Mensuration Notes

7/18/2019 Mensuration Notes

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MATHS 3.1

MENSURATION

3.1. AREA AND PERIMETER :

A surface is bounded by lines. A surface may be plane of curved. A solid figure is bounded by

plane or curved surfaces.

Plane figures like triangles, squares, etc. that are bounded by straight lines are called Rectilinear

figures

Mensuration means measurement of areas involve two dimensions, so area will be expressed in

square units, while volume (involving three discussions) will be expressed in cubic units.

AREA AND PERMIMETER OF PLANE FIGURES :

RECTANGLE :

ABCD is rectangle,

Having AB as length (=l ) and BC as breadth (= b)

Area = length × breadth = l × b

Perimeter = 2 (length + breadth) = 2 (l + b)

SQUARE :

In the given figure for a square, l = b.

Area = l × l = l 2

Perimeter = 4 l.

Notes : 1. It is correct to write “area = (10 × 10) sq. cm.”;

But it is wrong to write “area = 10 cm. × 10 cm.”.

2. 1 ft = 12 in., 1sq. ft. =122 or 144 sq. in.

1 m. = 10 dm., 1 sq. m = 102 or 100 sq., dm etc.

D C

BA l

b

BA

CD

b

l

STUDY NOTE-3



MATHS3.2

A

c

B Ca

c a2 2+

A DE

B a C

h

bc

F D E C

BA a

h b M E N S U R A T I O N



Example : To find Hypotenuse, when Base and Perpendicular are given :

By Pythagorus theorem, we know : square on AC = square on AB + square on BC

i.e., AC2 = c

2 + a

2 ; Hence, AC = 22 ac

Briefly, hypotenuse of right-angled triangle = 22 lar perpendicu base

Note : If any two sides are known, we may find the other side by the above formula.

Area = ½ × base × altitude = ½ × a × c

Perimeter = AB + BC + CA = c + a + 22 ca

Example : In the above figure, if AB = 3 cm. BC = 4 cm. find AC (hypotenuse), perimeter and

area of triangle.

AC = 22 BCAB = 22 43 = 169 = 25 = 5 cm

Perimeter = AB + BC + CA = 3 + 4 + 5 = 12 cm.

Area = ½ × base × altitude = ½ × BC × AB = ½ × 4 ×3 = 6 sq. cm.

Example : The sides of a right-angled triangle containing right angle are 9 and 12 cm. Find the

length of the perpendicular from the right angle to the hypotenuse.

Let, h = hypotenuse and required perpendicular = p.;

Now, h2 = 9

2 + 12

2; or, h = 15 cm , or, s = ½ (9 + 12 + 15) = 18

= 63918 = 54 sq. cm. Again , ½ × 15 p = 54 or, p = 7.2 cm.

ISOSCELES RIGHT-ANGLED TRIANGLE :

Here, Base = perpendicular

So, Hypotenuse = 22 ) base() base( = 2) base(2 = base 2

Area = ½ × base × base = ½ × (base)2

Perimeter = base + base + base 2 = 2 base + base 2 = base (2 + 2 ).

EQUILATERAL TRIANGLE :

Here, three sides are equal i.e., AB = BC = CA = a unit

Perpendicular AD bisects BC.

So, BD =

2

a unit.

A

B C

aa

Da/2

MATHS 3.3



Altitude, AD = 22 BDAB a2

3

4a

a2

2

i.e. Height of an equilateral triangle = side ×2

3

Area = ½ × base × altitude = ½ × a × 2

3

× a = 4

3

a

2

sq. units.

Perimeter = a + a + a = 3 a units.

Example : Find the area of an equilateral triangle of side 6 cm.

Area =4

3× 6

2 = 39 = 9 × 1.73 = 15.57 sq. cm.

Perimeter = 3 a = 3 × 6 = 18 cm.

Height of the triangle = side ×6

3 = 6

2

3 = 33 = 3 × 1.73 = 5.19 cm.

PARALLELOGRAM :

Area of ABCD = 2 × area of EBC = 2 × ½ × base × height = b × h sq. units.Therefore, Area of parallelogram = base × height

Example : The base of a parallelogram is 6 cm. and height is 1 cm. 2 dm. Find its area.

Here, base = 6 cm, height = 1.2 cm.

Therefore, area = base × height = 6 × 1.2 = 7.2 sq. cm.

Note. If all the sides are equal, then parallelogram becomes a Rhombus.

RECTANGLE :

ABCD is a rectangle. BC is base, AB is height. Area of any parallelogram = base × height.

Area of rectangle ABCD = BC × AB = length × breadth

Area of rectangle ABCD = BC × AB = length × breadth

Area = length × breadth

B b C

DEA

h

F

A D

CB

MATHS3.4

M E N S U R A T I O N



MATHS

Perimeter = AB + BC + CD + DA

= 2 BC + 2 AB = 2 (BC + AB) = 2 (length + breadth).

SQUARE :

In the rectangle ABCD, if all the sides are equal, i.e., AB = BC= DA (=l ), then it is a square.

Area = length × breadth = length × length = (length)2 sq. units.

Perimeter = l + l + l + l = 4 l units.

To find Area of Path (inside or outside of a rectangle) :

Inside : ABCD is a rectangle, a path of width x lies inside ABCD and enclosing EFGH (inside

ABCD), as shown in the given figure.

Area of the path = area ABCD – area EFGH = BC × AB — FG × EF

= BC × AB – (BC – 2x) (AB – 2x)

Outside : Area of the path = area EFGH – area ABCD = FG × EF – BC × AB

= (BC + 2x) (AB + 2x) – BC × AB.

Example : A path 4 ft. wide is to be laid round a rectangular garden of 60 ft. by 40 ft. on its inner

side. If the cost per sq. ft. is Rs. 2.50; find the total cost required for making the path.

Area of the path = 60 × 40 – (6 – 2 × 4) (40 – 2 × 4)

= 2400 – (60 – 8) (40 – 8) = 2400 – 52 × 32 = 2400 – 1664 = 736 sq. ft.

