NCERT Textbook: Mensuration | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download

 Page 1


MENSURATION  103
9.1  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 .
9.2  Area of a Polygon
W e split a quadrilateral into triangles and find its area. Similar methods can be used to find
the area of a polygon. Observe the following for a pentagon: (Fig 9.1, 9.2)
Mensuration
CHAPTER
9
By constructing one diagonal AD and two
perpendiculars BF and CG on it, pentagon ABCDE is
divided into four parts. So, area of ABCDE = area of
right angled ? AFB +  area of trapezium BFGC + area
of right angled ? CGD + area of ? AED. (Identify the
parallel sides of trapezium BFGC.)
By constructing two diagonals AC and AD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of  ? ABC + area of
? ACD + area of ? AED.
Fig 9.1
Fig 9.2
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Page 2


MENSURATION  103
9.1  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 .
9.2  Area of a Polygon
W e split a quadrilateral into triangles and find its area. Similar methods can be used to find
the area of a polygon. Observe the following for a pentagon: (Fig 9.1, 9.2)
Mensuration
CHAPTER
9
By constructing one diagonal AD and two
perpendiculars BF and CG on it, pentagon ABCDE is
divided into four parts. So, area of ABCDE = area of
right angled ? AFB +  area of trapezium BFGC + area
of right angled ? CGD + area of ? AED. (Identify the
parallel sides of trapezium BFGC.)
By constructing two diagonals AC and AD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of  ? ABC + area of
? ACD + area of ? AED.
Fig 9.1
Fig 9.2
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104  MATHEMATICS
TRY THESE
Fig 9.3
Fig 9.4
(i) Divide the following polygons (Fig 9.3) into parts (triangles and trapezium) to find
out its area.
FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR
(ii) Polygon ABCDE is divided into parts as shown below (Fig 9.4). Find its area if
AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm,
CH = 3 cm, EG = 2.5 cm.
Area of Polygon ABCDE = area of ? AFB + ....
Area of ? AFB = 
1
2
× AF × BF = 
1
2
× 3 × 2 = ....
Area of trapezium FBCH = FH ×
(BF CH)
2
+
= 3 × 
(2 3)
2
+
   [FH = AH – AF]
Area of  ?CHD = 
1
2
× HD× CH  = ....;   Area of  ?ADE = 
1
2
× AD × GE  = ....
So, the area of polygon ABCDE = ....
(iii) Find the area of polygon MNOPQR (Fig 9.5) if
MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm,
MA = 2 cm
NA, OC, QD and RB are perpendiculars to
diagonal MP .
Example 1: The area of a trapezium shaped field is 480 m
2
, the distance between
two parallel sides is 15 m and one of the parallel side  is 20 m. Find the other parallel side.
Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel
side be b, height h = 15 m.
The given area of trapezium = 480 m
2
.
Area of a trapezium =
1
2
h (a + b)
So    480 =
1
2
× 15 × (20 + b)     or     
480 2
15
×
 = 20 + b
or      64 = 20 + b   or   b = 44 m
Hence the other parallel side of the trapezium is 44 m.
Fig 9.5
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Page 3


MENSURATION  103
9.1  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 .
9.2  Area of a Polygon
W e split a quadrilateral into triangles and find its area. Similar methods can be used to find
the area of a polygon. Observe the following for a pentagon: (Fig 9.1, 9.2)
Mensuration
CHAPTER
9
By constructing one diagonal AD and two
perpendiculars BF and CG on it, pentagon ABCDE is
divided into four parts. So, area of ABCDE = area of
right angled ? AFB +  area of trapezium BFGC + area
of right angled ? CGD + area of ? AED. (Identify the
parallel sides of trapezium BFGC.)
By constructing two diagonals AC and AD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of  ? ABC + area of
? ACD + area of ? AED.
Fig 9.1
Fig 9.2
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104  MATHEMATICS
TRY THESE
Fig 9.3
Fig 9.4
(i) Divide the following polygons (Fig 9.3) into parts (triangles and trapezium) to find
out its area.
FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR
(ii) Polygon ABCDE is divided into parts as shown below (Fig 9.4). Find its area if
AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm,
CH = 3 cm, EG = 2.5 cm.
Area of Polygon ABCDE = area of ? AFB + ....
Area of ? AFB = 
1
2
× AF × BF = 
1
2
× 3 × 2 = ....
Area of trapezium FBCH = FH ×
(BF CH)
2
+
= 3 × 
(2 3)
2
+
   [FH = AH – AF]
Area of  ?CHD = 
1
2
× HD× CH  = ....;   Area of  ?ADE = 
1
2
× AD × GE  = ....
So, the area of polygon ABCDE = ....
(iii) Find the area of polygon MNOPQR (Fig 9.5) if
MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm,
MA = 2 cm
NA, OC, QD and RB are perpendiculars to
diagonal MP .
Example 1: The area of a trapezium shaped field is 480 m
2
, the distance between
two parallel sides is 15 m and one of the parallel side  is 20 m. Find the other parallel side.
Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel
side be b, height h = 15 m.
The given area of trapezium = 480 m
2
.
Area of a trapezium =
1
2
h (a + b)
So    480 =
1
2
× 15 × (20 + b)     or     
480 2
15
×
 = 20 + b
or      64 = 20 + b   or   b = 44 m
Hence the other parallel side of the trapezium is 44 m.
Fig 9.5
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MENSURATION  105
Example 2: The area of a rhombus is 240 cm
2
 and one of the diagonals is 16 cm.
Find the other diagonal.
Solution: Let length of one diagonal d
1
 = 16 cm
and length of the other diagonal = d
2
Area of the rhombus =
1
2
 d
1
 . d
2
 = 240
So,
2
1
16
2
d ·
 = 240
Therefore, d
2
 = 30 cm
Hence the length of the second diagonal is 30 cm.
Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 9.6). Aman and Ridhima
divided it in two different ways (Fig 9.7).
Find the area of this hexagon using both ways.
Fig 9.6
Fig 9.7
Solution: Aman’s method:
Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. Y ou can
verify it by paper folding (Fig 9.8).
Now area of trapezium MNQR = 
(11 5)
4
2
+
×
 = 2 × 16 = 32 cm
2
.
So the area of hexagon MNOPQR = 2 × 32 = 64 cm
2
.
Ridhima’s method:
? MNO and ? RPQ are congruent triangles with altitude
3 cm (Fig 9.9).
Y ou can verify this by cutting off these two triangles and
placing them on one another.
Area of ? MNO = 
1
2
 × 8 × 3 = 12 cm
2
 = Area of ? RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm
2
.
Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm
2
.
EXERCISE 9.1
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.
Fig 9.9
Fig 9.8
Aman’s method Ridhima’s method
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Page 4


MENSURATION  103
9.1  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 .
9.2  Area of a Polygon
W e split a quadrilateral into triangles and find its area. Similar methods can be used to find
the area of a polygon. Observe the following for a pentagon: (Fig 9.1, 9.2)
Mensuration
CHAPTER
9
By constructing one diagonal AD and two
perpendiculars BF and CG on it, pentagon ABCDE is
divided into four parts. So, area of ABCDE = area of
right angled ? AFB +  area of trapezium BFGC + area
of right angled ? CGD + area of ? AED. (Identify the
parallel sides of trapezium BFGC.)
By constructing two diagonals AC and AD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of  ? ABC + area of
? ACD + area of ? AED.
Fig 9.1
Fig 9.2
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104  MATHEMATICS
TRY THESE
Fig 9.3
Fig 9.4
(i) Divide the following polygons (Fig 9.3) into parts (triangles and trapezium) to find
out its area.
FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR
(ii) Polygon ABCDE is divided into parts as shown below (Fig 9.4). Find its area if
AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm,
CH = 3 cm, EG = 2.5 cm.
Area of Polygon ABCDE = area of ? AFB + ....
Area of ? AFB = 
1
2
× AF × BF = 
1
2
× 3 × 2 = ....
