Lesson: Mensuration
Exercise 11.1
Question 1
A square and a rectangular field with measurements as given in the figure
have the same perimeter.
Which field has a larger area?
Answer:
(a)
Solution:
Since, the perimeter of the square and the rectangle are the same. (given)
Perimeter of the square = Perimeter of the rectangle
Or, 4 × side = 2 (length + breadth)
Or, 4 × 60 m = 2 (80 m + breadth) or, 240 m = 2 (80 m + breadth)
Or, 80 m + Breadth =
240
2
m or, Breadth = (120 – 80) m = 40 m
Now, area of the square = (side)
2
= (60)
2
m
2
= 3600 m
2
and area of the rectangle
= length × breadth = 60 × 40 m
2
= 2400 m
2
Hence (a) has greater area.
Question 2
Mrs. Kaushik has a square plot with the measurement as shown in the figure.
She wants to construct a house in the middle of the plot.
A garden is developed around the house.
Find the total cost of developing a garden around the house at the rate of Rs.
55 per m
2
.
Answer:
Rs 17,875
Solution:
Given, side of the square plot = 25 m,
Area of the square plot = 25 × 25 = 625 m
2
.
The length of the house = 20 m and the breadth of the house = 15 m
Area of the house = length × breadth = 20 × 15 m
2
= 300 m
2
Area of the garden = Area of the square plot – Area of the house = (625 – 300) m
2
= 325 m
2
Now, the total cost of developing the garden = Rs 55 × 325 = Rs 17,875.
Question 3
The shape of garden is rectangular in the middle and semicircular at
the ends as shown in the diagram.
Find the area and the perimeter of this garden.
[Length of rectangle is 20 – (3.5 + 3.5) meters]
Answer:
Area = 129.5 m
2
; Perimeter = 48 m
Solution:
Given, the radius of the semicircular end =
7
2
m = 3.5 m
The length of the rectangle = (20 – (3.5 + 3.5)) m = (20 – 7) m = 13m
and the breadth of the rectangle = 7 m
Area of the rectangle = length × breadth = 13 × 7 m
2
= 91 m
2
Radius of the semi-circular portion = 3.5 m
Area of the semi-circular portion =
2
2
2 2 2 2
1 1 1 22 7 22 7 7 77
r (3.5)
2 2 2 7 2 4
2 7 2 2
m m m




Now, the perimeter of the semicircular portion
=
2 r 22 22 7
r 3.5
2 7 7 2
m
= 11 m
Also, the total area of the garden
= Area of the rectangular portion + 2 × Area of the semicircular portion.
Area of the garden = (91 + 2 ×
77
4
) m
2
= (91 +
77
2
) m
2
= (91 + 38.5) m
2
= 129.5 m
2
Perimeter of the garden
= 2 × length of the rectangular portion + 2 × perimeter of the semicircular portion
= 2 × 13 m + 2 × 11 m = 26 m + 22 m = 48 m
Question 4
A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm.
How many such tiles are required to cover a floor of area 1080 m
2
?
(If required you can split the tiles in whatever way you want to
fill up the corners).
Answer:
45000 tiles
Solution:
Area of 1 tile = base × height = 24 × 10 cm
2
= 240 cm
2
and the area of the floor = 1080 m
2
= 1080 × (100)
2
cm
2
(given)
= 10800000 cm
2
22
1 100
1 10000
m cm
m cm




We know that, the number of tiles
=
= 45000
Therefore, 45,000 tiles are required.
Question 5
An ant is moving around a few food pieces of different shapes scattered on the floor.
For which food-piece would the ant have to take a longer round? Remember,
circumference of a circle can be obtained by using the expression c = 2
r,
where r is the radius of the circle.
Answer:
(b)
Solution:
(a) Circumference of the semi-circular portion
=
2r
2
=
r =
×
2.8 22 14
cm cm
2 7 10




= 4.4 cm
Total perimeter = (2.8 + 4.4) cm = 7.2 cm.
