NCERT Textbook Chapter 11 - Mensuration , Class 8, Maths | EduRev Notes

Mathematics (Maths) Class 8

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Class 8 : NCERT Textbook Chapter 11 - Mensuration , Class 8, Maths | EduRev Notes

 Page 1


MENSURATION  169
11.1  Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
various plane figures such as triangles, rectangles, circles etc. W e have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
plane closed figures like quadrilaterals.
W e will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2  Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) What is the total length of the fence surrounding it? T o find the length of the fence we
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m
2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m
2
 area, how
many bags of cement would be required to construct the
cemented path?
We can say that the number of cement bags used = 
area of the path
area cemented by 1 bag
.
Area of cemented path = Area of park – Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m
2
.
That  is 28 × 18 m
2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
Page 2


MENSURATION  169
11.1  Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
various plane figures such as triangles, rectangles, circles etc. W e have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
plane closed figures like quadrilaterals.
W e will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2  Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) What is the total length of the fence surrounding it? T o find the length of the fence we
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m
2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m
2
 area, how
many bags of cement would be required to construct the
cemented path?
We can say that the number of cement bags used = 
area of the path
area cemented by 1 bag
.
Area of cemented path = Area of park – Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m
2
.
That  is 28 × 18 m
2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
170  MATHEMATICS
TRY THESE
Area of rectangular beds = ------------------
Area of park left after cementing the path = ------------------
Area covered by the grass = ------------------
We can find areas of geometrical shapes other than rectangles also if certain
measurements are given  to us . Try to recall and match the following:
Diagram Shape Area
rectangle a × a
square b × h
triangle pb
2
parallelogram
1
2
bh ×
circle a × b
Can you write an expression for the perimeter of each of the above shapes?
49 cm
2
77 cm
2
98 cm
2
(a) Match the following figures with their respective areas in the box.
(b) Write the perimeter of each shape.
Page 3


MENSURATION  169
11.1  Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
various plane figures such as triangles, rectangles, circles etc. W e have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
plane closed figures like quadrilaterals.
W e will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2  Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) What is the total length of the fence surrounding it? T o find the length of the fence we
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m
2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m
2
 area, how
many bags of cement would be required to construct the
cemented path?
We can say that the number of cement bags used = 
area of the path
area cemented by 1 bag
.
Area of cemented path = Area of park – Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m
2
.
That  is 28 × 18 m
2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
170  MATHEMATICS
TRY THESE
Area of rectangular beds = ------------------
Area of park left after cementing the path = ------------------
Area covered by the grass = ------------------
We can find areas of geometrical shapes other than rectangles also if certain
measurements are given  to us . Try to recall and match the following:
Diagram Shape Area
rectangle a × a
square b × h
triangle pb
2
parallelogram
1
2
bh ×
circle a × b
Can you write an expression for the perimeter of each of the above shapes?
49 cm
2
77 cm
2
98 cm
2
(a) Match the following figures with their respective areas in the box.
(b) Write the perimeter of each shape.
MENSURATION  171
EXERCISE 11.1
1. A square and a rectangular field with
measurements as given in the figure have the same
perimeter. Which field has a larger area?
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 ` 55 per m
2
.
3. The shape of a garden is rectangular in the middle and semi circular
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) metres].
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).
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 = 2pr, where r
is the radius of the circle.
(a) (b) (c)
(b) (a)
11.3  Area of Trapezium
Nazma owns a plot near a main road
(Fig 11.2). Unlike some other rectangular
plots in her neighbourhood, the plot has
only one pair of parallel opposite sides.
So, it is nearly a trapezium in shape. Can
you find out its area?
Let us name the vertices of this plot as
shown in Fig 11.3.
By drawing EC || AB, we can divide it
into two parts, one of rectangular shape
and the other of triangular shape, (which
is right angled at C), as shown in Fig 1 1.3. (b = c + a = 30 m)
Fig 11.3
Fig 11.2
Page 4