Therefore, required cost = 736 × 2.50 = Rs. 1,840.

Example : A path 3 ft. wide is to be laid outside, round a rectangular grass plot 100 yds. by 60

yds. Find the cost of macadamizing it at Rs. 1.15 per sq. yds.

Area of the path = (100 + 2) (60 + 2) – 100 × 60, as 3ft. = 1 yd.

CB

DA

l

l

A DH

x

GCB

F x

E

E HD

x

CGF

B x

A

MATHS 3.5



MATHS3.6

= 102 × 62 – 6000 = 6324 – 6000 = 324 sq. yds. Required cost = 324 × 1.15 = Rs. 372.60

RHOMBUS : A D

CB

O

In case of a rhombus diagonals bisect one another, at right angles.

i.e. AO = CO = ½ × AC

Area of rhombus = 2 × area of ABD

= 2 × ½ BD × AO = BD × AO = BD ×2

AC=

2

1× BD × AC

Therefore, area of rhombus = ½ × product of diagonals.

Example : The perimeter of a rhombus ABCD 48 cm. and one of the diagonals AC is 8 cm. Show

that the other diagonal is 216 cm. Find its area. [ICWA (F) June, 2004]

AB + BC + CD + DA = 4 BC = 48 or, BC = 12 cm.

Here, OC = ½ × 8 = 4 cm., BC = 12,

BO2 = BC

2 – CO

2 (as BOC = 90°) or, BO

2 = 144 – 16 = 128

BO = 128 = 28 BD = 2 × 28 = 216 cm.

Area = ½ × AC × BD = ½ × 8 × 216 = 264 (= 90.24) sq. cm

POLYGON :

A Polygon is a closed figure bounded by straight line.

Here ABCDEF is a polygon of 6 sides.

F E

D

CB

A

REGULAR POLYGON :

A regular polygon has all its sides equal and so also the angles.

Polygons are classified as follows : —

No. of sides 3 4 5 6 7 8 9 10

Name triangle quadrilateral Pentagon hexagon heptagon octagon novagon decagon




MATHSMATHS 3.7

O

B

A

r

O

A B

C

A B

O

E

D



or, AE2 = h(d – h) where, CE = h and CD = d

or,2

2AB

= h (d – h), or, AB

2 = 4h (d – h)

AB = 2 )hd(h

Fillet :

A circle is inscribed within a square touching the sides.

Then the shaded portions as shown in the figure represent the fillets.

Thus we get 4 fillets.

Area of 4 fillets = Area of square – Area of circle.

and Area of 1 fillet = ¼ (Area of square – Area of circle).

Circular ring (or Annulus) :

Two concentric circles form a circular ring.

Now area of circular ring (or annulus) = Area of outer circle – Area of inner circle

= R 2 – r

2 = (R + r) (R – r)

where, R = radius of outer circle. r = radius of inner circle.

Example : The inner diameter of a circular building is 20 m. and the thickness of the wall is 1 m;

what is the area occupied by the wall?

Inner diameter = 20 m.; outer diameter = 20 + 2 = 22 m.

Inner radius (r) = 10 m.; outer radius (R) = 11 m.

Area of wall = (R + r) (R – r) =7

22(11 + 10) (11 – 10) =

7

22× 21 ×1 = 66 sq. m.

O

r

R

MATHS3.8




MATHS

Example : The side of a square is 14 cm. Find the area of a fillet.

Area of 4 fillets = 142 – 7

2, as radius = ½ × 14 = 7 cm.

= 196 –722

× 7 × 7 = 196 – 154 = 42 sq. cm.

Area of a fillet = ¼ × 42 = 10.5 sq. cm.

SUMMARY :

Triangle () Area =21

× base × altitude ; () Area = )cs)( bs)(as(s

[Where, s =2

c ba ; and perimeter = a + b + c]

Equilateral Triangle : () area =4

3× a

2, perimeter = 3a.

Parallelogram : Area = base × height

Rectangle : Area = length × breadth, Perimeter = 2 (length + breadth)

Square : Area = a2, Perimeter = 4a

Rhombus : area = ½ × (product of diagonals)

Circles : Circumference = 2r, Area = r 2

SOLVED EXAMPLES :

1. The three sides of a triangle are 3, 4 and 5 cm. respectively; find its area.

S = ½ (3 + 4 + 5) = ½ × 12 = 6 cm.

Area of triangle = )cs)( bs)(as(s = )56)(46)(36(6 = 1236 = 6 sq. cm.

2. The perimeter of an isoosceles triangle is 544 cm. and each equal side is 5/6 of the base. Find

its area.

Let each of equal side = a cm ., base = b cm.

Now, a + a + b = 544. or, 2a + b = 544, or , 544 b6

b52

or, 544 b6 b10 , or, 544

6 b16

i.e. b =165446

= 204 cm.

a =65

× 204 = 5 × 34 = 170

Now, s = ½ × 544 = 272

)170272)(204272)(170272(272

10268102272 = 10210268684

= 2 × 68 × 102 = 13872 sq. cm.MATHS 3.9



MATHS3.10


3. The perimeter of a right-angled triangle is 30 cm.; its bypotenuse. Is 13 cm. Find the other

sides [ICWA (F) Dec. 2005]

Let the other sides be x and y now x + y 13 = 30 or, x + y = 17………(i)

Again 132 = x

2 + y

2 = x

2 + (17 – x)

2

or, 169 = 2x2 – 34 x + 289

or, x2 – 17x + 60 = 0 or, (x–5)(x–12) = 0

or, x = 5, 12,; y = 12, 5 from (i)

reqd. sides are 5 : 12 cm.

4. A piece of sheet steel weighing 7 lbs. per sq. ft. is cut in the form of a triangle, lengths of sides

37.5 inch. 10.5 inch and 36 inch. respectively.