(b) Perimeter of the given portion = Sum of the lengths of the three sides of the
rectangle + perimeter of the semi-circle
=
2r
( 2b)
2

= ((2.8 + 2 × 1.5) +
×
2.8
2
)
cm = ((2.8 + 3.0) +
22 14
7 10
) cm = (5.8 + 4.4) cm = 10.2 cm
(c) Perimeter of the given portion = sum of the lengths of the two sides of the
triangle + Perimeter of the semicircle
= (2 + 2) cm +
2r
2
= 4 cm +
22
7
×
2.8
2
cm
= (4 + 4.4) cm = 8.4 cm
Therefore, the ant would have to take a longer round for the food-piece (b).
Exercise 11.2
Question 1
The shape of the top surface of a table is a trapezium.
Find its area if its parallel sides are 1 m and 1.2 m and perpendicular
distance between them is 0.8 m.
Answer:
0.88 m
2
Solution:
Area of the trapezium shaped top surface of the table
=
1
2
× height × (sum of parallel sides) =
1
2
× 0.8 × (1 + 1.2) m
2
=
1
2
× 0.8 × 2.2 m
2
= 0.88 m
2
Question 2
The area of a trapezium is 34 cm
2
and the length of one of the parallel sides is 10 cm
and its height is 4 cm.
Find the length of the other parallel side.
Answer:
7 cm
Solution:
Let the other parallel side be x cm
Area of the trapezium =
1
2
× height × (sum of parallel sides)
Or, 34 =
1
2
× 4 × (10 + x)
Or, 34 = 2(10 + x)
Or,
34
2
= 10 + x or, 17 = 10 + x or, x = 17 – 10 i.e., x = 7.
Therefore, the other parallel side is 7cm.
Question 3
Length of the fence of a trapezium shaped field ABCD is 120 m. if BC = 48m,
CD = 17 m and AD = 40 m, find the area of this field. Side AB is
Perpendicular to the parallel sides AD and BC.
Answer:
660 m
2
Solution:
Perimeter of the trapezium shaped field ABCD is given as,
AB + BC + CD + DA = 120 m
AB + (48 + 17 + 40) m = 120 m
AB = (120 – 105) m = 15 m
Area of the trapezium shaped field ABCD =
1
2
× AB × (AD + BC)
=
1
2
× 15 × (40 + 48) m
2
=
1
2
× 15 × 88 m
2
= 660 m
2
Question 4
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars
dropped on it from the remaining opposite vertices are 8 m and 13 m.
find the area of the field.
Answer:
252 m
2
Solution:
Here, AC = 24 m, BE = 8 m, DF = 13 m.
Area of the quadrilateral shaped field ABCD
=
1
2
AC (BE + DF) =
1
2
× 24 × (8 + 13) m
2
= 12 × 21 m
2
= 252 m
2
Question 5
The diagonals of a rhombus are 7.5 cm and 12 cm.
Find its area.
Answer:
45 cm
2
Solution:
Area of the rhombus =
1
2
× Product of the diagonals =
1
2
× 7.5 × 12 m
2
= 45 m
2
Question 6
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm.
If one of its diagonals is 8 cm long, find the length of the other diagonal.
Answer:
24 cm
2
, 6 cm
Solution:
Area of the rhombus = Area of
ABD + Area of
CDB
=
1
2
× AB × DE +
1
2
× CD × DE = (
1
2
× 6 × 4 +
1
2
× 6 × 4) cm
2
= (12 + 12) cm
2
= 24 cm
2
Also, area of the rhombus =
1
2
× d
1
× d
2
Or, 24 cm
2
=
1
2
× 8 cm × d2 or, d
2
=
24 2
8
cm = 6 cm
Hence, the other diagonal is 6 cm
Question 7
The floor of a building consists of 3000 tiles which are rhombus shaped and diagonals
of each of these tiles are 45 cm and 30 cm in length.