MENSURATION  169
11.1  Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
various plane figures such as triangles, rectangles, circles etc. W e have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
plane closed figures like quadrilaterals.
W e will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2  Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) What is the total length of the fence surrounding it? T o find the length of the fence we
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m
2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m
2
 area, how
many bags of cement would be required to construct the
cemented path?
We can say that the number of cement bags used = 
area of the path
area cemented by 1 bag
.
Area of cemented path = Area of park – Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m
2
.
That  is 28 × 18 m
2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
170  MATHEMATICS
TRY THESE
Area of rectangular beds = ------------------
Area of park left after cementing the path = ------------------
Area covered by the grass = ------------------
We can find areas of geometrical shapes other than rectangles also if certain
measurements are given  to us . Try to recall and match the following:
Diagram Shape Area
rectangle a × a
square b × h
triangle pb
2
parallelogram
1
2
bh ×
circle a × b
Can you write an expression for the perimeter of each of the above shapes?
49 cm
2
77 cm
2
98 cm
2
(a) Match the following figures with their respective areas in the box.
(b) Write the perimeter of each shape.
MENSURATION  171
EXERCISE 11.1
1. A square and a rectangular field with
measurements as given in the figure have the same
perimeter. Which field has a larger area?
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 ` 55 per m
2
.
3. The shape of a garden is rectangular in the middle and semi circular
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) metres].
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).
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 = 2pr, where r
is the radius of the circle.
(a) (b) (c)
(b) (a)
11.3  Area of Trapezium
Nazma owns a plot near a main road
(Fig 11.2). Unlike some other rectangular
plots in her neighbourhood, the plot has
only one pair of parallel opposite sides.
So, it is nearly a trapezium in shape. Can
you find out its area?
Let us name the vertices of this plot as
shown in Fig 11.3.
By drawing EC || AB, we can divide it
into two parts, one of rectangular shape
and the other of triangular shape, (which
is right angled at C), as shown in Fig 1 1.3. (b = c + a = 30 m)
Fig 11.3
Fig 11.2
172  MATHEMATICS
DO THIS
TRY THESE
Area of ? ECD = 
1
2
h × c = 
1
12 10
2
××
 = 60 m
2
.
Area of rectangle ABCE = h × a = 12 × 20 = 240 m
2
.
Area of trapezium ABDE = area of ? ECD + Area of rectangle ABCE = 60 + 240 = 300 m
2
.
W e can write the area by combining the two areas and write the area of trapezium as
area of ABDE  =
1
2
h × c + h × a = h 
2
c
a
??
+
??
??
= h 
2
22
ca c a a
h
+++ ?? ? ?
=
?? ? ?
?? ? ?
=
()
2
ba
h
+
 = 
(sum of parallelsides)
height
2
By substituting the values of h, b and a in this expression, we find 
()
2
ba
h
+
 = 300 m
2
.
1. Nazma’ s sister also has a trapezium shaped plot. Divide it into three parts as shown
(Fig 11.4). Show that the area of  trapezium WXYZ 
()
2
ab
h
+
= .
 Fig 11.4
2. If h = 10 cm, c = 6 cm, b = 12 cm,
d = 4 cm, find the values of each of
its parts separetely and add to find
the area WXYZ. V erify it by putting
the values of h, a and b in the
expression 
()
2
ha b +
.
Fig 11.5
1. Draw any trapezium WXYZ on a piece
of graph paper as shown in the figure
and cut it out (Fig 11.5).
2. Find the mid point of XY by folding
the side and name it A (Fig 11.6).
Fig 11.6
Page 5


