 cm3 = 1l ] = 34.65 litres

The vessel can hold 34.65 litres of water.

Example11 : If the radius of the base of a right circular cylinder is halved, keepingthe height same, what is the ratio of the volume of the new cylinder to that of theoriginal cylinder.Sol. Let V1 be the volume of the original cylinder and V2 be the volume of the new cylinder.

Radius of original cylinder = r, radius of new cylinder = 2r

V1= r2h and V2 =

2r h2

1

2

VV =

2r h2r h4

= 21

V4 11 V 4

V2 : V1 = 1 : 4

R + r = 92r + 2.5 = 4.5

r = 4.5 – 2.5 = 2 cm

384 UNIT-16

Example 12 : A solid cylinder has a TSA of 462 cm2. Its CSA is one - third of its TSA.Find the volume of the cylinder.

Sol. CSA of cylinder= 13

TSA of the cylinder 2rh= 13

× 2r (h + r) = 13

462154 2 cm

2rh = 154 cm2

TSA of a cylinder = 2r (h + r) 462 cm2= 2rh + 2r2 = 154 cm2 + 2r2

462cm2 – 154 cm2 = 2r2 308 cm2= 22227

r 308

14 7

72 22

= r2

r2 = 49 r = 7 cmCSA of cylinder = 154 cm2 2rh= 154 cm2

222

77cm h = 154 cm2 h=

1547

442 =

72 cm

Volume of the cylinder = r2h = 2 322 77 cm7 2 =

11 227

749

23cm = 539cm3

EXERCISE 16.1

1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm.Find its (i) CSA (ii) TSA.

2. The CSA of a cylindrical pipe is 550 sq.cm. If the height of the pipe is 25 cm, find thediameter of the base.

3. An iron pipe 20cm long has external radius equal to 12.5cm and internal radiusequal to 11.5cm. Find the TSA of the pipe.

4. The radii of two right circular cylinders are in the ratio 2 : 3 and the ratio of theircurved surface areas is 5 : 6. Find the ratio of their heights.

5. Find the ratio between TSA of a cylinder to its CSA given its height and radius are7 cm and 3.5 cm respectively.

6. The inner diameter of a circular well is 2.8m. It is 10 m deep. Find its inner curvedsurface area. Also find the cost of plastering this curved surface at the rate of ̀ . 42per m2?

7. Craft teacher of a school taught the students to prepare cylindrical pen holders outof card board. In a class of strength 42, if each child prepared a pen holder of radius5 cm and height 14 cm, how much cardboard was consumed?

8. A solid cylinder has a total surface area of 462cm2. If its curved surface area is onethird of its total surface area, find the radius and height of the cylinder.

9. A cylindrical vessel without lid has to be tin coated on its outside. If the radius ofthe base is 70 cm and its height is 1.4m, calculate the cost of tin coating at the rateof ̀ . 3.50 per 1000 cm2.

10. The diameter of a garden roller is 1.4 m and is 2 m long. How much area will it coverin 5 revolutions?

Mensuration 385

11. Find the volume of a right circular cylinder whose radius is 10.5 cm and height is16 cm.

12. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a massof 0.6 gm.

13. The lateral surface area of a cylinder of height 5 cm is 94.2 cm2. Find (i) radius of itsbase (ii) Volume of the cylinder.

14. Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Findthe ratio of their radii.

15. A rectangular sheet of paper, 44 cm × 20 cm is rolled along its length to form acylinder. Find the volume of the cylinder so formed.

16. The trunk of a tree is cylindrical. Its circumference is 176 cm. If the length of thetrunk is 3m, find the volume of the timber that can be obtained from the trunk.

17. A well of diameter 14 m is dug 8m deep. The earth taken out of it has been evenlyspread all around it to a width of 21m to form an embankement. Find the height ofthe embankment.

18. In a village fair, a stall keeper has kept a large cylindrical vessel of base radius15 cm filled upto a height of 32 cm with orange juice. He sells them at `. 3 per glassin a cylindrical glass of radius 3 cm and height 8 cm. How much money does thestall keeper earn by selling thejuice completely.

ConeObserve the following objects.

They are conical in shape. Theycan be hollow or solid.

Observe that cones have a circular base and a curved surface which tapers fromthe base to an end point.

Compare cones with pyramids. Cones are sort of pyramids withcircular bases.

A cone is a solid object that tapers smoothly from a flat circularbase to a point called vertex.

A right circular cone has the circular base and its axis passes through the centre ofthe base and is perpendicular to the base.

Activity: Take a thick paper and cut a right angled triangle ABC, with B = 900. Paste along thick string along one of the perpendicular sides, say AB. Hold the string with thehands on either side of the triangle androtate the triangle about the string.

Can you recognise the shape formed?The shape formed is a right circular

cone. Therefore we can conclude that,If a right angled triangle is revolved about one of the side containing the right angle,the solid formed is called a right circular cone.

A

BC

A

BC DB

CD

A

Cone Pyramid

Ice-creamcone

Birthdaycap

Roof of the hut

Heap of sand

386 UNIT-16

Observe the figure of cone.

• The length AB or the distance from the centre of the circular base tothe vertex is called the height (h) of the cone. It is called the axis ofthe cone.

• The length BC is the radius of the base (BC = r).

• The length AC or the distance from the vertex to any point on thecircular base is called the slant height (l) of the cone. (AC = AD = l)Is there any relation between the height (h) and slant height (l) of a cone?

Consider the right angled triangle ABC in the figure

We have, AC2= AB2 + BC2 (Pythagoras theorem) l2 = h2 + r2

where l is the slant height, h is the height, r is the radius of the base of the cone.

l = 2 2h r or h = 2 2l r or r = 2 2l hNow, let us consider a cone having the following properties.

• It has a circular plane as its base.

• The axis of the cone and the slant height intersect at a point called vertex.

• It has a curved surface which connects the edge of circular base and the vertex.

• The line joining the vertex and the centre of the circular base is perpendicular tothe base.

Such a cone is called right circular cone. As in case of cylinder, since we will bestudying about only right circular cones in this unit, by "cone" we shall mean "rightcircular cone".

Surface area of a coneObserve the given below figures:

The surface of a cone consists of acurved surface made by bending a sectorsuch that its arc touches with the edge of the circular base.The area of the sector which forms the cone is called thecurved surface area or the lateral surface area of the cone.

If the base of the cone is closed with a circular piece ofcardboard, then we consider the Total surface area of a cone (TSA) by adding the curvedsurface area (CSA) with the area of the circular base.

i. Curved surface area of coneIf a cone made up of cardboard is cut and spread out, we get a sector as shown below.

h l l

The area of the sector is the curved surface area of the cone.

Sector: is a part ofcircular region boundedby two radii and an arc.

A

B CD

h l

Mensuration 387

How to find the area of the sector?