Find (i) the area of triangle,

(ii) the weight of the triangle,

(iii) the length of perpendicular to the longest side from the opposite angle.

s = ½ (37.5 + 10.5 + 36) = 42 in.

(area) (42 4.5 31.5 6) = 189 sq. in =144

189sq. ft. = 1.313 sq.ft

Hence, Weight = 1.313 × 7 = 9.191 = 9.2 lbs (app.)

Taking p as required perpendicular,2

1 × base × perpendicular

or, 189 = ½ × 37.5 × p or, 10.1 inch (app)

5. The sides of a triangle are in the proportionof 9 : 10 : 11. The perimeter is 300 cm. Find its area.

9030011109

9

a , ;10030030

10

b

1103003011

c

s = ½ (90 + 100+ 110) = 150

Area = )110150)(100150)(90150(150 = 3000 2 sq. cm. (on reduction)

6. The diagonal of a rectangle is 5 cm. If the sum of its length and breadth is 7 cm, find its area.

[ICWA (F) June, 2000]

Given, length + breadth = l +b = 7, …………(i), here diagonal represents the hypotenuse of the

right-angled triangle having length (l ), breadth (b), so that :

l 1

+ b2 + = 5

2 or, l

2 + (7 – 1)

2 = 25 or, l

2 – 7l + 12 = 0 or, (l –3) (l –4) = 0

i.e. 1 = 3,4

and b = 4, 3 Hence, the sides are 4 cm. and 3 cm.

Hence, area of rectangle = length × breadth = 4 × 3 = 12 sq. cm.

7. The sides of an equilateral triangle are shortened by 1 cm. 1 cms. and 3 cms. respectively, then

a right angled triangle is formed. Find the length of each side of the equilateral tringle.

[ICWAI(Prel.) June 1988]

Let length of each side of equilateral triangle = x cm.



MATHS

After shortened, the sides will be : (x –1), (x –2) cms. respectively.

Again, (x –1)2 = (x – 2 )

2 + (x – 3)

2, as the square on the hypotenuse is equal to the sum of quares

of other two sides.

Now, x2 – 2x + 1 = x

2 – 4x + 4 + x

2 – 6x + 9

or, x2 – 8x + 12 = 0 or, x

2 – 2x – 6x + 12 = 0

or, x(x–2) – 6(x–2) = 0 or, (x – 6) (x –2) = 0 i.e. x = 6, 2 (inadmissible)

Required length is 6 cm.

8. A bicycle wheel makes 5,000 revolutions in moving 11km. Find the diameter of the wheel.

Let radius of the wheel = r cm., circumference = 2 r

Now, 2 r × 5000 = 11 × 1000 × 100, as. 11km = 11× 1000 × 100 cm.

or, 2 ×722

× r × 5000 = 1100000. or, r = 35 cm.

Hence, diameter = 2r = 70 cm.

9. A road which is 7 meters wide surrounds a circular park whose circumference is 352 metres.

Find the area of the road.

Circumference of park = 2 r = 352 [r = radius in meters] or, 2 ×7

22 × r = 352

r = 56 m. as width = 7 m. so, radius R of outer circle (including road) = 65 + 7 = 63 m

Hence, area of road = (R 2 – r

2) =

722

× (R – r) (R – r) =722

(63 + 56) 63 – 56)

=722

× 119 × 7 = 22 × 119 = 2618 sq. m.

10. Find the area of the road which is 7m. Wide and is around but outside a circular park, whose

circumference is 88 metre. What would be the total cost to develop the road of at the rate of

Rs. 100 per sq. metre. and to develop the park at the rate of Rs. 150 per sq. metre?

[ICWA (F) June, 2007]

circumference of circular park = 2 r = 2.7

22. r = 88 or, r = 14 metre. Circular Road is 7

metre wide, so outer radius (R)= 14 + 7 = 21 m.

Area of circular road = (R 2 – r

2) =

7

22 (21

2 – 14

2) =

7

22× (21 + 14) (21 – 14) =

7

22× 35 × 7

= 770 sq. m.

Total cost = 770 × 100 = Rs. 77,000

Again total cost of park = × 142× 150 =

722

× 14×14×150 = Rs 92,400

11. A number of circular pieces of 0.25 cm. radius is to be cut from a metal sheet of dimension

11 cm by 2 cm. Find the possible number of such pieces. [ICWA (F) June, 2006]

Area of circular piece = r 2 =

722

×

4

1 2 =

722

×161

sq. cm.

Area of metal sheet = 11 × 2 sq. cm.MATHS 3.11



MATHS3.12


reqd. no of pieces =

161

722

211

= 11 × 2 ×22

7×

161

= 112.

SELF EXAMINAION QUESTIONS :

1. The perimeter of a right angle triangle measures 234 cm and the hypotenuse measures 97 cm:

find the other two sides. [Ans. 72 cm; 65 cm]

2. Find the area of a triangle whose sides are 3 cm. 4 cm. and 5 cm. [Ans. 6 sq. cm]

3. The sides of a triangle are in the ratio of 4 : 6 : 7 and the perimeter is 34 cms. Find its area.

[Ans. 41.91 sq. cm.]

4. The are of rectangle is 102 sq. ft and its perimeter is 4l ft. What are its dimensions?

[Ans. 12 ft. 8.5 ft]

5. A rectangle lawn 20 m long and 12 m wide is bordered by a path 2 m wide. Find area of the

path. [Ans. 144 sq.m]

6. Find the area of a right angled isosceles triangle whose hypotenuse is d cm.

[ICWA (F) June ’ 89]

[Ans.22

da sq. cm, where a = length of equal side.]

7. The radius of a circle is 3 m. What is the circumference of another circle, whose area is 40

times that of the first ? [Ans. 132 m]

8. How many times will the wheel of a car rotate in a journey of 1925 m., if it is known that the

radius of the wheel is 49 cm? [Ans. 625]

9. A circle of diameter 14 meter is inscribed within a square touching the sides. Find the area of a

fillet thus formed, and cost of developing one fillet at the rate of Rs. 200 per sq. m

[ICWA (F) Dec. 2006][Ans. 10.5 sq. cm; Rs. 2100]

3.2 VOLUME, SURFACE AREA OF SOLD FIGURES :

A space bounded by one or more surfaces (plane or curved) is called a solid figure (or solid). If we

measure the areas of all such surfaces, their sum is called the surface area of the solid figure.

