STUDY NOTES FOR BANK AND OTHER EXAM

MENSURATION

Introduction- Mensuration is a topic in Geometry which 3D Shape -

is a branch of mathematics. Mensuration deals with A 3D shape is a shape that is bounded by a number of
length, area and volume of different kinds of shape- both surfaces or planes. These are also referred to as solid
2D and 3D. So moving ahead in the introduction to shapes. These shapes have height or depth unlike 2D
Mensuration, let us discuss what are 2D and 3D shapes shapes; they have three dimensions- length, breadth and
and the difference between them. height/depth and are therefore called 3D figures. 3D
shapes are actually made up of a number of 2D shapes.
Also, known as solid shapes, for 3D shapes we measure
Volume (V), Curved Surface Area (CSA), Lateral
Surface Area (LSA) and Total Surface Area (TSA).

2D Shape -
A 2D shape is a shape that is bounded by three or more
straight lines or a closed circular line in a plane. These
shapes have no depth or height; they have two Important Terms
dimensions- length and breadth and are therefore called Before we move ahead to the list of important
2D figures or shapes. For 2D shapes, we measure area and mensuration formulas, we need to discuss some important
perimeter. terms that constitutes these mensuration formulas.
Area (A) – The surface occupied by a given closed shape
is called its area. It is represented by the alphabet A and
is measured in unit square- m2/ cm2.

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
Perimeter (P) – The length of the boundary of a figure is 1. Triangle -
called its perimeter. In other words, it is the continuous
line along the periphery of the closed figure. It is
represented by the alphabet P and is measures in cm/ m.
Volume (V) – The space that is contained in a three-
dimensional shape is called its volume. In other words, it
is actually the space that is enclosed in a 3D figure. It is
represented by the alphabet V and is measured in cm3/ m3. The sides AB, BC and AC are respectively denoted by c,
Curved Surface Area (CSA) – In solid shapes where a and b.
there is a curved surface, like a sphere or cylinder, the 1
total area of these curved surfaces is the Curved Surface Area of a triangle =  base × height
2
Area. . The acronym for this is CSA and it is measured in
m2 or cm2. Area by Hero’s formula = s ( s  a )( s  b)( s  c )
Lateral Surface Area (LSA) – The total area of all the abc
s
lateral surfaces of a given figure is called its Lateral 2
Surface Area. Lateral Surfaces are those surfaces that Right Angled Triangle -
surround the object. The acronym for this is LSA and it is A triangle having one of its angles equal to 90° is called
measured in m2 or cm2. a right-angled triangle. The side opposite to the right
Total Surface Area (TSA) – The sum of the total area of angle is called the hypotenuse.
all the surfaces in a closed shape is called its Total Surface
Area. For example, in a cuboid when we add the area of
all the six surfaces we get its Total Surface Area. The
acronym for this is TSA and it is measured in m2 or cm2.
Square Unit (m2/ cm2) – One square unit is actually the
area occupied by a square of side one unit.
Cube Unit (m3/ cm3) – One cubic unit is the volume
occupied by a cube of side one unit.
AB = Perpendicular (P)
BC = Base (B)
AC = Hypotenuse (H)
In a right angled triangle-
(Hypotenuse)2 = (Perpendicular )2 + (Base)2
(Pythagoras theorem)
1
Area of right angled triangle =  P  B
2
Perimeter of right angled triangle = P + B + H
TWO - DIMENSIONAL FIGURE(2D)
Equilateral Triangle-
Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org
 STUDY NOTES FOR BANK AND OTHER EXAM
A triangle whose all sides are equal is called an equilateral
triangle.

1 2
Area of Isosceles Right-angled Triangle = a
2

Perimeter of Isosceles Right-angled Triangle = 2a +

o
All angle = 60 2 a = 2 a( 2 + 1) = h( 2 + 1)

Area of equilateral triangle =

3 2
a Rectangle
4 A four-sided shape that is made up of two pairs of parallel
Perimeter of equilateral triangle = 3a lines and that has four right angles.
3
Altitude (Height) = a
2
Isosceles Triangle
A triangle whose two sides are equal is an isosceles
triangle.
The diagonals of a rectangle bisect each other and are
equal.
Area of rectangle = length x breadth = l x b

Area of rectangle = l d 2  l 2

Area of rectangle = b d 2  b2
Perimeter (P) of rectangle = 2 (length + breadth)=2(l +
b).
Area of isosceles triangle =
b
4a 2  b 2 Square
4
A four-sided shape that is made up of four straight sides
1
Height of an isosceles triangle = 4a 2  b 2 that are the same length and that has four right angles.
2
Perimeter of an isosceles triangle = (2a+b)
Isosceles Right-angled Triangle-
An isosceles right-angled triangle has two sides (a) equal
with equal sides making 90° to each other.