Let 'r' be the radius of the cone, 'l ' be the slant height of the cone and'' be the central angle of the sector.Here, radius 'r' of the sector is same as the slant height 'l' of the cone. r = l

Area of the sector = 2360

l .....(1)

Let 'L' denote the length of the arc of the sector.

Then,2 360lL

Length of the arc L 2360º

l .....(2)

In the figure,Length of the arc of the sector = Circumference of the circular base of the cone L = 2r ....(3)

From (2) and (3), we get 2r = 2360

l rl

= 360

....(4)

Let 'A' be the area of the sector. Then, 2l

A=

360

A= 2360

l = 2rll

= rl

The curved surface area of the cone = rl sq. units

Since rl can be written as 1

22

rl and 2r is the circumference of base of the

cone, l is the slant height. We have,

Curved surfacearea of the cone

circumference of thebase of the cone × Slant height

12=

Curved surface area (CSA) of the cone is half the product of the circumferenceof its base (2r) and slant height (l)

(ii) Total surface area of the right circular cone

Total surface areaof the cone

Curved surface areaof the cone

Area ofthe base= +

=rl + r2

=r (l + r)

The total surface area of the cone = r (l + r) sq. units.Volume of coneLet us consider the experiment conducted to find the volume of a cone. Study theexperiment and repeat it in small groups.

h l l l

2 r

rl

r2

388 UNIT-16

Take a hollow cylinder with one base closed and a hollow cone of same radius andsame height. It is useful, if they are made of transparent material. Let their thickness beas less as possible and almost negligible.

Observe the given figure to easily check theirradius and height.

Cone is filled with sand or water and poured intothe cylinder. The cylinder is completely filled bypouring sand or water three times by the cone.

Volume of a cylinder = r2h, where 'r' is the radius of its circular base and 'h' is theheight.

From the above experiment, we can conclude that,

Volume of cone = 13

× Volume of cylinder

Volume of cone = 13 r

2h cubic units

So far we have learnt about the surface area and volume of cylinder and cone.It is interesting to note the change in the surface area and volume of cylinder andcone when their radius and height are either doubled or halved. Study the followingtable.

Sl. Solids Cylinder Cone

No. LSA TSA V LSA TSA V

1.whenr = h 2r

2 4r2 r3 – –13r3

2.r 2rh h 4rh 4r(2r + h) 4r

2h - -43r2h

3.r rh 2h 4rh 2r(r + 2h) 2r

2h - -23r2h

4.r2r

h hrh

rr h2

2r h4

- -2r h

12

5. h2

rh

rrh

h2 r r

2

2r h2 - -

2r h6

6.r2

h2

rh

rh2

r hr2 2

2r h8

- -2r h

24

Mensuration 389

ILLUSTRATIVE EXAMPLES

Example 1 : The diameter of a cone is 14 cm and its slant height is 10 cm. Find itscurved surface area.

Sol. d = 14 cm r = 142 = 7 cm, l = 10 cm

Curved surface area of a cone = rl = 227

7 cm 10cm = 220 cm2

Example 2 : Find the TSA of a cone, whose slant height is 9m and radius of the base is14 m.Sol. l = 9 m, r = 14 m

TSA of a cone= r (r + l ) = 227

142m (14 9) m = 244 23 m = 1012m2

TSA of the cone= 1012 m2

Example 3 : How many metres of cloth 5m wide will be required to make a conicaltent, the radius of the base is 7m and whose height is 24 m?Sol. r = 7m, h = 24m, l = ?

l2= r2 + h2 = 72 + 242 = 49 + 576 = 625 l = 625 = 25 mArea of canvas used = CSA of conical tent

CSA of a cone = rl = 227

7m 25m = 550 m2

Area of canvas = 550m2

length × width = 550m2 length = 2550m

width=

2550m5

= 110 m

Length of canvas used = 110 mExample 4 : The CSA of a cone is 4070 cm2 and its diameter is 70 cm. What is its slantheight.Sol. d = 70 cm r= 35 cm CSA of a cone = rl.

4070 cm2= 227

355cm l

37 407 011 0 = l l = 37 cm

Slant height of the cone = 37 cmExample 5 : The slant height and diameter of the base of a conical tomb are 25m and14 m respectively. Find the cost of white washing its CSA at the rate of ` 210 per100 m2.

Sol. l = 25m, d= 14 m r = 142 = 7m

CSA of a cone = rl = 227

7m 25m CSA of conical tomb = 550 m2

Cost of white washing 100 m2 = ` 210

Cost of white washing 550 m2 = 210`

10 0 2550

m2m = ̀ 1155

390 UNIT-16

Example 6 : The diameters of 2 cones are equal. If their slant heights are in the ratio5 : 4, find the ratios of their curved surfaces.Sol. Let S1 be the CSA of 1st cone and S2 be the CSA of 2nd cone

Let their slant heights be 5l and 4l, respectively. CSA of a cone = rl

1

2

SS =

r 5lr 4l S1 : S2 = 5 : 4

Example 7 : Find the volume of a right circular cone 90 cm high if the radius of thebase is 21 cm.

Sol. r = 21 cm, h = 90 cm Volume of a cone= 13r2h =

13

227

2131 321 90cm

Volume of the cone= 41,580 cm3

Example 8 : The volume of a cone is 18480 cm3. If the height of the cone is 40 cm, findthe radius of its base.

Sol. Volume of a cone = 213

r h 18480= 21 22

403 7

r

r2= 21 795 1848 0 3 7

22 4 0 = 441 cm2 r = 2441cm = 21 cm

Example 9 : A cone of height 24 cm has a CSA of 550 cm? Find its volume.Sol. h = 24 cm, CSA = 550 cm2

We know that, l2 = r2 + h2 = r2 + 242 = r2 + 576 l = 2r 576CSA of the cone = 550 cm2 rl = 550

222 r r 5767

= 550 2r r 576 = 550

257

22 2r r 576 = 175

Square on both sides2

2r r 576 = (175)2 r2 (r2 + 576) = 30625r4 + 576r2– 30625 = 0 Factorising this equation we get,(r2 + 625) (r2– 49) = 0

If r2 – 49 = 0, r2 = 49, r = 49 r = 7 cm

Volume of of a cone = 21

r h3

= 13

227

7 7 248 3cm

Volume of the cone = 1232 cm3

Example 10 : Find the weight of a solid cone whose base is of diameter 14 cm andvertical height 51cm, if the material of which it is made weighs 10 gm/cm3.

Sol. d = 14 cm = r = 142 = 7 cm, h = 51 cm

Volume of a cone= 21 r h3

= 13

227

7 7 5117 3cm = 2618 cm3

Volume of the cone = 2618 cm3

Weight= volume × density = 2618 cm3 × 10

1000kg/cm3 = 26.18 kg

Mensuration 391

Example 11 : A conical flask is full of water. The flask has base radius 3 cm and height15 cm. The water is poured into cylindrical glass tube of uniform inner radius 1.5 cm,placed vertically and closed at lower end. Find the height of water in the glass tube.