Surface area is expressed in square units.

Now the total space occupied by a solid is called its volume. It is the number of cubic units o

measure contained in the three dimensional space occupied by the solid.

The plane or curved surfaces bounded bounding a solid are called Faces. Intersection of two

adjacent faces is called an Edge, the points where the edges meet are its Vertices. A diagonal is the

line joining any two vertices (not in the same face). If all the faces of a solid are plane, it is called

polyhedron.



MATHSMATHS 3.13

PRISM :

A prism is a polyhedron bounded by two parallel congruent plane faces (called ends or bases) and

parallelogram side faces. If the side faces are rectangles, then the prism is called a right prism.

A prism is triangular, rectangular or polygonal, if the ends of the prism are triangles, rectangles or

polygons respectively.

Rectangle ABCD is the base of the right prism. If AB (length) = a, BC (breadth) = b. the area of

base = ab.

If again BF (height) = h, then Volume (V) of the prism = ab × h

i.e. Volume = (area or base) × height

Now the surface of the solid consists of six rectangles, so total surface (S) = 2 (ab + bh + ah).

Area of lateral surface = 2 (a + b) h = perimeter of base × height

Example : The length, breadth and height of a rectangular prism are 6, 4 and 2 ft. respectively.

Find the surface and the length of diagonal of the prism.

V = 6 × 4 × 2 = 48 cu. ft.

S = 2 (6×4 + 4×2 + 2×6) = 88 sq. ft.

Length of diagonal = 222 246 = 7.483 ft.

Example : A right prism has a triangular base whose sides are 13 cm. and 21 cm. If the altitude

of the prism is 9 cm. find its volume. [ICWA, Dec. ’75]

.cm272

54

2

212013

2

c bas

)2127()2027()1327(27)cs)( bs)(as(s = .cm.sq126671427

Volume = are of base × altitude = 126 ×9 1134 cm.

RECTANGULAR PARALLELOPIPED :It is a prism which has six faces all rectangles.

Volume = length + breadth × height = abh

Surface area = 2 (ab + bh + + ah)

CUBOID :

A cuboid is a rectangular solid having six faces all of which are rectangles.

Rectangular parallelopiped is a cuboid.

So, volume of a cuboid = length × breadth × height = abh

Surface area = 2 (ab + bh + ah)

H

E

D

A a B

C

b

h

F

G



MATHS3.14

Where, r = radius of circular base

Area of side-faces = perimeter of base × height = 2 r × h

Total surface area = area of side faces + area of two ends = 2 rh + 2 r 2 = 2 r(h + r)

a

a

a

C B

AD

h

r




MATHSMATHS 3.15

Example : The curved surface of cylinder is 1000 sq. cm. and the diameter of the base is 20 cm.

find the volume of cylinder.

Curved surface = 2 rh, here .cm102

20r

Now, 1000 = 2 ×

7

22× 10 × h ; or, h cm

10222

10007

Again, Volume = r 2 h =

722

× 102 ×

7 10005000 cu.cm.

2 220

HOLLOW CIRCULAR CYLINDER :

For R = external radius r = internal radius of hollow cylinder, and h = height.

Vol. of hollow cylinder = Vol. of outer cylinder – Vol. of inner cylinder.

= R 2h – r

2 h = (R

2 – r

2) h ………….(i)

Thus, volume = area of base × height

Curved surface area = area of outer surface + area of inner surface = 2 R h + 2 r × h

= 2 (R + r) h………(ii)

Example : Find the volume and surface of hollow cylinder height 10 in, internal and external radii

of base 4 in. and 6 in. respectively.

Volume = (R 2 – r

2) h =

722

(36 – 16). 10 = 628.32 sq. in.

Surface =22

2 (6 4) 10 628.32 sq.in7

SOLVED EXAMPLES :

1 : The total surface of a cube is3

233 m

2. Find its volume. [ICWA(F) Dec. 2000]

Let x be the edge, then 6 x2 =

398

; or, x =37

Volume = x3 =

3

3

7

=

27343

cu. m.

2 : Three cubes whose edges are 6 cm., 8 cm. and 10 cm. respectively, are melted without any loss

of metal into a single cube. Find the edge of new cube. [ICWA (F) June 2000]

Volume of new cube = Volume of three cubes or, x

3

= 6

3

+ 8

3

+ 10

3

= 216 + 512 + 1000 = 1728

R

r



or, x3 = 12

3 hence, x = 12 [where, x = edge of new cube]

i.e. required edge = 12 cm.

3 : A cubical shaped water tank made of steel has its edge 2.2 m. in length. The outer surface of

tank are painted at the rate of Rs. 25.50 per sq. m. How much will be the total cost?

ICWA(F) June, ’98]

Total surface with lid = 6l 2 = 6 × (2.2)

2 = 29.04 cu. m [where, l = edge = 2.2 m]

Total cost = 29.04 × 25.50 = Rs. 740.52

If the volume of a right circular cylinder is numerically equal to its lateral area, find the area of its

base. [ICWA(F) June ’89]

Volume = r 2 h [where, r = radius of base, h = height]

Surface (or lateral) area = 2 r h

Now, r 2 h = 2 r h or, r = 2

Area of base = 22

= 722

×4 = 74

12788

sq. units.

5 : The circumference of the base of a cylinder is 44 cms. and its height is 20 cms. Find the

volume of the cylinder. [ICWA(Prel.) June 1988]

2 r = 44 or, 44r 722

2 or, r = 7 cm.

Volume = r 2 h =

722

× 49 × 20 = 3080 cu. cm.