MENSURATION  169
11.1  Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
various plane figures such as triangles, rectangles, circles etc. W e have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
plane closed figures like quadrilaterals.
W e will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2  Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) What is the total length of the fence surrounding it? T o find the length of the fence we
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m
2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m
2
 area, how
many bags of cement would be required to construct the
cemented path?
We can say that the number of cement bags used = 
area of the path
area cemented by 1 bag
.
Area of cemented path = Area of park – Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m
2
.
That  is 28 × 18 m
2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
170  MATHEMATICS
TRY THESE
Area of rectangular beds = ------------------
Area of park left after cementing the path = ------------------
Area covered by the grass = ------------------
We can find areas of geometrical shapes other than rectangles also if certain
measurements are given  to us . Try to recall and match the following:
Diagram Shape Area
rectangle a × a
square b × h
triangle pb
2
parallelogram
1
2
bh ×
circle a × b
Can you write an expression for the perimeter of each of the above shapes?
49 cm
2
77 cm
2
98 cm
2
(a) Match the following figures with their respective areas in the box.
(b) Write the perimeter of each shape.
MENSURATION  171
EXERCISE 11.1
1. A square and a rectangular field with
measurements as given in the figure have the same
perimeter. Which field has a larger area?
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 ` 55 per m
2
.
3. The shape of a garden is rectangular in the middle and semi circular
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) metres].
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).
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 = 2pr, where r
is the radius of the circle.
(a) (b) (c)
(b) (a)
11.3  Area of Trapezium
Nazma owns a plot near a main road
(Fig 11.2). Unlike some other rectangular
plots in her neighbourhood, the plot has
only one pair of parallel opposite sides.
So, it is nearly a trapezium in shape. Can
you find out its area?
Let us name the vertices of this plot as
shown in Fig 11.3.
By drawing EC || AB, we can divide it
into two parts, one of rectangular shape
and the other of triangular shape, (which
is right angled at C), as shown in Fig 1 1.3. (b = c + a = 30 m)
Fig 11.3
Fig 11.2
172  MATHEMATICS
DO THIS
TRY THESE
Area of ? ECD = 
1
2
h × c = 
1
12 10
2
××
 = 60 m
2
.
Area of rectangle ABCE = h × a = 12 × 20 = 240 m
2
.
Area of trapezium ABDE = area of ? ECD + Area of rectangle ABCE = 60 + 240 = 300 m
2
.
W e can write the area by combining the two areas and write the area of trapezium as
area of ABDE  =
1
2
h × c + h × a = h 
2
c
a
??
+
??
??
= h 
2
22
ca c a a
h
+++ ?? ? ?
=
?? ? ?
?? ? ?
=
()
2
ba
h
+
 = 
(sum of parallelsides)
height
2
By substituting the values of h, b and a in this expression, we find 
()
2
ba
h
+
 = 300 m
2
.
1. Nazma’ s sister also has a trapezium shaped plot. Divide it into three parts as shown
(Fig 11.4). Show that the area of  trapezium WXYZ 
()
2
ab
h
+
= .
 Fig 11.4
2. If h = 10 cm, c = 6 cm, b = 12 cm,
d = 4 cm, find the values of each of
its parts separetely and add to find
the area WXYZ. V erify it by putting
the values of h, a and b in the
expression 
()
2
ha b +
.
Fig 11.5
1. Draw any trapezium WXYZ on a piece
of graph paper as shown in the figure
and cut it out (Fig 11.5).
2. Find the mid point of XY by folding
the side and name it A (Fig 11.6).
Fig 11.6
MENSURATION  173
DO THIS
TRY THESE
3. Cut trapezium WXYZ into two pieces by cutting along ZA. Place ? ZY A as shown
in Fig 11.7, where A Y is placed on AX.
What is the length of the base of the larger
triangle? Write an expression for the area of
this triangle (Fig 11.7).
4. The area of this triangle and the area of the trapezium WXYZ are same (How?).
Get the expression for the area of trapezium by using the expression for the area
of triangle.
So to find the area of a trapezium we need to know the length of the parallel sides and the
perpendicular distance between these two parallel sides. Half the product of the sum of
the lengths of parallel sides and the perpendicular distance between them gives the area of
trapezium.
Find the area of the following trapeziums (Fig 11.8).
(i) (ii)
Fig 11.7
Fig 11.8
In Class VII we learnt to draw parallelograms of equal areas with different perimeters.
Can it be done for trapezium? Check if the following trapeziums are of equal areas but
have different perimeters (Fig 11.9).
Fig 11.9
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