Find the total cost of polishing the floor, if the cost per m
2
is Rs 4.
Answer:
Rs. 810
Solution:
Total number of tiles = 3000
Area of 1 rhombus shaped tile =
1
2
× d
1
× d
2
=
1
2
× (45) × (30) cm
2
= 675 cm
2
Total floor Area = Number of tiles × Area of one tile
= 3000 × 675 cm
2
= 2025000 cm
2
=
2025000
10000
m
2
= 205.5 m
2
Total cost of polishing the floor = Rs 4 × 202.5 = Rs 810.0 = Rs 810
Question 8
Mohan wants to buy a trapezium shaped field.
Its side along the river is parallel to and twice the side along the road.
If the area of this field is 10500 m
2
and the perpendicular distance
between the two parallel sides is 100 m, find the length of the side
along the river.
Answer:
140 m
Solution:
Let the length of the side along road be x m.
So, length of the side along river is 2x m.
Area of the trapezium shaped field =
1
2
× height × (sum of parallel sides)
10500 =
1
2
× height × (sum of parallel sides)
Or, 10500 =
1
2
× 100 × (x + 2x) or,
10500 2
100
= 3x
Or, 210 = 3x or, x =
210
3
= 70
Length of the side along the river = 2x m = 140 m.
Question 9
Top surface of a raised platform is in the shape of a regular octagon as shown in
the figure.
Find the area of the octagonal surface.
Answer:
119 m
2
Solution:
Area of the octagon ABCDEFGH
= Area of the trapezium ABCH + Area of the rectangle CDGH + Area of the
trapezium DEFG
= (
1
2
× 4 × (5 + 11) + (5
× 11) +
1
2
× 4 × (5 + 11)) m
2
= (32 + 55 + 32) m
2
= 119 m
2
Question 10
There is a pentagonal shaped park as shown in the figure.
For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways.
Can you suggest some other way of
finding its area?
Answer:
Area using Jyoti’s way =
22
1
2 30 15 7.5 337.5
2
mm
,
Area using
Kavita’s way =
22
1
( 15 15 15 15) 337.5
2
mm
Solution:
Jyoti’s Diagram
Area of the park = Area of the trapezium ABCF
+ Area of the trapezium AEDF
= (
1
2
× 7.5 × (30 + 15) +
1
2
× 7.5 × (30 + 15)) m
2
=
1
2
× 7.5 × (45 + 45) m
2
= 337.5 m
2
Kavita’s Diagram
Area of park = Area of ABE + Area of square BCDE
=
1
2
× 15 × 15 + (15 × 15)
=
225
2
+ 225 = 225
1
1
2



= 225 ×
3
2
=
675
2
m
2
= 337.5 m
2
Area can also be calculated in the following way:
Area of the park = Area of the
ABE + Area of the
BCD + Area of the
BED
=
1
2
× BE × AF +
1
2
× CD × BC +
1
2
× BE × DE
= (
1
2
× 15 × 15 +
1
2
× 15 × 15 ×
1
2
× 15 × 15) m
2
=
2 2 2
225 225 225 225 225 225 675
()
2 2 2 2
mm
2
m

= 337.5 m
2
Question 11
Diagram of the adjacent picture frame has outer dimensions 24 cm × 8cm.
and inner dimensions 16 cm × 20 cm.
Find the area of each section of the frame, if the width of each section is
same.
Answer:
80 cm
2
, 96 cm
2
, 80 cm
2
, 96 cm
2
Solution:
Let the width of each section be ‘x’ cm.
Each of the frames has a shape of a trapezium.
As per the figure,
24 = 16 + x + x
Or, 24 – 16 = 2x or, 2x = 8
Or, x = 4
Two opposite sections have the same area.
The area of the section trapezium
ABFE =
1
2
× 4 × (16 + 24) cm
2
= 2 × 40 = 80 cm
2
Area of the opposite section (Trapezium CDHG) = 80 cm
2
(Trapezium BCGF) =
1
2
× 4 × (28 + 20) cm
2
= 96 cm
2
Area of the opposite section (Trapezium ADHE) = 96 cm
2
.