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
The diagonals of a square are equal and bisect each other 
Length of arc AB(l)= 2 r
at 900. 360

d2 
Area of a square = a 2  Area of Sector AOB =  r 2
2 360
Perimeter (P) of a square = 4a Ring or Circular Path:
R=outer radius
Length (d) of the diagonal of a square = a 2
r = inner radius
Circle

A circle is the path traveled by a point which moves in

such a way that its distance from a fixed point remains
constant.

Area = π (R2-r2)
Perimeter = 2π(R + r)
Radius = r
Diameter (d)= 2r
The fixed point is known as center and the fixed distance
is called the radius.
Circumference(C) or perimeter of circle = 2 r
Area of circle =  r 2 Parallelogram
Sector- A quadrilateral in which opposite sides are equal and
parallel is called a parallelogram. The diagonals of a
A sector is a figure enclosed by two radii and an arc lying parallelogram bisect each other.
between them.
Area of a parallelogram = base × altitude = b × h

In given figure AOB is a sector

In a parallelogram, the sum of the squares of the diagonals
= 2 (the sum of the squares of the two adjacent sides).
Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org
 STUDY NOTES FOR BANK AND OTHER EXAM

i.e., d12  d 2 2  2(a 2  b 2 ) Area (a) of a trapezium =

1
× (sum of parallel sides)×
2
Perimeter (P) of a parallelogram = 2 (a+b), 1
height = (a  b)h
Where a and b are adjacent sides of the parallelogram. 2
Rhombus Perimeter of trapezium = AB+ BC + CD + DA

Rhombus is a quadrilateral whose all sides are equal. Pathways Running across the middle of a rectangle:

The diagonals of a rhombus bisect each other at 900

Area of a rhombus = a × h
1
= d1  d2
2 w is the width of the path
Perimeter of Rhombus = 4a
Area of path= (l + b - w )w
Relation between side and diagonal of Rhombus;
perimeter= 2(l + b - 2w)
2
d  d 
2 Outer Pathways:
a2   1    2 
 2  2 
Trapezium (Trapezoid)
A trapezium, also known as a trapezoid, is a quadrilateral
in which a pair of sides are parallel, but the other pair of
opposite sides are non-parallel.

Area = (l+b+2w)2w
Perimeter = 4(l+b+2w)
Inner Pathways:

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
 Volume of cuboid = length × breadth × height=
l×b×h
 Lateral surface area = 2h(l + b)
 Total surface area = 2 ( lb + bh + hl)

 Longest diagonal = l 2  b2  h2
Right Circular Cylinder

Area = (l+b-2w)2w
Perimeter = 4(l+b-2w)
Three - Dimensional figure (3D)
Cube
 Volume of Cylinder = π r2 h
 Lateral Surface Area (LSA or CSA) = 2π r h
 Total Surface Area = TSA = 2 π r (r + h)
Right Circular Cone

 a = side
 Volume: V = a3
 Lateral surface area = 4a2
 Surface Area (S) = 6a2  l2 = r2 + h2
1
 Diagonal (d) = a√3  Volume of cone =  r 2 h
3
Cuboid
 Curved surface area(CSA)= πrl
 Total surface area (TSA) = πr(r + l )
Frustum of a Cone

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
 r = Top radius,  Volume = Base area x height
 R = base radius,
 h = height, l = slant height
1
 Volume: V   (r 2  rR  R 2 )h
3
 Surface Area
S   ( R  r )l   r 2   R 2   ( R  r )l   (r 2  R 2 )
Sphere
 Lateral Surface area = perimeter of the base x
height
 Total Surface area = LSA + 2(Area of base)
Pyramid

 r = radius
4
 Volume: V =  r 3
3
 Curved Surface Area = Total Surface Area =
4π2
Hemisphere

 Volume of a right pyramid

1
=  area of the base  height
3

2
 Area of the lateral faces of a right pyramid
 Volume =  r 3
3 1
=  perimeter of the base  slant height
 Curved surface area = 2 π r2 2
 Area of whole surface of a right pyramid =
 Total surface area = 3πr2 Area of the lateral faces + Area of the base.
Prism