6. One cubic foot of brass is drawn into a cylindrical wire one tenth of an inch in diameter; find

the length of the wire. [ICWA (Prel.) June 1986]

Diameter of cylindrical wire =101

inch =120

1ft.

Radius of cylindrical wire =2401

12021

ft.

Volume of cylindrical wire = 1h240

1

240

1

7

22hr 2

or, .ft27.1832722

2402407h [as volume of cube = 1 cu. ft.]

7 : A solid cylindrical rod of length 80 cms. radius, 15 cms. is melted and made into a cube. Find

the side of the cube. [ICWA(prel.) Dec. 1997]

Volume of cube = a3 cu. cm. [where a is the side of cube]

Volume of cylindrical rod = r 2 h =

722

× 152 × 80 = a

3, by question.

Or , a3 = 56571.428 or, a = 3 428.56571 cm.

8 : A right circular cylinder whose base has a radius of 5cms. is equivalent to a cube having an

edge of 15 cms. Find the altitude of the cylinder.

Volume of cube = a3 = 15

3 cu. cm.

MATHS3.16




MATHSMATHS 3.17

A

B COr

hl



If the base of a cone is a circle and the perpendicular drawn from the vertex a to the base passes

through the centre of the base, then the cone is called a right circular cone.

Right cone also can be defined as a right pyramid having a circular base.

So preceding rules for volume and surface of a pyramid are used.

Volume = 31 (area of base) × height = 31 r 2 h

Where, r = radius of base, which is a circle, h = height (or altitude)

Lateral (or slant) surface =2

1× circumference of base × slanting height =

2

1× 2 r × l = r l

Now, l 2 = h

2 + r

2

Total surface = r l + r 2 = r (l + r)

Note : For an oblique cone, Volume = ½ (area of base) × height.

Example : Find the volume of the largest right circular cone that can be cut out a cube the edge of

the which is 42 cm. (take = 22/7)

Here the base of the cone will be circular and this base will be inscribed in one side of the cube.

Again the height of the cone will be equal to one edge of the cube.

Edge of the cube = 42 cm.

So, height of cone = 42 cm.

And. Radius of cone = ½ × 42 cm. = 21 cm.

Volume of cone =31 r

2 h = .cm.cu1940442)2121(

722

31

Example : If the radius of base of a right circular cone is 6 cm. and its slant height is 10 cm. Find

its volume. [ICWA (Prel.) June 1989]

We know, l 2 = h

2+ r

2 [Where, l = slant height, h = altitude, r = radius]

Or, 102 = h

2 + 6

2

h2 = 10

2 – 6

2= 64 h = 8 cm.

Volume of cone =31 r

2 h =

31

×722

× 62 × 8 = 301.71 sq. cm.

Example : A rectangular metallic block of dimension 44 cm × 27 cm × 7 cm. was melted to cast a

right circular cone, radius of whose base was 21 cm. What was the height of the right circular cone

if there was no loss of metal in the process of casting?

Volume of rectangular block = 44 × 27 × 7 c.c. Volume of cone =31

r 2 h = h21

722

31 2

By statement, h2121722

31 = 44 × 27 × 7 i.e. h = 18 cm.

Example : A right circular cylinder has base radius ‘r’ units and height ‘h’ units. Two third of the

cylinder is filled with water. A right circular cone of base radius ‘r’ units and height ‘h’ units is

placed in the cylinder. How much of the water will flow out of the cylinder?MATHS3.18




MATHS

Volume of water in the cylinder .cc3

h2r 2

Volume of empty space in the cylinder hr 3

1hr

3

2hr 222

Volume of cone hr

3

1 2

The cone will replace water equal to its on volume which again is equal to the volume of empty

space (above water level) in the cylinder.

So, no water will come out of the cylinder.

Example : A conical tent is required to accommodate 11 people, each person must have 14 sq. ft.

of space on the ground and 140 cubic ft. of air to breathe. Give the vertical height, slant height and

the width of the tent. [ICWA (F) June 2005]

R= radius of base (say), area of base = r 2

Now r 2 = 14 × 11 ……. (i). Again vol. of base hr 31 2

So 11140hr 3

1 2 ….. (ii) Dividing (ii) by (i),1114

11140

r

hr 3

1

2

2

or, h = 30 ft.

From (i), r 2 = 49

22

71114 or, r = 7 ft l

2 = h

2 + r

2 = 30

2 + 7

2 = 900 + 49 = 949,

.ft80.30949 l

Width = 2r = 2 × 7 = 14 ft.

SELF EXAMINATION QUESTIONS :

1. Find the volume of a right circular cone of slant height 5 cms. and base radius 4 cm. ( 3.14)

[Ans. 50.24 cu. cm]

2. The radius of the base of conical tent is 5 mts. If the height of the tent is 12 mts. Find the area of

the canvas used in making the tent. [Ans. 204.29 sq. mts.]

3. The circumference of the base of cone is 44 cm. and the slant height is 25 cm. Find the volume

and curved surface of the cone. [ICWA (F) Dec. 2003]

[Ans. 1232 cu. cm. ; 550 sq. cm.]

4. The radius and height of a cone are in the ratio 3 : 4. If the volume is 301.44 cu. cm., what is its

radius? What is its slant height ? ( = 3. 14) [ICWA (F) June 2000]

[Hints : 301.44 .etc&4

3

h

r as;hh

4

314.3

3

1 2

] [Ans. 6 cm, 10 cm]

5. The diameter of the base of a conical tank is 28 m and its height is 18 m. How much water does

the tank hold? [ICWA (F) June 98] [Ans. 3696 cu. m]

6. A conical tent is required to accommodate 5 people ; each person must have 16 sq. ft. of space on the

ground and 100 cu. ft. of air to breathe. Find the vertical height, slant height and the width of the tent.

[ICWA (F) Dec. 97] [Ans. 18.75 ft., 31.61 ft 50.90 ft]MATHS 3.19



MATHS3.20


[Hints : ;5001005hr 3

1;80165r 22 1

2 = h

2 + r

2 ; width = 2r & etc.]