Exercise 11.3
Question 1
There are two cuboidal boxes as shown in the adjoining figure.
Which box requires the lesser amount of material to make?
Answer:
(a)
Solution:
(a) Total surface area of the cuboidal box
= 2 (lb + bh + hl)
= 2 (60 × 40 + 40 × 50 + 50 × 60) cm
2
[ l = 60 cm, b = 40 cm, h = 50 cm]
= 2 (2400 + 2000 + 3000) cm
2
= 2 × 7400 cm
2
= 14800 cm
2
(b) Total surface area of the cuboidal box
= 2 (lb + bc + hl)
= 2(50 × 50 + 50 × 50 + 50 × 50) cm
2
[ l = 50 cm, b = 50 cm, h = 50 cm]
= 2 (2500 + 2500 + 2500) cm
2
= 2 × 7500 cm
2
=
15000 cm
2
The cuboid (a) requires lesser amount of material.
Question 2
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth.
How many meters of tarpaulin of width 96 cm is required to cover 10 such suitcases?
Answer:
144 m
Solution:
Given, length, l = 80 cm, breadth, b = 8 cm and height, h = 4 cm.
Total surface area of one suitcase = 2 (lb + bh + hl)
= 2 (80 × 48 + 48 × 24 + 24 × 80) cm
2
= 2 × 6912 cm
2
= 13824 cm
2
and the total surface area of
100 suitcases = 13824 × 100 cm
2
Width of the tarpaulin = 96 cm
Length of the tarpaulin =
1382400
96
cm = 14400 cm
=
14400
100
m = 144 m.
Question 3
Find the side of a cube whose surface area is 600 cm
2
.
Answer:
10 cm
Solution:
Given, surface area of the cube = 600 cm
2
.
But surface area of cube = 6l
2
Or, 600 cm
2
= 6l
2
or, l
2
= 100 cm
2
Or, l = 10 cm
Therefore, the side of the cube is 10 cm.
Question 4
Rukhsar painted the outside of the cabinet of measure 1 m × 2m × 1.5 m.
How much surface area did she cover if she painted all except the bottom of the
cabinet.
Answer:
11 m
2
Solution:
Total surface area of the cuboidal cabinet box = 2 (lb + bh + hl)
But the bottom of the box was not painted, so total surface area of the cabinet painted
= 2 (lb + bh + hl) – l × b = (2 (1 × 2 + 2 × 1.5 + 1.5 × 1) – 1 × 2) m
2
= (2 (2 + 3 + 1.5) – 2) m
2
= (2 × 6.5 – 2) m
2
= (13 – 2) m
2
= 11 m
2
Question 5
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and
height of 15 m, 10 m and 7 m respectively.
From each can of paint 100 m
2
of area is painted.
How many cans of paint will she need to paint the room?
Answer:
5 cans
Solution:
Length of the class room (l) = 15 m
Breadth of the class room (b) = 10 m
Height of the class room (h) = 7 m
Total area to be painted = lb + 2bh + 2hl (excluding the floor)
= (15 × 10 + 2 × 10 × 7 + 2 × 7 × 15) m
2
= (150 + 140 + 210) m
2
= 500 sq.m.
Area covered by 1 can of paint = 100 m
2
The total number of cans of paint =
500
5
100
Question 6
Describe how the two figures at the right are alike and how they are
different.
Which box has larger lateral surface area?
Answer:
Similarity Both have same heights.
Difference one is a cylinder, the other is a cube.
The cube has larger lateral surface area
Solution:
Similarity: Both have same heights.
Difference: One is a cylinder, the other is a cube.
Lateral surface area of the cylindrical box = 2
rh
= 2 ×
22 7
72
× 7 cm
2
= 154 cm
2
Lateral surface area of the cubical box = 4l
2
cm
2
= 4(7)
2
cm
2
= 4 × 49 cm
2
= 196 cm
2
The cube has larger lateral surface area.