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM

QUESTION WITH SOLUTION

Q.1. Three cubes of metal whose edges are in ratio Q.6. Find the volume of a sphere whose radius is √7
of 3:4:5, are melted and one new cube is formed. times the radius of another sphere which exactly
If the diagonal of the cube is 18√3 cm, then find fits in a cube of side 9 cm?
the edge of the largest among three cubes. (A) 2673√7 cm3 (B) 4876√7 cm3
(A) 18 cm (B) 24 cm (C) 2486√7 cm3 (D) 3286√7 cm3
(C) 15 cm (D) 12 cm Q.7. The outer radius of a hemispherical iron tank is
Q.2. What is the area of the largest triangle that can 14 cm and the in radius is 13.3 cm. It is to be
be fitted into a rectangle of length 'a' units and painted inside as well as outside. Find the cost
width 'b' units? of painting it at the rate of ₹15 per 28 cm².
(A) Rs. 1855.65 (B) Rs. 1255.65
(A) unit2 (B) unit2 (C) Rs. 1355.65 (D) Rs. 1755.65
Q.8. Find the area of a trapezium whose parallel sides
(C) unit2 (D) unit2 are length 58 meters and 42 meters and whose
Q.3. Find the perimeter and area of an isosceles non-parallel sides are equal each being 20m.
triangle whose equal sides are 5 cm and the (A) 300√21 (B) 264√21
height is 4 cm. (C) 232√21 (D) 200√21
(A) 24 cm2 , 13 cm (B) 18 cm2 , 16 cm
(C) 12 cm2 , 13 cm (D) 12 cm2 , 16 cm Q.9. What is the area of the sector whose central
Q.4. At each corner of a triangular field of sides 20m, angle is 90° and radius of the circle is 14 cm?
28m, and 36m, a cow is tethered by a rope of (A) 308 sq cm (B) 77 sq cm
length 7m. Find the approximate area not grazed (C) 154 sq cm (D) 231 sq cm
by the cow. Q.10. If the ratio of Area and diameter of a circle is
(A) 77 m2 35 : 14 then find the circumference of the circle.
(B) 200.2 m2 (A) 7 (B) 10
(C) 157.5 m2 (C) 12 (D) 18
(D) 172.8 m2 Q.11. A parallelogram has area A sq. mts. A second
(A) π/2 (B) 2 π/ 3 parallelogram is formed by joining the mid-
(C) π/3 (D) π/4 points of its sides. A third parallelogram is
formed by joining the mid-poinds of the sides of

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
the second parallelogram. This process is
continued upto infinite. What is the sum of area (A) (B)
(in sq. mts) of all the parallelograms so formed?
(C) (D) 15
(A) A (B) Q.23. A hollow iron pipe is 21 cm long and its exterior
diameter is 8 cm. If the thickness of the pipe is
(C) 2A (D) 1 cm and iron weights 8 gm/cm3, then the weight
Q.13. From the four corners of a rectangular sheet of of the pipe is-
(A)3.696 kg (B)3.6 kg
dimensions 25 cm x 20 cm, square of side 2 cm
(C) 36 kg (D) 36.9 kg
is cut off from four corners and a box is made.
Q.24. Water flows at the rate of 10 meters per minute
The volume of the box is-
(A) 828 cm3 (B) 672 cm3 from cylindrical pipe 5 mm in diamter how long
(C) 500 cm3 (D) 1000cm3 it will take to fill up a conical versel whose
diameter at the base is 30 cm and depth 24 cm ?
Q.17. The base of a pyramid is a rectangle 30 m × 20 (A) 28 minutes 48 seconds
(B) 51 minutes 12 seconds
m and its slant height to the smaller side of the
(C) 51 minutes 24 seconds
base is 17 metre. Find its volume of pyramid ?
(A) 12 (B) 9/2 (D) 28 minutes 36 seconds
(C) 4 (D) 9
Q.18. The radius of wheel is 21 cm. How many rounds Q.25. The radii of the base of two cylinders A and B
will it take to cover the distance of 792? are in true ratio 3 : 2 and their height in the ratio
(A) 600 (B) 200 x : 1. If the volume of cylinder A is 3 times that
(C) 300 (D) 400 of cylinder B, the value of x is-
Q.19. If the base of a right pyramid is triangle of sides
5 cm, 12 cm, and 13 cm and its volume is 330 (A)
cm, then its height (in cm) will be-
(A) 33 (B) 32 (B)
(C) 11 (D) 22
Q.21. A rectangular paper sheet of dimensions 22 cm (C)
× 12 cm is folder in the form of a cylinder alongs
its length. What will be the volume of this (D)
cylinder. Q.26. If the radius of a right circular cylinder is
(A) 460 cm3 (B) 462 cm3
increased by 50% and its height is increased by
(C) 624 cm3 (D) 400 cm3
60%. Its volume will be decreased by ?
Q.22. A copper rod of 1 cm diameter and 8 cm length
(A) 100% (B) 60%
is drawn into a wire of uniform diamter and 18 (C) 40% (D) 20%
m length. The radius (in cm) of the wire is-
Q.27. Side of a tetrahedron is Cm. Find its
volume.
(A) 12 Cm3 (B)15 Cm3
(C) 9 Cm3 (D) 6 15 Cm3