7. If the ratio of the volumes of two cones with equal radius of the base is ½ . What is the ratio of

their heights? [Ans. ½ ]

8. The circumference of the base of cone is 44 cm and the slant height is 25 cm. Find the volume

and curved surface of the cone. [Ans. 1232 cu. cm ; 550 sq. cm.]

9. Find the curved surface of the cone whose radius of the base is 5 cm ; slant height is 7 cm.

[Ans. 110 sq. cm.]

10. How many solid circular cylinders each of length 8 cm. and diameter 6 cm. can be made out of

the material of solid cone of height 12 cm. and whose base is of radius 6 cm. [Ans. 2]

PYRAMID :

It is a solid bounded by plane face one of which is a plane rectilinear figure called the base, and

the others are triangles with a common vertex lying outside the base.

Tetrahedron

If the base is a triangle, the pyramid is called a tetrahedron and for a polygon, we base will find a

polygon pyramid. If tetrahedron ABCD is bounded by four equilateral triangle, then it is called a

right tetrahedron.

In the above figure, ABCD is a regular tetrahedron. The height of pyramid is the perpendicular

distance from the vertex (or apex) to the base. Here AO is the height. The altitude of any

triangular side-faces is called slant height. AP is slant height.

Now, AP2 = AO

2 + OP

2

RIGHT PYRAMID :

A

B C

O

PD

D

P

O

CB

L

A



MATHSMATHS 3.21

If the perpendicular drawn from a vertex of a pyramid to the base passes through the centre of the

base (centre of its inscribed or circumscribed circle) then the pyramid is called a right pyramid .

Further a right pyramid is regular, if the base is regular rectilinear figure.

SLANT PYRAMID :

Slant Surface : The slant (or lateral) surface of a pyramid is the sum of areas of the triangles,

which form the faces of the solid. Slanting height PL (= l) is the same for all faces, provided a

circle can be traced touching each side of the base.

PL2 = LO

2 + PO

2

or, l 2 = r

2 + h

2 [where, LO = radius]

Again, Slant surface area = ½ × (perimeter of base) × slant height

And, Volume3

1 (area of base) × height.

SOLVED EXAMPLES :

1. Find the volume and total surface area of a square pyramid, side of base 4 ft., height 6 ft.

area of base = 4 × 4 = 16 sq. ft.; Volume = 6163

1 = 32 cu. ft.

Slant height = .ft3.64036462 22

Perimeter of base = 4 × 4 = 16 ft.

Slant surface area = = ½ × 16 × 6.3 = 50.4 sq. ft.

Therefore, total surface area = 50.4 + 16 = 66.4 sq. ft.

2. A right pyramid stands on a base 16 cm square and its height is 15 cm. Find its slant surfaces

and the volume. [ICWA (F) June 98]

l = slant height, h = height, r = ½ × base

l 2 = h

2 + r

2 = 15

2 + 8

2 = 225 + 64 = 289 ; i.e., l = .17289

Slant surface area = ½ × perimeter of base × l , = ½ × 64 × 17 = 544 sq. cm.

Volume 3

1 area of base × height .cm.cu12801516

3

1 22

3. A right pyramid of height 12 cm, stands on a square base whose side is 10 cm. Find –

[ICWA (F) Dec. 2005](i) the slant edge. (ii) slant surface, and (iii) the volume of the pyramid.

Here, AB = BC = 10

AC2 = AB

2 + BC

2 = 100 + 100 = 200

.cm210200AC OA = 25 cm.

Slant edge AP ,OPAO 22 PO = 12 cm (given)

19414450 = 13.9 cm

PE

2

= PO

2

+ OE = 144 + 25 = 169, PE = 13



MATHS3.22


Slant surface = 4 × area of PAD

cm.sq2601310214

Volume3

1 × (area of base) × ht. 121010

3

1 = 400 cu. Cm.

4. The base of a right pyramid is an equilateral triangle of side 310 cm. each. The area of the

total surface of the pyramid is 3270 cm2. Find the slant height of the pyramid [ICWA (F) Dec.

2006]

Area of base (of equilateral triangle)

.cm.sq375310.4

3 2

Slant surface area 2

1 perimeter of base × h, h = height of pyramid

cm.sqh315h31032

1

total surface area 3270h315375 (given)

or, .cm13315

3195h,or 31953753270h315


1. The altitude of a pyramid having square base is 6 ft. Each side of base is 4 ft. Find the

volume. [Ans. 32 cu. Ft.]

2. The altitude of a rectangular hexagonal pyramid is 8 ft. Each side of the base is 6 ft, find the

volume. 73.13take [Ans. 249.12 cu.ft.]

3. A pyramid has a square base of area is 16 sq. cm. and each of the edges passing through the

vertex is 24 cm. in length. Find the volume of pyramid. [Ans.

3

121 cu. cm.]

( (310

2

P

D C

BA

OE



MATHSMATHS 3.23

r

O

Now, Surface area (S) = 4 r 2 = d

2

Volume (V) 3r 3

4

[The ratio of V to S is ;3

r so if S is multiplied by ,

3

r V can be obtained]

HOLLOW SPHERE :

Let, R = external radius, r= internal radius, R – r = thickness = t (say)

Now, external volume 3R 3

4 · Internal volume 3r

3

4

Therefore, Volume of hollow sphere 33 r R 3

4 .