Area of trapezium FBCH = FH ×
(BF CH)
2
+
= 3 × 
(2 3)
2
+
   [FH = AH – AF]
Area of  ?CHD = 
1
2
× HD× CH  = ....;   Area of  ?ADE = 
1
2
× AD × GE  = ....
So, the area of polygon ABCDE = ....
(iii) Find the area of polygon MNOPQR (Fig 9.5) if
MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm,
MA = 2 cm
NA, OC, QD and RB are perpendiculars to
diagonal MP .
Example 1: The area of a trapezium shaped field is 480 m
2
, the distance between
two parallel sides is 15 m and one of the parallel side  is 20 m. Find the other parallel side.
Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel
side be b, height h = 15 m.
The given area of trapezium = 480 m
2
.
Area of a trapezium =
1
2
h (a + b)
So    480 =
1
2
× 15 × (20 + b)     or     
480 2
15
×
 = 20 + b
or      64 = 20 + b   or   b = 44 m
Hence the other parallel side of the trapezium is 44 m.
Fig 9.5
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MENSURATION  105
Example 2: The area of a rhombus is 240 cm
2
 and one of the diagonals is 16 cm.
Find the other diagonal.
Solution: Let length of one diagonal d
1
 = 16 cm
and length of the other diagonal = d
2
Area of the rhombus =
1
2
 d
1
 . d
2
 = 240
So,
2
1
16
2
d ·
 = 240
Therefore, d
2
 = 30 cm
Hence the length of the second diagonal is 30 cm.
Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 9.6). Aman and Ridhima
divided it in two different ways (Fig 9.7).
Find the area of this hexagon using both ways.
Fig 9.6
Fig 9.7
Solution: Aman’s method:
Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. Y ou can
verify it by paper folding (Fig 9.8).
Now area of trapezium MNQR = 
(11 5)
4
2
+
×
 = 2 × 16 = 32 cm
2
.
So the area of hexagon MNOPQR = 2 × 32 = 64 cm
2
.
Ridhima’s method:
? MNO and ? RPQ are congruent triangles with altitude
3 cm (Fig 9.9).
Y ou can verify this by cutting off these two triangles and
placing them on one another.
Area of ? MNO = 
1
2
 × 8 × 3 = 12 cm
2
 = Area of ? RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm
2
.
Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm
2
.
EXERCISE 9.1
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.
Fig 9.9
Fig 9.8
Aman’s method Ridhima’s method
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106  MATHEMATICS
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.
3. Length of the fence of a trapezium shaped field ABCD is 120 m. If
BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side
AB is perpendicular to the parallel sides AD and BC.
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.
5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find
its area.
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm.
If one of its diagonals is 8 cm long, find the length of the other diagonal.
7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of
its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor,
if the cost per m
2
 is ` 4.
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.
9. T op 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.
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?
11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm
and inner dimensions 16 cm × 20 cm. Find the area of each section of
the frame, if the width of each section is same.
9.3  Solid Shapes
In your earlier classes you have studied that two dimensional figures can be identified as
the faces of three dimensional shapes. Observe the solids which we have discussed so far
(Fig 9.10).
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Page 5


MENSURATION  103
9.1  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 .
9.2  Area of a Polygon
W e split a quadrilateral into triangles and find its area. Similar methods can be used to find
the area of a polygon. Observe the following for a pentagon: (Fig 9.1, 9.2)
Mensuration
CHAPTER
9
By constructing one diagonal AD and two
perpendiculars BF and CG on it, pentagon ABCDE is
divided into four parts. So, area of ABCDE = area of
right angled ? AFB +  area of trapezium BFGC + area
of right angled ? CGD + area of ? AED. (Identify the
parallel sides of trapezium BFGC.)
By constructing two diagonals AC and AD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of  ? ABC + area of
? ACD + area of ? AED.
Fig 9.1
Fig 9.2
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104  MATHEMATICS
TRY THESE
Fig 9.3
Fig 9.4
(i) Divide the following polygons (Fig 9.3) into parts (triangles and trapezium) to find
out its area.
FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR
(ii) Polygon ABCDE is divided into parts as shown below (Fig 9.4). Find its area if
AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm,
CH = 3 cm, EG = 2.5 cm.