Question 7
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal.
How much sheet of metal is required?
Answer:
440 m
2
Solution:
Radius of the cylindrical tank (r) = 7m and height (h) = 3m
Total surface area of the cylindrical tank
= 2
r (h + r) = 2 ×
22
7
× 7 × (3 + 7) m
2
= 440 m
2
Hence, the area of the metal sheet required is 440 m
2
.
Question 8
The lateral surface area of a hollow cylinder is 4224 cm
2
.
It is cut along its height and formed a rectangular sheet of width 33cm.
Find the perimeter of rectangular sheet?
Answer:
322 cm
Solution:
Lateral surface area of the hollow cylinder = 4224 cm
2
Height of the cylinder = 33 cm
Lateral surface area of the cylinder = 2
rh or, 4224 cm
2
= 2
r × h
Or, 4224 cm
2
= 2
r × 33 cm or, 2
r = 128 cm
Circumference (2
r) of the circular end of the cylinder becomes the length of the
rectangular sheet.
the
perimeter of the rectangular sheet = 2(l + b) = 2(128 + 33) cm = 2 ×
161 cm = 322 cm
Question 9
A road roller takes 750 complete revolutions to move once over to level a road.
Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Answer:
1980 m
2
Solution:
Height of the road roller (h) = 1 m
Radius of the road roller (r) = 84/2 cm = 42 cm = 0.42 m
Curved surface area of the road roller = 2
rh = 2 ×
22
7
× 0.42 × 1 m
2
= 44 × 0.06 m
2
= 2.64 m
2
Area of the road covered in one revolution = 2.64 m
2
Total area of the road covered in 750 revolutions = 2.64 × 750 m
2
= 1980 m
2
Question 10
A company packages its milk powder in cylindrical container whose base has
Company diameter of 14 cm and height 20 cm.
Places a label around the surface of the container (as shown in the figure).
If the label is placed 2 cm from top and bottom, what is the area of the label.
Answer:
704 cm
2
Solution:
For cylindrical container:
Diameter = 14 cm, Radius (r) =
14
2
cm = 7 cm, Height (h) = 20 cm
For label: Radius of the label (r) = 7 cm
Height of the label (h) = (20 – (2 × 2)) cm = 16 cm
Total curved surface area of the label = 2
rh = 2 ×
22
7
× 7 × 16 cm
2
= 704 cm
2
Exercise 11.4
Question 1
Given a cylindrical tank, in which situation will you find surface area and in which
situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Answer:
(a) Volume (b) Surface area (c) Volume
Solution:
(a) Volume of the cylinder
(b) Total surface area of cylinder
(c) Volume of the cylinder
Question 2
Diameter of cylinder A is 7 cm, and the height is 14 cm.
Diameter of cylinder B is 14 cm and the height is 7 cm.
Without doing any calculations can you suggest whose volume is greater?
Verify it by finding the volume of both the cylinders.
Check whether the cylinder with greater volume also has greater surface
area?
Answer:
Volume of the cylinder B is greater; Surface area of the cylinder B is greater.
Solution:
For cylinder A: Diameter (d) = 7 cm and height (h) = 14 cm
For cylinder B: Diameter (d) = 14 cm and height (h) = 7 cm
If the radius is doubled and the height is halved, the new volume will
be
22
h
(2r) 2 r h
2
,
Which is clearly greater than the original volume
2
rh
Therefore, volume of the cylinder B is more than that of B.
Volume of the cylinder A =
2 3 3
22 7 7
r h 14cm 539cm
7 2 2
Volume of the cylinder B =
2 3 3
22
r h 7 7 7cm 1078cm
7
,
Which is greater than volume of the cylinder A.
Hence, verified.