SUMMARY :

Cube :

Surface area = 6a3, Volume = a

3, Diagonal of solid = .3a

Right circular cylinder :

Surface area = perimeter of base × height (= 2 r h)

Volume = area of base × height = r 2 h



MATHS3.24


Hollow circular cylinder :

Surface area = area of outer surface + area of inner surface (= 2 R0h + 2rh)

Volume = Volume of outer cylinder – Volume of inner cylinder

= ( R 2 h – r

2h)

Cone : Lateral surface area = ½ × (circumference of base × slant height)= ½ × 2 r × l = r l

l 22 r h

Volume3

1 (area of base × altitude) hr

3

1 2

Pyramid : Slant surface area = ½ × perimeter of base × slant height

Volume3

1 × area of base × height

Sphere : Surface area = 4 r 2 · Volume = 4/3 r 3

Hollow sphere : Volume 33 r R 3

4

SOLVED EXAMPLES :

1. If the diameter of a sphere is 6 cm, find :_

(i) Its volume (ii) its surface area ( = 22/7)

(i) Volume 33 37

22

3

4r

3

4 [as, r 3

2

6

2

d ]

798893

722

34 = 113.14 sq. cm.

(ii) Surface area = 4 r 2 23

7

224 = 113.14 sq. cm.

2. If the volume of a sphere is numerically equal to its surface area, find its diameter.

Volume = 4/3 r 3, Surface area = 4 r

2

Now, 23 r 4r 3

4 or, 1r

3

1 i.e., r = 3

Hence, diameter = 6 cm.3. A solid sphere of radius 6 cm. is melted and recast into three solid spheres of radii 3 cm.,

x cm., and 5 cm. Find x.

Here the volume of original sphere will be equal to the sum of volumes of the other three

spheres, so that–

36.3

4 333333 5x3

3

45.

3

4x.

3

43.

3

4

or, 63 = 3

3 + x

3 + 5

3

or, 216 = 27 + x

3

+ 125 or, x

3

= 64 = 4

3

i.e., x = 4 cm.



MATHSMATHS 3.25

4. The radius of a sphere is 3 cms. It is melted and drawn into a wire of radius 0.1 cm. Find the

length of the wire.

Volume of sphere 33 33

4r

3

4

Let l = length of wire, so that volume of wire = (0.1)2 1

Now, ,l.01.027..3

4 or, l = 3600 cm = 36 m.

5. How will the volume and the surface area of a sphere changes, if the diameter of the sphere is

halved?

Diameter = 2 radius For, ,2

dr

Volume

8

d

3

4

2

d

3

433

Surface area 222

d4

d4

2

d4

For, ,4

dr diameter (d) is halved, radius (r) becomes

4

d

2

d

2

1

Volume8

1

64

d

3

4

4

d

3

433

of previous volume.

Surface area41

4d

4d4

22

of previous surface area.

6. How many solid cylindrical iron rods of 3.5 cm. radius and 1.6 dm length may be moulded by

melting three solid iron spheres each of 14 cm. diameter? [ICWA (F) June 2007]

Vol. of 3 spheres ,747.3

43 33 vol, of iron rod

2

162

cu. cm. = 196 cu. cm.

Reqd. no. .7196

74 3

7. A solid metal cylinder with height 45 cm and diameter 4 cm is melted to get a number of

spheres with diameter 6 cm. Find the number of such spheres. [ICWA (F) June. 2003]

Volume of cylinder = 22. 45 sq. cm, here 2

2

4r

Volume of sphere 34.3 ,

3 here 3

2

6r

Now, 45.2.3.3

4.n 23 where n = no, of spheres or, n = 5.



MATHS3.26


1. The volume of a sphere as 4312 cm3, find diameter. [Ans. 14 cm]

2. Find the surface area of a sphere having diameter 14 cm. [Ans. 616 cm2]

3. The surface of a sphere is 154 sq. cm ; find radius [Ans. 3.5 cm.]

4. If the surface area of a sphere is numerically thrice its volume, find the radius of the sphere.

[ICWA (F) Dec. 87] [Ans. 3.35 cm]

5. Three solid gold spherical beads of radii 3, 4, 5 cm., respectively are melted into one solid

spherical bead. Find its radius. [ICWA (F) Dec. 97] [Ans. 6 cm]

6. Given that volume of a metal sphere is 4851 cu. cm. Find its radius and surface area

7

22. [ICWA (F) Dec. 2000] [Ans. 10.5 cm. ; 1386 sq. cm.]

7. A cone is 8.4 cm high and radius of base is 2.1 cm. It is melted into a sphere. Determine the

radius of the sphere. [ICWA (F) Dec. 2000] [Ans. 2.1 cm]

[Hints : .etc&R 344.81.2

3

32 ]

8. A solid metal cylinder of radius 14 cm. and height 21 cm. is melted down and recast into

spheres of radius 3.5 cm. Calculate the number of spheres that can be made, (7

22 ) [Ans. 72]

9. The volume of two spheres are in the ratio 64 : 27. Find their radii if the sum of their radii is

21 cm. [ICWA (F) June 2000] [Ans. 12, 9 cm]

[Hints : ;27

64

r 3

4

r 3

4

3

2

31

r 1 + r 2 = 21 & etc. ]

10. A hollow cylinder of iron with height 32 cm, internal and external radii 4 cm and 5 cm.

respectively, is melted to form a sold sphere. Find the radius of the sphere.

[ICWA (F) June 2006] [Ans. 6]

[Hints : 6r ,3245r 3

4 223 ]

P

D´

A´B´

C´

CD

A B




MATHSMATHS 3.27

the base and the parallel plane is called a Frustum of Cone.

A1

O1

B1

AO P B

r

In the adjacent figure, ABB1 A1 is the frustum of a cone. The perpendicular distance of two

parallel ends is the height of the frustum, i.e., O1 O = h, where O1 is the centre of the upper end

and O is the centre of the base.

O1B1 = r, OB = R, PB = OB – OP = R – r.

Now, slanting height (1) 22221 r R hPBPB

Slant Surface = ½ × (sum of circumferences) × slant height



MATHS3.28

It is a part of the sphere lying between two parallel planes.

The perpendicular distance between the parallel planes forming the frustum is called the

thickness of the frustum.

The parallel planes may lie in the same side or opposite sides of the centre.

In the adjacent figure ABCD is the frustum.