Sol. Volume of water in cone = Volume of water in cylinder 2 c1 r h3 c

= 2 cyr hcy

31 22 3 3 15cm3 7

= cy22

1.5 1.5 h7

hcy =

13

227

32

32

155

227

1.5 1.5 = 20 cm

height of water in the glass tube = 20 cmExample 12 : A solid metallic right circular cylinder 1.8m high with diameter of itsbase 2 m is melted and recast into a right circular cone with base of diameter 3 m.Find the height of the cone.

Sol. Cylinder: h= 1.8 cm, d = 2m r = 22 = 1m Cone: d= 3m, r =

32 = 1.5m, h = ?

Volume of cylinder = Volume of cone 2 cyr hcy = 2

c1

r h3 c

322 1 1 1.8m7

= 31 22 1.5 1.5 m3 7 c

h

hc =

227

1 1 1.8

1 223 7

1.5 1.5=

1.86

32

1.5 6 1.5=

125

= 2.4

height of the cone = 2.4m

EXERCISE 16.2

1. Find the curved surface area of a cone, if its slant height is 60 cm and the radius ofits base is 21 cm.

2. The radius of a cone is 7cm and area of curved surface is 176 cm2. Find its slantheight.

3. The area of the curved surface of a cone is 60cm2. If the slant height of the cone is8 cm, find the radius of the base.

4. Curved surface area of a cone is 308cm2 and its slant height is 14cm. Find theradius of the base and total surface area of the cone.

5. A clown's cap is in the form of a right circular cone of base radius 7cm and height24 cm. Find the area of the sheet required to make 10 such caps.

6. Find the ratio of the curved surface areas of two cones if their diameters of thebases are equal and slant heights are in the ratio 4 : 3.

392 UNIT-16

7. A cylinder and a cone have equal radii of their bases and equal heights. If theircurved surface areas are in the ratio 8 : 5, show that the ratio of the radius of eachto the height of each is 3 : 4.

8. Find the volume of a right circular cone with (i) radius 5 cm, height 7 cm (ii) radius10.5 cm, height 20 cm (iii) height 21 cm, slant height 28 cm.

9. Two cones have their heights in the ratio 1 : 3 and the radii of their bases in theratio 3 : 1. Find the ratio of their volumes.

10. A right angled triangle of which the sides containing the right angle are 6.3 cm and10 cm in length, is made to turn around on the longer side. Find the volume of thesolid thus generated.

11. A right circular cone of height 81 cm and radius of base 16 cm is melted and recastinto a right circular cylinder of height 48 cm. Find the radius of the base of thecylinder.

12 A right circular cone is of height 3.6cm and radius of its base is 1.6 cm. It is meltedand recast into a right circular cone with radius of its base 1.2cm. Find the heightof the cone so formed.

13. A conical flask is full of water. The flask has base radius 'a' and height '2a'. The

water is poured into a cylindrical flask of base radius 23a

. Find the height of water

in the cylindrical flask.14. A tent is of the shape of a right circular cylinder upto a height of 3 m and then

becomes a right circular cone with a maximum height of 13.5m above the ground.Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per sq m,if the radius of the base is 14 m.

Frustum of a coneObserve the shape of the objects that weuse in our daily life.

These are neither cylinders nor cones.What are these shapes called? How are they obtained?Consider the following activity to answer these questions.

Activity: Take some clay or plasticine and form a right circular cone. Cut it with a knifeparallel to its base. You will get two solids as shown in the figure given below.

If we remove the smaller cone obtainedat the top, we are left with a solid object at thebottom. This part of the cone containing itsbase is called the frustum of the cone.

Tumbler Bucket Cap

Know this!'Frustum' is a Latin word meaning "piececut off" and its plural form is 'Frusta'

Mensuration 393

The frustum of a cone has two circular bases, one at the bottom and the other at thetop of it with different radii.

From the above activity, we can conclude that,

If a plane cuts a right circular cone parallel to its base and the upper smallercone is removed, then the remaining part of the cone containing the base is called"Frustum of the cone".

Now, can you name the shape of tumbler,bucket and cap?

Why do you call it so? Discuss in groups.

Is there any relationship between the dimensionsof the original cone and the smaller one cut off ? Observe the figures given below.

A

B

C

D

O

A

B C

DE F

O

LH

R

A

FE D

l h

r

Let the base radius, height and the slant height of the original cone be denoted byR, H and L; and those of the smaller cone be r, h and l.

Observe that the two triangles ABO and AED are similar.

Hence, their corresponding sides are proportional.

i.e. EDBO

= AD AE

=AO AB

rR

= hH L

l

We can conclude that,

If from the top of a cone, a smaller cone is cut off by slicing parallel to the base,then the base radius, height and the slant height of the two cones (smaller coneobtained and the original cone) are proportional.

How is the slant height of the frustum related to its height and radii?

Let 'l ' be the slant height of the frustum, 'h' be the height, 'R' be the radius of thebottom base, and r be the radius of the top base.

The relatonship is given by l2 = h2 + (R – r)2 or l = 2 2h (R r)

Surface area and volume of frustum of a cone.We know that frustum of a cone is a part of the cone ABC

obtained by cutting off the smaller cone A'B'C parallel to the base.

Discuss :Compare a frustum with cone andcylinder. What similarities anddifferences do they have?

r

h

R

A

C

BA

B

l

394 UNIT-16

Hence,

Lateral surface area = Lateral surface – Lateral surfaceof frustum area of cone ABC area of cone A'B'C Total surface area of frustum = LSA of frustum + Area of the two circular bases

Volume of frustum = Volume of original Cone ABC – Volume of smaller Cone A'B'CSurface area and volume of a frustum of a cone can be found by using the above

method only when the radius of the base and slant height of both the original (bigger)cone and the smaller cone are given.

Then, how to find the surface area and volume of a frustum when onlythe heights and radii of the two bases of the frustum are given?

Let h be the height, l be the slant height and r1, and r2 be the radii ofthe two bases (r1> r2) of the frustum of a cone.

Then, we can directly find the surface area and volume of the frustum by using theformulae given below:

* Lateral surface area of the frustum of the cone

= (r1+r2)l , where, l = 2 2

1 2h +(r r )

* Total surface area of the frustum of the cone = (r1+r2)l +r12+r2

2

= {(r1+r2)l +r12+r2

2}, where, l = 2 21 2h +(r r )

* Volume of the frustum of the cone = 13h (r1

2+r22+r1r2)

[ Note : These formulae can be derived using the area of similarity of triangles but weshall not be doing derivations here. We will use the formulae to solve problems.]

ILLUSTRATIVE EXAMPLES

Example 1 : From the top a conical shaped jaggery of base diameter 10 cm and height10 cm, a small cone is cut off by slicing parallel to the base, 4 cm from the vertex.Find the surface area and volume of the frustum so obtained.