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
Q.28. A hollow cylindrical iron pipe is of length
28 cm. It's outer diameter is 8 cm. and =
thickness of the metal is 2 cm. Find the
weight of the pipe, if 1 cm.3 of iron weights = 22 × 7 × 0.5 = 77
7 gm. Required Area = 277.2 - 77 = 200.2 m2
(A) 13.248 kg. (B) 13.648 kg. Q.6.(A) Radius of sphere situated in cube = 9/2 cm
(C) 16.528 kg. (D) 17.248 kg.
Q.29. A field is in triangular shape whose sides
Radius of new sphere =
are 26m, 28m, and 30 m. At each corner a
4 3 4 22 729
cow is tied by 7m rope. Find ungrazed area. Volume = r    7 7 =
3 3 7 8
(A) 259 m.2 (B) 154 m.2 2673 7 cm3
(C) 306 m.2 (D) 186 m.2 Q.7.(B) Total Surface Area
Q.30. ABCD is a rectangle of dimensions 8 units
and 6 units. AEFC is a rectangle drawn in
= = 2343.88
such a way that diagonal AC of the first
Total cost = 2343.88 × (15/28) =Rs.
rectangle is one side and side opposite to it,
is touching the first rectangle at D. What is
1255.65
Q.8.(D)
the ratio of the area of rectangle ABCD to
that of AEFC?
(A) 2 : 1 (B) 3 : 2
(C) 1 : 1 (D) 9
Solution

Q.1.(C) Total volume = (3x)3+(4x)3+(5x)3 = 216x3

Then side of new cube/= 6x h = 202 - 82
Then diagonal of new cube = 6x√3 → 18√3 h = 4√21
x=3 Area = Area of parallelogram + area of
Then side of larger cube = 5×3 = 15 triangle = 42 × 4√21 + (1/2) ×16×4√21
Q.2.(B) = 200√21
Q.9.(C) Required area

Area of triangle = (1/2) base × height

= (1/2) ab unit2 = 154 sq. cm.
Q.3.(D) 12 cm2 , 16cm Q.10.(B)
Q.4.(B) Area of triangle = 84√11 = 84 × 3.3 = 277.2
Area covered by cow

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM

Hence circuference of circle Q.21.(B)

= cm

Q.11.(C)

Q.22.(C)
sum of areas

Q.23.(A)
Q.13.(B) Volume of the box
= 16×21×2
= 672 cm3
Q.17.(D)

Q.24.(A) Volume of water flowing from the pipe in 1

Q.18.(A) Circumference/परिध = 2πr minute

=2 = 132

Number of revalution = = 600

Q.19.(A)
28 minutes 48 seconds

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org

 STUDY NOTES FOR BANK AND OTHER EXAM
Q.25.(C) Q.30.(C)

In ΔADC
AC2 = 62 + 82 = 36 + 64 = 100
AC = 10
Now Let ED = x and DF = 10-x
Q.26.(B) Radius 2→1 [As AC = EF]
Height 5→ 8 Using pythagoras theorem in ΔAED
Volume 20→ 8 and ΔDFC
62 - x2 = 82 - (10 - x)2
36 - x2 = 64 - 100 - x2 + 20x
20x = 72
x = 3.6 units
Q.27.(C) AE2 = 36 - 12.96 = 23.04
AE = 4.8 units
Q.28.(D) Volume of cylindrical pipe
Required ratio = 8 × 6 : 10 × 4.8
= 48 : 48 = 1 : 1
=
= 2464 cm.3
1 cm.3= 7 gm.
Weight of pipe = 2464 × 7
= 17248 gm.
= 17.248 gm.
Q.29.(A)

Write us : content@mahendras.org www.mahendras.org www.mahendraguru.com myshop.mahendras.org