NCERT Textbook- Heron's Formula Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Created by: Indu Gupta

Class 9 : NCERT Textbook- Heron's Formula Class 9 Notes | EduRev

 Page 1


CHAPTER 12
HERON’S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles  and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
 
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
 Area of a triangle =
1
2
 × base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
Page 2


CHAPTER 12
HERON’S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles  and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
 
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
 Area of a triangle =
1
2
 × base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
area of ? ABC is given by
1
2
 × base × height = 
1
2
 × 12 × 5 cm
2
, i.e., 30 cm
2
Note that we could also take 5 cm as the base and 12 cm as height.
Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height of
this triangle?
Let us recall how we find its height when we
know its sides. This is possible in an equilateral
triangle. Take the mid-point of QR as M and join it to
P. We know that PMQ is a right triangle. Therefore,
by using Pythagoras Theorem, we can find the length
PM as shown below:
PQ
2
 =PM
2
 + QM
2
i.e., (10)
2
 =PM
2
 + (5)
2
, since QM = MR.
Therefore, we have PM
2
 = 75
i.e., PM = 75 cm = 53 cm.
Then area of ? PQR = 
1
2
 × base × height = 
2
1
10 5 3 cm 25 3
2
×× = cm
2
.
Let us see now whether we can calculate the area of an isosceles triangle also
with the help of this formula. For example, we take a triangle XYZ with two equal
sides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we draw
a perpendicular XP to side YZ. You can see that this perpendicular XP divides the
base YZ of the triangle in two equal parts.
Therefore, YP = PZ = 
1
2
 YZ = 4 cm
Then, by using Pythagoras theorem, we get
XP
2
 =XY
2
 – YP
2
=5
2
 – 4
2
 = 25 – 16 = 9
So, XP = 3 cm
Now, area of ? XYZ = 
1
2
 × base YZ × height XP
 = 
1
2
 × 8 × 3 cm
2
 = 12 cm
2
.
Fig. 12.2
Fig. 12.3
Page 3


CHAPTER 12
HERON’S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles  and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
 
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
 Area of a triangle =
1
2
 × base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
area of ? ABC is given by
1
2
 × base × height = 
1
2
 × 12 × 5 cm
2
, i.e., 30 cm
2
Note that we could also take 5 cm as the base and 12 cm as height.
Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height of
this triangle?
Let us recall how we find its height when we
know its sides. This is possible in an equilateral
triangle. Take the mid-point of QR as M and join it to
P. We know that PMQ is a right triangle. Therefore,
by using Pythagoras Theorem, we can find the length
PM as shown below:
PQ
2
 =PM
2
 + QM
2
i.e., (10)
2
 =PM
2
 + (5)
2
, since QM = MR.
Therefore, we have PM
2
 = 75
i.e., PM = 75 cm = 53 cm.
Then area of ? PQR = 
1
2
 × base × height = 
2
1
10 5 3 cm 25 3
2
×× = cm
2
.
Let us see now whether we can calculate the area of an isosceles triangle also
with the help of this formula. For example, we take a triangle XYZ with two equal
sides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we draw
a perpendicular XP to side YZ. You can see that this perpendicular XP divides the
base YZ of the triangle in two equal parts.
Therefore, YP = PZ = 
1
2
 YZ = 4 cm
Then, by using Pythagoras theorem, we get
XP
2
 =XY
2
 – YP
2
=5
2
 – 4
2
 = 25 – 16 = 9
So, XP = 3 cm
Now, area of ? XYZ = 
1
2
 × base YZ × height XP
 = 
1
2
 × 8 × 3 cm
2
 = 12 cm
2
.
Fig. 12.2
Fig. 12.3
HERON’S FORMULA 199
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Now suppose that we know the lengths of the sides of a scalene triangle and not
the height. Can you still find its area? For instance, you have a triangular park whose
sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want
to apply the formula, you will have to calculate its height. But we do not have a clue to
calculate the height. Try doing so. If you are not able to get it, then go to the next
section.
12.2 Area of a Triangle — by Heron’s Formula
Heron was born in about 10AD possibly in Alexandria in
Egypt. He worked in applied mathematics. His works on
mathematical and physical subjects are so numerous and
varied that he is considered to be an encyclopedic writer
in these fields. His geometrical works deal largely with
problems on mensuration written in three books. Book I
deals with the area of squares, rectangles, triangles,
trapezoids (trapezia), various other specialised
quadrilaterals, the regular polygons, circles, surfaces of
cylinders, cones, spheres etc. In this book, Heron has
derived the famous formula for the area of a triangle in
terms of its three sides.
The formula given by Heron about the area of a triangle, is also known as Hero’ s
formula. It is stated as:
Area of a triangle = ()( )( ) ss a sb sc - -- (II)
where a, b and c are the sides of the triangle, and s = semi-perimeter i.e. half the
perimeter of the triangle = 
2
ab c + +
,
This formula is helpful where it is not possible to find the height of the triangle
easily. Let us apply it to calculate the area of the triangular park ABC, mentioned
above (see Fig. 12.5).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s = 
40 24 32
2
++
 m = 48 m.
Fig. 12.4
Heron (10AD – 75 AD)
Page 4