Sol. Diameter of the cone ABO = 10cm radius r1 = 102 = 5cm

Height of the cone ABO = h1 = 10cm

Height of the cone AI BI O = h2 = 4cm

First, let us find the other required data.

Radius of cone AI BI O = r2Slant height of cones ABO and AI BI O , l1 and l2 respectively.

h

P BA

a

r1

r2

l

r

P

5cm

AI

O

BA

B

4cm

10cm

I

Mensuration 395

1 1

2 2

r hr h 2

5 10r 4 r2 =

5 410

= 2cm

l12 = h1

2+ r12 = 102+ 52 = 100 + 25 = 125 l1

= 125 5 5 cm

l22 = h2

2+ r22 = 42 + 22 = 16 + 4 = 20 l2

= 20 2 5 cm

(i) Lateral surface area of cone ABO = r1l1 = 5 5 5 25 5

LSA of cone ABO = 25 5 sq.cm

Lateral surface area of cone AI BI O = r2l2 = 2 2 5 4 5

LSA of cone AI BI O = 4 5 sq.cm

Lateral surface area of frustum = LSA of cone ABO – LSA of cone AI BI O

= 25 5 4 5 21 5 3 21 225

7 1 = 66 5 sq. cm

LSA of frustum = 66 5 sq. cm. or 147.84 sq.cm

(ii) Total surface area of cone ABO = r1 (r1+l1) 5 5 5

TSA of cone ABO = 25 5 sq.cm

Lateral surface area of cone AI BI O = r2l2 = 2 2 5 4 5

LSA of cone AI BI O = 4 5 sq.cm

Total surface area of frustum = TSA of cone ABO - LSA of cone AI BI O + r22 5 - 4 5 + 4

25 25 5 4 5 4

29 21 5

22 75.957

TSA of frustum = 238.70 sq.cm

(iii) Volume of cone ABO = 13 r1

2 h1 = 13 5 5 10

Volume of cone ABO = 2503

cu.cm

Volume of cone AI BI O = 13 r2

2 h2= 13 2 2 4

396 UNIT-16

Volume of cone AI BI O = 163 cu.cm

Volume of the frustum = Volume of cone ABO Volume of cone AI BI O

= 2503

– 163 =

2343

Volume of the frustum = 78 cu.cm or 245.14 cu.cmWe can also directly find the lateral surface area, total surface area and volume ofthe frustum by using the formulae. Use the formulae and verify.

Example 2 : The radii of two circular ends of a frustum shaped dust bin are 15cm and8cm. If its depth is 63cm, find volume of the dust bin.Sol. Given r1=15cm, r2 = 8cm, h = 63 cm

Volume of the dustbin (frustum) = 13h (r1

2+r22+r1r2)

= 1 22 633 7

(152+82+15×8)

= 1

13

2271

639

3

(225+64+120)

= 66 × 409 = 26,994 Volume of the dustbin = 26,994 cu.cm

Example 3 : From the top of a cone of base radius 12cm and height 20cm, a small coneof base radius 3cm is to be cut off. How far down the vertex is the cut to be made? Findthe volume of the frustum so obtained.Sol. Given, r1 = 12cm, h1= 20cm, r2 = 3cm, h2 =?

we know, 1 1

2 2

r hr h 2

12 203 h h2 =

3 20 5 cm12

The cut to be made at 5cm from the vertex.

Volume of the frustum = 13h (r1

2+r22+r1r2)

= 1 22

153 7

(122+32+12×3) = 110

7×(144+9+36)

= 110

7×189 = 2,970

Volume of the frustum = 2,970 cu.cmExample 4 :The slant height of the frustum of a cone is 4cm, and the perimeter of itscircular bases are 18cm and 6cm respectively. Find the curved surface area and totalsurface area of the frustum.Sol. Given, 2r1 = 18cm, 2r2 = 6cm, l = 4cm

r1 = 9

cm r2 = 3

cm

63 c

m

8 cm

15 cm

Mensuration 397

Curved surface area of the frustum = (r1+r2)l

= 9 3

4= 12 4

CSA of the frustum = 48 sq.cm Total surface area of the frustum =(r1+r2)l +r1

2+r22

= 489 9 3 3

48 + 81

+ 9

= 48 + 81 7 9 7

22 22 = 48 + 25.8 + 2.9

TSA of the frustum = 76.7 sq.cm

EXERCISE 16.3

1. A flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44cm and 8.4cm. If the depth is 14 cm, find how much water it can hold?

2. A bucket is in the shape of a frustum with the top and bottom circles of radii15cm and 10cm. Its depth is 12 cm. Find its curved surface area and total surfacearea. (Express the answer in terms of

3. From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height17cm is cut off. What is the volume of the remaining frustum of the cone?

4. A vessel is in the form of a frustum of a cone . Its radius at one end is 8 cm and the

height is 14 cm. If its volume is 5676

3cm3, find the radius of the other end.

5. A container, opened from the top and made up of a metal sheet, is in the form of afrustum of a cone of height 16cm with radii of its lower and upper ends as 8 cm and20 cm, respectively. Find the cost of the milk which can completely fill the container,at the rate of ` 40 per litre. Also find the cost of metal sheet used to make thecontainer, if it costs ` 8 per 100 sq cm (take = 3.14)

SphereLet us conduct an activity as done in the previous section. Take acircular disc and paste a string along one of its diameter. Rotate itand observe the new solid formed.

What does it look like? It is called a sphere.If a circular disc is roated about one of its diameters, the

solid thus generated is called sphere.The centre of the circle, when it forms a sphere on rotation becomes the centre of

the sphere. This centre point will be equidistant from all points in the space which haveformed the sphere.

A

B

398 UNIT-16

Therefore, a sphere is a three dimesnional figure or solid figure, which is made upof all points in the space, which lie at a constant distance called the radius and from afixed point called the centre of the sphere.

The word solid sphere is used for the solid whose surface is a sphere.

It has both surface area and volume.

There are many things around us which are

sphere - shaped. Observe some examples.

How many surfaces do you see in the sphere - shapedobjects?

There is only one which is curved.

Now, let us assume that a solid sphere like a ball be sliced exactlythrough the middle with a plane that passes through its centre.

The sphere gets divided into two equal parts. Each part of the sphereis called a hemisphere.Surface area of sphere and hemisphere

The formula used to find the surface area of a sphere can be obtained through aninteresting activity. You are familiar with the game played using a top, by winding a thickthread around it. The following activity where a thread is wound around the sphericalobject will help you to generate the formula for finding the surface area of a sphere. Studythe activity and try to conduct it in groups.

• Take a plastic ball.

• Fix a pin at the top of the ball.

• Wind a uniform thread over the ball so as to coverthe whole curved surface area. Use pins to fix thethread in place.