The total surface area of the cylinder A
= (2 ×
22
7
×
7
2
× 14 + 2 ×
22
7
×
7
2
×
7
2
cm
2
= 308 cm
2
+ 77 cm
2
= 385 cm
2
The total surface area of the cylinder B
= 2
r (r + h) = 2 ×
22
7
× 7 × (7 + 7)
= 44 × 14 = 616 cm
2
Yes, cylinder with greater volume has greater surface area.
Question 3
Find the height of a cuboid whose base area is 180 cm
2
and volume is 900 cm
3
?
Answer:
5 cm
Solution:
Let the height of the cuboid be h m.
Base area of the cuboid = 180 cm
2
Volume of the cuboid = 900 cm
3
Volume of the cuboid = Area of the base × height
900 = 180 × h or, h =
900
180
= 5.
Hence, the height of the cuboid is 5 cm.
Question 4
A cuboid is of dimensions 60 cm × 54 cm × 30 cm.
How many small cubes with side 6 cm can be placed in the given cuboid?
Answer:
450
Solution:
Volume of the cuboid = 60 × 54 × 30 cm
3
= 97200 cm
3
Volume of l small cube = (side)
3
= (6)
3
cm
3
= 216 cm
3
No. of small cubes =
Volumeof thecuboid 97200
Volumeof1cube 216
= 450
Question 5
Find the height of the cylinder whose volume is 1.54 m
3
and diameter of the base is
140 cm?
Answer:
1 m
Solution:
Volume of the cylinder = 1.54 m
3
(Given)
Diameter of the base = 140 cm (Given)
Radius of the base =
140
2
cm = 70 cm = 0.70 m.
Let height of the cylinder be h m.
Now, Volume of the cylinder =
r
2
h
or 1.54 =
22
7
× 0.7 × 0.7 × h or, h =
1.54 7
22 0.7 0.7

= 1m
Hence, the height of the cylinder is 1 m.
Question 6
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m.
Find the quantity of milk in liters that can be stored in the tank?
Answer:
49500 l
Solution:
Radius of the cylindrical tank (r) = 1.5 m (Given)
Length of the cylindrical tank (l) = 7 m (Given)
Volume of the tank =
22
7
× 1.5 × 1.5 × 7m3
= 49.50 m
3
= 49.50 × 100 × 100 × 100 cm
3
= 49500000 cm
3
=
49500000
1000
liters = 49500 liters [
1000 cm
3
= 1 liters.]
Hence, the volume of the milk that can be stored in the tank is 49500 liters.
Question 7
If each edge of a cube is doubled,
(i) How many times will its surface area increase?
(ii) How many times will its volume increase?
Answer:
(i) 4 times (ii) 8 times
Solution:
Let each edge of the cube be x
Surface area of the cube = 6x
2
, Volume of the cube = x
3
When each edge of the cube is 2x, then,
the surface area of the new cube = 6 (side)
2
= 6 (2x)
2
= 6 × 4x
2
= 24x
2
Now, the volume of the new cube = (side)
3
= (2x)
3
= 8x
3
(i)
Surfaceareaof the newcube
Surfaceareaof theinitialcube
2
2
24x
= = 4
6x
Surface area of the new cube = 4 × (surface area of the initial cube)
(ii)
Volumeof newcube
8
Surfaceareaof theinitialcube
3
3
8x
=
x
Volume of the new cube = 8 × Volume of the initial cube
Question 8
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute.
If the volume of reservoir is 108 m
3
, find the number of hours it will take to fill
the reservoir.
Answer:
30 hours
Solution:
Given, volume of the reservoir = 108 m
3
The rate of pouring of water = 60 liters per minute
= 60 × 1000 cm
3
per min. [1 liter = 1000 cm
3
]
=
60 1000
60
cm
3
per sec [1 min. = 60 sec]
= 1000 cm
3
per sec =
1000
100 100 100
m
3
per sec =
1
1000
m
3
per sec
Time required to fill the reservoir =
Volumeof the reservoir 108
Rateof waterpouring 1/1000
sec
= 108 × 1000 sec. =
108 1000
60 60
hr = 30 hr.