Let, r = radius of the sphere, r 1 = radius of circle CD

R 2 = radius of circle AB, h = thickness of the frustum

Lateral surface area of the frustum = 2 rh

Whole surface area of frustum 22

21 r r rh2

P

O

D

BA

C

h

Q




MATHS

Volume of frustum .hr 3r 36

h 222

21

SEGMENT :

It is a portion of a sphere cut off by a plane. In the corresponding figure, APB is a segment

(here it is minor). Another segment AQB (it is major).

A segment may also be considered as a frustum with the top of a point or taking r 2 being zero.

Now putting r 2 = 0 in the above formula, we find––

Internal surface area of segment = 2 h, Area of whole surface = 2 rh + r 12

Volume ,hr 36

h 221

r 1 = radius of circle AB, h = height of segment i.e., PQ.

Alternative Formula : Volume

3

hr h2

r = radius of sphere

h = height of segment.

Example : The radius of the sphere is 5 cm. Two parallel planes 1 cm apart are drawn at

distances 2 cm and 3 cm, respectively from the centre. Find the volume of the sphere between the

two planes.

Let radius of the slice (near the centre) = r

Now, r 2 = 5

2 – 2

2 = 25 – 4 = 21

Again, Radius of the other end be R

So, R 2 = 52 – 32 = 25 – 9 = 16 ; h = 1

Volume 11632136

1hR 3r 3

6

h 222

67.587

22

3

56

3

56 cu. cm.

Example : Find the whole surface (including the two plane surface) and volume of a spherical

frustum, the diameter of the sphere being 20 cm. and the distance of the plane ends from the

centre (on the same side of it) being 6 cm. and 8 cm.

Radius of the plane (nearer to centre) = r

cm86436100610 22

and, radius of other plane = R = 366410081022 = 6 cm

curved surface area = 2.10.2 = 40 sq. cm.

area of two ends = . 82 + . 6

2 = 64 + 36 = 100 sq. cm.

Therefore, whole surface area = 40 + 100 = 140 sq. cm

.cm.sq44022207

22140

MATHS 3.29



MATHS3.30


Volume 3

30426.38.3

6

2hR 3r 3

6

h 222222

.cm.cu5.318.cm.cu476.3187

22.

3

304

Example : A vessel is in the form of a frustum of a cone. The height of the frustum is 30 cm.

The diameters of the top and the bottom are 32 cm. and 24 cm. respectively. How many litres can

it hold when full?

R = 16 cm, r = 12 cm. h = 30 cm

Volume 121612163

30Rr r R

3

h 2222

7

225920592019214425610

= 18605.71 cc = 18.60571 litres (as 1 litre = 1000 cc) = 18.6 litres.


1. Two brackets have the same depth of 84 cm. each. One is cylindrical with 75 cm as its

diameter while the other is a conical frustum, the diameters of its ends being 85 cm. and 65

cm. respectively, which one will hold more water and by how much?

[Hints : Find the volume in each case and compare] [Ans. conical frustrum ; 2200

2. Find the volume of a segment of a sphere, the radius of the base being 10.2 cm. and the

radius of the sphere 12 cm. [Ans. 1076 cu.

[Hints : Use formula as shown before.]

3. The radii of the two circles of a spherical zone are 12.5 cm. 4.25 cm. and the thickness of

the zone is 6 cm. What is its volume, its convex surface and its total surface?

[Ans. 1756 cu. cm. ; 1118.04 sq.

[Hints : r 1 = 4/15. R 2 = 12 5, h = 6, r = radius of sphere, r 2 = x

2 + (12.5)

2.

Again, r 2 = (x + 6)

2 + (4. 25)

2

x = distance from centre to nearer circle of zone.

Solving, we get = 8.52 So, r 2 = (8.52)

2 + (12.52)

2 = 228.84 ; or, r = 15.13 (using log),

hence and volume and total surface area.]

OBJECTIVE QUESTIONS :

1. Find the area of an equilateral triangle of side 4 cm. [Ans. 34 sq.

2. The base of a parallelogram is 5 cm and its height is 1 cm. 2 dm. Find its area. [Ans. 6 sq.

3. The side of a square is 14 cm. Find the area of fillets. [Ans. 42 sq. cm]

)3

304



MATHSMATHS 3.31

4. The sides of aright angled triangle are 3, 4 and 5 cm respectively. Find the length of the

perpendicular on the largest side from the vertex of the greatest angle.

[ICWA (F) June ’07] [Ans. 2. 4

5. A bicycle wheel makes 1000 revolutions is moving 4.4 km. Find the radius of the wheel.

[ICWA (F) June 2006] [Ans. 70 c

6. The circumference of base of a cylinder is 66 cm and its height is 10 cm. Find the volume

of the cylinder. [Ans. 3465 cu. c

7. A solid cube has volume 125 cu. cm. Find the surface area of the cube

[ICWA (F) June 2007] [Ans. 30 cu. c

8. The circumference of a circle is 88 cm. Find the length of perimeter of its semi-circle.

[ICWA (F) Dec. 2006] [Ans. 44

9. The diameter of the base of a cylindrical pillar is 7 m and its height is 18 cm. Find the

volume of the pillar. [Ans. 693 cu.

10. The external and internal radii of a hollow sphere are 6 cm and 3 cm. respectively. Find the

volume of the sphere. [Ans. 792 cu.

11. A right pyramid stands on a square base of side 16 cm and has height 15 cm. Find the

volume of the pyramid. [Ans. 1280 cu.

12. A path of 4 ft. wide is to be laid outside round the square garden of 60 ft by 60 ft. Find the

area of the path. [ICWA (F) Dec. 2005] [Ans. 1024 sq.

13. Base radius of a conical tent is 5 m and its height is 10m. Find the area of the canvas of the

tent. [Ans. 176 sq. cm (ap

14. The perimeter of an isosceles triangle is 544 cm. and each equal side is6

5 of the base. Find

the area. [Ans. 13872 sq. c

BOOKS FOR REFERENCE :

1. Basic Mathematics and Statistics – N.K. Nag

2. ICSE Mathematics – O. P. Sinhal.



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