• Mark the starting and finishing points on the thread.

• Slowly unwind the thread and measure the length of the thread used to wind aroundthe surface area of the sphere.

• Cut the thread into four equal parts.

• Measure the diameter of the ball and find its radius.

• Take a sheet of paper and draw four circles with radius equal to the radius of theball.

• Fill each circle with one piece of the thread as shown in the figure.

What do you observe and conclude from this activity?

The thread which had covered the surface area of the sphere (ball) has completelyfilled the four regions of the circle, all of the same radius as of the sphere.

This suggests that surface area of the sphere is four times the surface area ofcircles, where both the sphere and the circles have same radius (r).

Ball Marble

Mensuration 399

Curved surface area of the sphere = 4 × Area of the circle = 4 × r2 = 4r2

The curved surface area of the sphere = 4r2 sq. units

2The surface area of the sphere = 4 r sq.units

It is interesting to note that the surface area of a sphere is equal to the curvedsurface area of a cylinder just containing it.

Observe the figure

A sphere with radius (r) is placed inside a cylinder whose radius isalso 'r' and height is '2r'.

Surface area of sphere = 4r2

Curved surface area of cylinder = 2rh = 2r × 2r = 4r2

Surface area of the sphere = Curved surface area of the cylinder,

where, radius of the sphere and the cylinder are equal and height of the cylinder isequal to the diameter of the sphere.

Now, let us find the surface area of hemisphere.

We know that hemisphere is exactly half of a sphere. Let us usethis idea and find surface area of hemisphere.

(i) Curved surface area of solid hemisphere

C.S.A of solid hemisphere = CSA of solid sphere

2 = 24 r

2 = 2r2

CSA of solid hemisphere = 2r2 sq. units(ii) Total surface area of solid hemisphere

TSA of solid hemisphere = CSA of hemisphere + surface area of base circle

= 2r2 + r2 = 3r2

T.S.A of solid hemisphere = 3r2 sq. unitsYou might have observed the spherical object used to decorate

christmas trees, during birthdays etc.

This sphere is made up of several cones.

Imagine a sphere which is formed by using several tiny coneswhose radius is 'R' and height is 'h'

It can be considered that a sphere is made up miniature coneswhose height 'h' is equal to the radius 'r' of the sphere and each having a circular base.

Volume of each cone = 13

× Area of base × height

Volume of cone 1 = 13

× B1 × r Volume of cone 2 = 13

× B2 × r

Volume of cone 3 = 13

× B3 × r Volume of cone n = 13

× Bn × r Volume of sphere = Sum of the volumes of all cones

r

Solid Hemisphere

400 UNIT-16

= 13

B1r + 13

B2r + 13

B3r+........+13

Bnr = 13

r(B1 + B2 + B3 + .......Bn)

We observe that the sum of base areas of all the cones which have formed thesphere is equal to the surface area of the sphere. Since, the surface area of the sphere is4r2 , we get

Volume of sphere = 13

r(4r2) = 34 r3

Volume of sphere = 34 r3 cubic units

There is an interesting relationship between volume of a sphere and a cone whoseradii (r) are same.

Study the relationship.

If a cone is inscribed within the sphere as shown in the figure,

Then volume of the sphere is four times the volume of the cone.

Volume of sphere = 4 × volume of cone

= 4 × 13r2h = 4 ×

13r2 × r = 4 ×

13r2 =

43r3

Volume of sphere = 43r3

Volume of a solid hemisphere

Volume of solid hemisphere = 12 × volume of solid sphere =

3 31 4 2r r2 3 3

ILLUSTRATIVE EXAMPLES

Example 1 : Find the surface area of a sphere of radius 21 cm.

Sol. Surface area of a sphere = 4r2 = 2247

213 221cm

Surface area of the sphere = 5544cm2

Example 2 : Find the CSA and TSA of a solid hemisphere of radius 14 cm.

Sol. CSA of a hemisphere = 2r2 = 22

27

142 214cm

CSA of the hemisphere = 1232 cm2

TSA of a hemisphere = 3r2 = 22

37

142 214cm

TSA of the hemisphere = 1848 cm2

Example 3 : Find the volume of a sphere of radius 3 cm.

Sol. Volume of a sphere = 34 r3

= 43

22 37

33 3cm = 113.14 cm3

Mensuration 401

Example 4 : A hemispherical bowl has inner diameter 9 cm. Find the volume of milkit can hold.

Sol. d = 9 cm r = 92 cm

Volume of a hemisphere = 32

r3

= 23

22 97

3

239 9 cm

2 2 = 35346 190.92cm

28 Volume of hemispherical bowl = 190.92 cm3

Volume of milk it can hold = 190.92 ml [1cm3 = 1ml]

Example 5 : A hemispherical bowl of internal radius 18 cm is full of fruit juice. Thejuice is to be filled into cylinderical shaped bottles each of radius 3 cm and height 9cm. How many bottles are required to empty the bowl?

Sol. Hemisphere: r = 18 cm, V = ? Volume of a hemisphere = 32 r3

Volume of hemispherical bowl = 32 22

18 18 18cm3 7

Cylinder: r = 3, h = 9 Volume of a cylinder = r2h

Volume of 1 bottle = 322 3 3 9cm7

No. of bottles = Volume of fruit juice in hemispherical bowlVolume of fruit juice in 1cylindrical bottle

=

23

227

186

2

186

182

227

3 3 9 = 48

Number of bottles required = 48

Example 7 : The diameter of a metallic sphere is 4.2 cm. It is melted and recast intoa right circular cone of height 8.4 cm. Find the radius of the base of the cone.

Sol. Sphere: d = 4.2 cm r = 4.22 = 2.1 cm Cone: h = 8.4 cm

Volume of sphere = Volume of cone

3s4 r3

= 2c c1 r h3

2Cr =

43

3Sr

13 C

h = 4 2.1 2.1 2.1

8.4 2.1= 2.1 × 2.1

rC= 2.1 2.1 = 2.1 cm

radius of base of the cone = 2.1 cmExample 8 : The internal and external diameters of a hollow hemispherical shell are6cm and 10 cm respectively. It is melted and recast into a solid cone of base diameter14 cm. Find the height of the cone so formed.

Sol. Hollow hemisphere: External radius, R = 102 = 5cm, Internal radius , r =

62 = 3 cm

402 UNIT-16

Cone: r = 142 = 7 cm, h = ?

Volume of hollow hemisphere = Volume of cone

3 32

(R r )3

= 21

r h3

h=

23

3 3(R r )

13

3 3

22

2(R r )rr

= 3 3

2

2(5 3 ) 2(125 27)497

= 98

22

49 = 4 cm

height of the cone of formed = 4cmExample 9 : The radii of two solid metallic spheres are r1 and r2. The spheres aremelted together and recast into a solid cone of height (r1 + r2). Show that the radius ofthe cone is 2 21 2 1 22 r r r r .