Area of Polygon ABCDE = area of ? AFB + ....
Area of ? AFB = 
1
2
× AF × BF = 
1
2
× 3 × 2 = ....
Area of trapezium FBCH = FH ×
(BF CH)
2
+
= 3 × 
(2 3)
2
+
   [FH = AH – AF]
Area of  ?CHD = 
1
2
× HD× CH  = ....;   Area of  ?ADE = 
1
2
× AD × GE  = ....
So, the area of polygon ABCDE = ....
(iii) Find the area of polygon MNOPQR (Fig 9.5) if
MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm,
MA = 2 cm
NA, OC, QD and RB are perpendiculars to
diagonal MP .
Example 1: The area of a trapezium shaped field is 480 m
2
, the distance between
two parallel sides is 15 m and one of the parallel side  is 20 m. Find the other parallel side.
Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel
side be b, height h = 15 m.
The given area of trapezium = 480 m
2
.
Area of a trapezium =
1
2
h (a + b)
So    480 =
1
2
× 15 × (20 + b)     or     
480 2
15
×
 = 20 + b
or      64 = 20 + b   or   b = 44 m
Hence the other parallel side of the trapezium is 44 m.
Fig 9.5
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MENSURATION  105
Example 2: The area of a rhombus is 240 cm
2
 and one of the diagonals is 16 cm.
Find the other diagonal.
Solution: Let length of one diagonal d
1
 = 16 cm
and length of the other diagonal = d
2
Area of the rhombus =
1
2
 d
1
 . d
2
 = 240
So,
2
1
16
2
d ·
 = 240
Therefore, d
2
 = 30 cm
Hence the length of the second diagonal is 30 cm.
Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 9.6). Aman and Ridhima
divided it in two different ways (Fig 9.7).
Find the area of this hexagon using both ways.
Fig 9.6
Fig 9.7
Solution: Aman’s method:
Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. Y ou can
verify it by paper folding (Fig 9.8).
Now area of trapezium MNQR = 
(11 5)
4
2
+
×
 = 2 × 16 = 32 cm
2
.
So the area of hexagon MNOPQR = 2 × 32 = 64 cm
2
.
Ridhima’s method:
? MNO and ? RPQ are congruent triangles with altitude
3 cm (Fig 9.9).
Y ou can verify this by cutting off these two triangles and
placing them on one another.
Area of ? MNO = 
1
2
 × 8 × 3 = 12 cm
2
 = Area of ? RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm
2
.
Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm
2
.
EXERCISE 9.1
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.
Fig 9.9
Fig 9.8
Aman’s method Ridhima’s method
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106  MATHEMATICS
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.
3. Length of the fence of a trapezium shaped field ABCD is 120 m. If
BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side
AB is perpendicular to the parallel sides AD and BC.
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.
5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find
its area.
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm.
If one of its diagonals is 8 cm long, find the length of the other diagonal.
7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of
its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor,
if the cost per m
2
 is ` 4.
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.
9. T op 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.
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?
11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm
and inner dimensions 16 cm × 20 cm. Find the area of each section of
the frame, if the width of each section is same.
9.3  Solid Shapes
In your earlier classes you have studied that two dimensional figures can be identified as
the faces of three dimensional shapes. Observe the solids which we have discussed so far
(Fig 9.10).
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MENSURATION  107
DO THIS
Observe that some shapes have two or more than two identical (congruent) faces.
Name them. Which solid has all congruent faces?
Soaps, toys, pastes, snacks etc. often come in the packing of cuboidal, cubical or
cylindrical boxes. Collect, such boxes (Fig 9.11).
Fig 9.11
Fig 9.10
All six faces are rectangular,
and opposites faces are
identical. So there are three
pairs of identical faces.
Cuboidal Box Cubical Box
All six faces
are squares
and identical.
One curved surface
and two circular
faces which are
identical.
Cylindrical Box
Now take one type of box at a time. Cut out all the faces it has. Observe the shape of
each face and find the number of faces of the box that are identical by placing them on
each other. Write down your observations.
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