Sol. Volume of (sphere 1 + sphere 2) = Volume of cone

3 31 24 4

r r3 3

= 2

C 1 21

r (r r )3

3 31 24 r r3

= 2C 1 21 r (r r )3

rC=

43

3 31 2r r

13 1 2

r r = 1 24 r r

2 21 1 2 2

1 2

r r r r

r r

2Cr = 2 2

1 1 2 24 r r r r

rC= 2 2

1 1 2 24 r r r r = 2 21 1 2 22 r r r r

radius of the base of cone = 2 21 1 2 22 r r r r

Example 10 : A solid sphere of radius 1 cm is melted to stretch into a wire of length100 cm. Find the radius of the wire.Sol. Sphere r = 1 cm

Wire is a cylinderCylinder h = 100 cm, r = ?Volume of sphere = Volume of wire (cylinder)

3S4 r3

= 2cy cyr h

2Cyr =

43

3sr

cyh=

3s

cy

4r3 h =

4 31 1 1cm3 100 25cm

= 21

cm75

rCy= 1 1 cm75 5 3

= 0.11 cm

radius of the wire = 0.11 cm

Mensuration 403

EXERCISE 16.4

1. Find the surface area of a sphere of radius (i) 14 cm (ii)2.8 cm (iii) 6.3cm

2. Find the TSA of a hemisphere of radius 5 cm.

3. A hemispherical bowl made of wood has inner diameter of 10.5 cm. Find the cost ofpainting it on the inside at the rate of `12 per 100 sq. cm.

4. Calculate the surface area of the largest sphere that can be cut out of a cube of side15 cm.

5. The surface area of a solid hemisphere is 432 cm2. Calculate its radius.

6. A hemispherical bowl made of steel is 0.25 cm thick. The inner radius of the bowl is5 cm. Find the outer curved surface area of the bowl.

7. Find the volume of a sphere whose radius is (i) 7 cm (ii) 10.5 cm

8. The diameter of a metal ball is 3.5cm. What is the mass of the ball, if the density ofthe metal is 8.9g/cm3 [Hint: Mass = Volume × density)

9. Find the volume of a sphere whose surface area is 154 cm2.

10. The inner radius of a hollow sphere is 10 cm. Find its volume.

11. The volume of a solid hemisphere is 1152cm3. Find its curved surface area.

12. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How muchmedicine is needed to fill this capsule?

13. A solid hemisphere of wax of radius 12 cm is melted and made into a cone of baseradius 6 cm. Calculate the height of the cone.

14. A right circular metallic cone of height 20 cm and base radius 5 cm is melted andrecast into a sphere. Find the radius of the sphere.

15. The diameter of a metallic sphere is 18 cm. It is melted and drawn into a wirehaving diameter of cross section 0.4 cm. Find the length of the wire.

16. The internal and external radii of metallic sherical shell are 4 cm and 8 cmrespectively. It is melted and recast into a solid right circular cylinder of height

193

cm. Find the diameter of the base of the cylinder.

17. A solid metallic sphere of diameter 28 cm is melted and recast into a number of

smaller cones each of diameter 2

43

cm and height 3cm. Find the number of cones

so formed.

Combination of solidsYou are familiar with solid objects like cube, cuboid, prism, pyramid, cylinder, cone andsphere. You know how to find their surface area and volume. It is common to identifyobjects from our surroundings with these solid shapes.

404 UNIT-16

Now, observe the following objects and their shapes. Discuss in groups and identifythe shape of each one of them.

Toy Oil tanker Rocket model Green-house shed

In each case, the object is made of two or more solid objects. Hence, each one ofthem is a combination of solids.Toy Oil tank

Cone + Hemisphere

Rocket Green-house

Cone + CylinderCuboidHalf of cylinder +

How to find the surface area and volume of such solid objects? Let us learn to findthe surface area and volume of combination of solids.

Surface area of combination of solidsConsider the example of the toy. We observe that the total surface area

of the toy is sum of the curved surface area of the hemisphere and curvedsurface area of the cone.

TSA of the toy = CSA of the hemisphere + CSA of the cone

CSA of the toy = CSA of the hemisphere + CSA of the cone

Similarly, for the other examples we can find the total surface area and curvedsurface area as follows.

TSA of oil tank = CSA of hemisphere + CSA of cylinder + CSA of hemisphereCSA of oil tank = CSA of hemisphere + CSA of cylinder + CSA of hemisphere

TSA of rocket model = CSA of cylinder + Area of its base + CSA of coneCSA of rocket model = CSA of cylinder + CSA of cone

Hemisphere + Cylinder + Hemisphere

Toy

Mensuration 405

TSA of shed = LSA of cuboid + Area of its base + Half of TSA of cylinder

LSA of shed = LSA of cuboid + Half of TSA of cylinder

From the above examples, we can conclude that, total surfacearea of the combination of solids may or may not be the sum ofcurved surface areas of the solids which are combined together.

We can note the following observations.

• The total surface area of a combiantion of solids is not simply the sum of totalsurface area of solids combined together, because some part of the total surfacearea disappears in the process of joining them.

• It is also not simply the sum of curved surface area of solids joined together becausesome part other than the curved surface area has to be added to its total surfacearea.

• The curved surface area of a combination of solids may be equal to the sum ofcurved surface areas of the solids combined together.

• The curved surface area of a combination of solids may not be equal to the sum ofthe curved surface areas of solids combined togehter, because some part other thanthe curved surface area has to be added.

Now let us solve some problems based on the surface area of combination of solids.

ILLUSTRATIVE EXAMPLES

Example 1 : A cylindrical container of radius 6 cm and height 15 cm is filled with icecream. The whole ice cream has to be distributed to 10 children in equal cones withhemispherical tops. If the height of the conical portion is 4 times the radius of itsbase, find the radius of the ice cream cone.

4r r r

6 cm

15 c

m

Sol. Volume of cone with hemispherical top

= 2 31 2

r h r3 3

= 2 31 2

r 4r r3 3

= 3 34 2

r r3 3

= 6

2

33r = 2r3

Volume of 10 cones with hemispherical tops = 10 × 2r3 = 20r3

Volume of ice-cream in cylindrical container = r2h × 6cm × 6cm ×15cm = 540 cm3

Volume of 10 cones with hemispherical tops=Volume of cylinder containing ice-cream

20r3 = 540 r3 = 540

27

20 = 27 r = 3 27

r = 3 cm

406 UNIT-16

Example 2 : A solid is in the form of a cone mounted on a right circular cylinder, bothhaving same radii of their bases. Base of the cone is placed on the top base of thecylinder. If the radius of the base and height of the cone be 7 cm and 10 cm respectivly,and the total height of the solid be 30 cm, find the volume of the solid.Sol. Cylindrical part: r = 7 cm, h = 30 – 10 = 20 cm

Conical part: r = 7 cm, h = 10 cmVolume of the solid = Volume of cylinder + Volume of cone

= r2h + 21

r h3

= 227

7 27 20cm + 1 223 7

7 7 10

= 3080 cm3 + 31540

cm3

= 3080 + 513.33 cm3

Volume of the solid = 3593.33 cm3

Example 3 : A toy is in the form of a cone mounted on a hemisphere with the sameradius. The diameter of the conical portion is 6 cm and its height is 4 cm. Determinethe surface area and volume of the solid.Sol. Conical portion Spherical portion

r = 3cm, h = 4 cm r = 3cml2 = r2 + h2 = 32 + 42 = 9 + 16l2 = 25

l = 25 = 5 cm

Surface area of the toy = CSA of hemisphere + CSA of cone

= 2r2 + rl = 2 222 222 3 3cm 3 5cm7 7

= 2 2396 330cm cm7 7

= 2 2726 cm 103.71cm7

Surface area of the toy = 103.71 cm2

Volume of the toy = Volume of hemisphere + Volume of cone

= 3 22 1r r h3 3

= 23

22 37

3 13 3cm3

22 37

33 4cm

= 3 3 3396 264 660

cm cm cm7 7 7

Volume of the toy = 94.28 cm3

Example 4 : A circus tent is made of canvas and is in the form of a right circular cylinderand a right circular cone above it. The diameter and height of the cylindrical part of thetent are 126 m and 5 m respectively. The total height of the tent is 21m. Find the totalcost of the canvas used to make the tent when the cost per m2 of the canvas is ` 15.

Sol. Cylindrical part: r = 126

2 = 63m, h = 21 – 5 = 16 m

A

B C

D E

30 c

m

10 cm

10 cm10 cm

7 cm7 cm

20 c

mA B

C

4 cm

3 cm3 cm

Mensuration 407

l = 2 2r h = 2 263 16

= 3969 256 = 4225

l = 65 m

Total canvas used = CSA of cylindrical part + CSA ofconical part

= 2rh + rl = 22

27

639 25m +

227

639 265m

= 1980 m2 + 12870 m2 = 14850 m2

Total canvas used = 14,850 m2

Cost of canvas at the rate of ` 16 per m2 = 14,850 × `15 = `2,22,750

Cost of canvas = `2,22,750

Example 5 : A solid is composed of a cylinder with hemispherical ends. If the wholelength of the solid is 104 cm and the radius of each hemispherical end is 7 cm, findthe cost of polishing its surface at the rate of ` 4 per 100 cm2

Sol. Hemispherical part: r = 7 cmCylindrical part: r = 7cm, h = 104 – (7 + 7) = 90 cmTotal surface area of the solid

= CSA of cylindrical part

+ 2 × CSA of hemispherical part

= 2rh + 2 × 2r2 = 2r (h + 2r)

=22

27

7 290 2(7) cm

= 44 × 104 cm2 = 4576 cm2

TSA of the solid = 4.576 cm2

Cost of polishing at the rate of ` 4 per 100 cm2

= 4576 × 4

100`

= ̀ 183.04

Cost of polishing the solid = ̀ 183.04

EXERCISE 16.5

1. A petrol tank is in the shape of a cylinder with hemispheres of same radius attachedto both ends. If the total length of the tank is 6m and the radius is 1m, what is thecapacity of the tank in litres.

2. A rocket is in the shape of a cylinder with a cone attached to one end and ahemisphere attached to the other. All of them are of the same radius of 1.5m. Thetotal length of the rocket is 7m and height of the cone is 2m. Calculate the volumeof the rocket.

126 m

21 m

16m

30 m 30 mO5 m

104

cm

90 cm

7 cm

7 cm

7 cm

7 cm

408 UNIT-16

3. A cup is in the form of a hemisphere surmounted by a cylinder. The height of thecylindrical portion is 8 cm and the total height of the cup is 11.5cm. Find the TSA ofthe cup.

4. A storage tank consists of a circular cylinder with a hemisphere adjoined on eitherends. The external diameter of the cylinder is 1.4m and length is 8m, find the costof painting it on the outside at the rate of ` 10 per m2.

5. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter ofthe base of the cone is 16 cm and its height is 15 cm. Find the cost of painting thetoy at 7 per `100 cm2.

6. A circus tent is cylindrical upto a height of 3m and concial above it. If the diameterof the base is 105 m and the slant height of the conical part is 53m, find the totalcost of canvas used to make the tent if the cost of the canvas per sqm is `10.

Scale drawing :We are familiar with the formulae used to find area of plane geometrical figures. Observethe figures given below and recall their formulae to find areas.

.

D b C

h

A

B

S

P

R

b

Ql

K

N

L

h

Mb Triangle Rectangle Parallelogram

A = 12 × base × height A = length × breadth A = base × height

A = 1/2 bh A = lb A = bh

A B

EdF

CD

h1h2

G

D E

Fb

a

h

Quadrilateral TrapeziumA = 1/2 × diagonal × sum of heights A = 1/2 × height × sum of parallel sides

A = 1/2 × d (h1+ h2) A = 1/2 h (a + b)

Now, observe the following figures

B

CA

E D

E D

C

BA

F G

A BC

DE

F

How are the areas of these figures found?

Mensuration 409

In our daily life we also come across situations where the figures formed have irregularshapes. Observe the figures given below.

In all such cases, where the figures are not having triangle or quadrilateral shapesand figures having irregular shapes, the simplest technique is to divide the given figureinto triangles and quadrilaterals and then find their areas.

Such figures can be divided into several familiar shapes as follows. study them.

Observe that each figure is divided into triangles and trapeziums.

How do you think that the total area of each figure is calculated by using the areas ofdifferent triangles and trapeziums formed? Discuss in the class in groups.

Do the following activity to understand this.

Activity 1:Take a graph sheet.

Draw a pentagon A C B E D.

Join AB, which forms the base line.

Draw perpendiculars from C, D and E to AB.

Marks the points F, G and H

Identify the triangles and trapeziums formed.

Read the measure of the length of their sides.

Calculate the area of each triangle ADF, ACBand EHB and area of the trapezium FHED.

Find the sum of these areas.

You have found the area of the pentagon. Repeat this activity for the other figuresdiscused earlier.

O 1–1–2–3–4–5X

Y

A

D

E

B

G

F

H

C

2 3 4 5

1

2

3

4

5

410 UNIT-16

Now consider the example of the piece of land. Suppose a farmer owns it and a landsurveyor wants to find its area. How is it done?

The procedure followed by surveyors while measuing and calculating areas of irregularshaped lands are as follows.

Irregular shaped field is divided into known geometrical shaped fragments.

Measurements of the sides are made using Guntor’s chain.

Measurements are recorded in the surveyor’s field book .

The sketch or figure repesenting the irregular shaped field is drawn to scale.

Area of each fragment is calculated.

Total area of the field is found by finding the sum of the areas of the fragments.

How are the measurements taken and recorded in the field book?

How is scale drawing done?

Do the following activity to understand this .

Activity 2Take a graph sheet.

Draw an irregular shape as shown in the figure.

Mark the extreme end points, whereverpossible.

First mark the top and bottom points to obtainthe base line by joining them. AB is the baseline.

Now, mark the other end points on either sideof the base line, let the points be C,D, E and F.

Draw perpendiculars from C,D,E and F to thebase line AB.

Let the perpendicular lines be CG, DJ, EK,and FB.

Join A and C, C and F, D and E, B and E and D and A.

Now you have obtained a regular shape ACFBED.

Read the measurements of lines formed and make entries in a table as given below.

Use the values in the table and find the area as discussed in activity 1.

The land surveyor’s method of recording the measurements of a land will be the same asdiscussed above. But the measurements are usually in meters. Hence, while drawingthe rough sketch, a convenient scale is taken and all the original measurements areconverted according to the scale.

Now study the following examples.

O X

Y

A

D

E

B

G

J

K

C

F1

2

3

4

5

1 2 3 4 5

Mensuration 411

ILLUSTRATIVE EXAMPLES

Example 1: Draw a plan and calculate the area of a level ground using the informationgiven below. Sol. Take suitable scale, say 20m = 1cm.

Convert the given measurements

200 m =120 × 200 = 10 cm

140 m = 120 ×140 = 7 cm

120 m = 120 × 120 = 6cm

40 m =120 × 40 = 2 cm

60 m =120 × 60 = 3cm

50m = 120 × 50 = 2.5cm

30 m = 120 × 30 = 1.5 cm

Draw the vertical base line AD = 10cm (200 m)

Mark points P, Q and R on AD such that AP = 2cm (40m),

AQ = 6cm (120 m) or PQ = 4cm ( 6-2 is taken)

AR = 7cm (140cm) or QR = 1cm ( 7-6 is taken)

Draw perpendiculars from P,Q and R with the given measurements.

PB = 1.5 cm (30m) QE = 3 cm (60m) RC = 2.5 cm (50m)

Join the points to get the figure ABCDE

Record the measurements.

Calculate the area of ABCDE.

Area of ABCDE = Area of ABP + Area of trap . PBCR+ Area of CRD+ Area of DEQ +Area of EQA.

=12 × AP × PB +

12 ×PR(PB+RC) +

12 × RC× RD+

12 × EQ × QD +

12 × EQ × QA

= 12 × 40×30 +

12 ×100(30 + 50) +

12 × 50 × 60 +

12 ×60×80+

12 × 60 × 120

= 600 + 4000 + 1500 + 2400 + 3600

Area of ABCD = 12,100 sq.m.

D

C

BP

A

E

R

Q

to D (meter)

200

140

120

40

From A

50 to C

30 to B To E 60

412 UNIT-16

Example 2: Plan out and find the area of the field from the following notes from thefield book.

Metre to D150100 70 to C

To E 80 8030 40 to BFrom A

Sol. Scale 20 m = 1 cmObserve that

AM = 30 m.

AN = 80 m.

MP = (100 – 30) = 70 m.

ND = (150 – 80) = 70 m.

PD = (150 – 100) = 50 m.

(1) Area of ABM = 1 1bh 30 40 600 sq.m.2 2

(2) Area of Trapezium MBCP = 1 1h a b 70 70 40 3850 sq.m.2 2

(3) Area of DPC = 1 50 70 1750 sq.m.2

(4) Area of DEN = 1 80 70 2800 sq.m.2

(5) Area of NEA = 1 80 80 3200 sq.m.2

EXERCISE 16.6

1. Draw a plan and calculate the area of a level ground using the information givenbelow.

Metre to C220

To D 120 210120 200 to B

To E 180 80From A

E80

P

20

7050

5040

30

A

M B

D

C

N

Mensuration 413

2. Plan out and find the area of the field from the data given from the Surveyor's fieldbook

Metre to E350

To D 100 300 150 to FTo C 75 250

150 100 to GTo B 50 50

From A

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

Metre to D225

To E 90 175125 20 to C

To F 60 100To G 15 80

60 70 to BFrom A

4. Calculate the area of the field shown in the diagram below:[Measurements are in metre]

D

40 30

C 35F

E

25

G

30

70

B

A

35

414 UNIT-16

ANSWERSEXERCISE 16.1

1] (i) 176cm2 (ii) 201.14 cm2 2] 7cm 3] 3168cm2 4] 5:4 5] 3:2 6] 88m2, ̀ 36967] 21,780cm2 8] r = 7 cm, h = 3.5 cm 9] ̀ 269.50 10] 44 m2

11] 5,544 cm3 12] 3,432 g 13] (i) 2.99cm` (ii) 141.42 cm3 14] 2 :115] 3,080 cm3 16] 7,39,200 cm3 17] 0.53m 18] ̀ 300

EXERCISE 16.21] 3960 cm2 2] 8 cm 3] 7.5 cm 4] 7cm and 462 cm2 5] 5,500 cm2 6] 4:3

8] (i) 183.33 cm3 (ii) 2,310 cm3 (iii) 7,546 cm3 9] 3:1 10] 415.8 cm3

11] 12 cm 12] 6.4cm 13] 3 a2 14] ̀ 2,068

EXERCISE 16.31] 1,408.58 cm3 2] CSA = 1,021.42 cm2, TSA = 1,335.71cm2 3] 26,148.57cm3

4] 5 cm 5] ̀ 418, ̀ 156.75

EXERCISE 16.41] (i) 2,464 cm2 (ii) 98.56 cm2 (iii) 498.96 cm2 2] 235.71cm2 3] ̀ 20.794] 707.14 cm2 5]12 cm 6] 173.25 cm2 7] (i) 1437.33 cm3 (ii) 4851 cm3

8] 199.8g 9] 179.66 cm3 10] 2095.24 cm3 11] 905.14 cm2 12] 22.45 mm3

13] 96 cm 14] 5 cm 15] 243 m 16] 16 cm 17] 672EXERCISE 16.5

1] 16,761.9l 2] 36.53m3 3] 253 cm2 4] ̀ 413.6 5] ̀ 58.08 6] ̀ 97,350EXERCISE 16.6

1] 49,300m2 2] 50,625m2 3] 15,250m2 4] 4,600m2

Cylinder

Statistics

Cone Frustum ofa cone

Sphere Hemisphere

Combinationof solids

Scale drawing

2 r(h+r)

r h2

2 rh rl

r (l+r)

r h213

(r +r )1 2

{(r +r )l+1 2r +r1 2

2 2

}

2 2

h(r +r +r +r )1 2 1 2

13

4 r2

r343

2 r2

3 r2

r323

LSA

TSA

Mensuration