CHAPTER 12
HERON’S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles  and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
 
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
 Area of a triangle =
1
2
 × base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
area of ? ABC is given by
1
2
 × base × height = 
1
2
 × 12 × 5 cm
2
, i.e., 30 cm
2
Note that we could also take 5 cm as the base and 12 cm as height.
Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height of
this triangle?
Let us recall how we find its height when we
know its sides. This is possible in an equilateral
triangle. Take the mid-point of QR as M and join it to
P. We know that PMQ is a right triangle. Therefore,
by using Pythagoras Theorem, we can find the length
PM as shown below:
PQ
2
 =PM
2
 + QM
2
i.e., (10)
2
 =PM
2
 + (5)
2
, since QM = MR.
Therefore, we have PM
2
 = 75
i.e., PM = 75 cm = 53 cm.
Then area of ? PQR = 
1
2
 × base × height = 
2
1
10 5 3 cm 25 3
2
×× = cm
2
.
Let us see now whether we can calculate the area of an isosceles triangle also
with the help of this formula. For example, we take a triangle XYZ with two equal
sides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we draw
a perpendicular XP to side YZ. You can see that this perpendicular XP divides the
base YZ of the triangle in two equal parts.
Therefore, YP = PZ = 
1
2
 YZ = 4 cm
Then, by using Pythagoras theorem, we get
XP
2
 =XY
2
 – YP
2
=5
2
 – 4
2
 = 25 – 16 = 9
So, XP = 3 cm
Now, area of ? XYZ = 
1
2
 × base YZ × height XP
 = 
1
2
 × 8 × 3 cm
2
 = 12 cm
2
.
Fig. 12.2
Fig. 12.3
HERON’S FORMULA 199
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Now suppose that we know the lengths of the sides of a scalene triangle and not
the height. Can you still find its area? For instance, you have a triangular park whose
sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want
to apply the formula, you will have to calculate its height. But we do not have a clue to
calculate the height. Try doing so. If you are not able to get it, then go to the next
section.
12.2 Area of a Triangle — by Heron’s Formula
Heron was born in about 10AD possibly in Alexandria in
Egypt. He worked in applied mathematics. His works on
mathematical and physical subjects are so numerous and
varied that he is considered to be an encyclopedic writer
in these fields. His geometrical works deal largely with
problems on mensuration written in three books. Book I
deals with the area of squares, rectangles, triangles,
trapezoids (trapezia), various other specialised
quadrilaterals, the regular polygons, circles, surfaces of
cylinders, cones, spheres etc. In this book, Heron has
derived the famous formula for the area of a triangle in
terms of its three sides.
The formula given by Heron about the area of a triangle, is also known as Hero’ s
formula. It is stated as:
Area of a triangle = ()( )( ) ss a sb sc - -- (II)
where a, b and c are the sides of the triangle, and s = semi-perimeter i.e. half the
perimeter of the triangle = 
2
ab c + +
,
This formula is helpful where it is not possible to find the height of the triangle
easily. Let us apply it to calculate the area of the triangular park ABC, mentioned
above (see Fig. 12.5).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s = 
40 24 32
2
++
 m = 48 m.
Fig. 12.4
Heron (10AD – 75 AD)
200 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
s – a = (48 – 40) m = 8 m,
s – b = (48 – 24) m = 24 m,
s – c = (48 – 32) m = 16 m.
Therefore, area of the park ABC
= ()( )( ) -- - ssa s b s c
= 
22
48 8 24 16 m 384m ×× × =
We see that 32
2
 + 24
2
 = 1024 + 576 = 1600 = 40
2
. Therefore, the sides of the park
make a right triangle. The largest side i.e. BC which is 40 m will be the hypotenuse
and the angle between the sides AB and AC will be 90°.
By using Formula I, we can check that the area of the park is 
1
2
 × 32 × 24 m
2
= 384 m
2
.
We find that the area we have got is the same as we found by using Heron’s
formula.
Now using Heron’s formula, you verify this fact by finding the areas of other
triangles discussed earlier viz;
(i) equilateral triangle with side 10 cm.
(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm.
You will see that
For (i), we have s = 
10 10 10
2
+ +
 cm = 15 cm.
Area of triangle  = 15(15 10) (15 10) (15 10) -- - cm
2
                         =
22
15 5 5 5 cm 25 3 cm ×× × =
For (ii), we have s = 
85 5
cm 9 cm
2
++
=
.
Area of triangle = 9(98)(95)(9 5) -- - cm
2 
= 
22
9 1 4 4 cm 12 cm . ×× × =
Let us now solve some more examples:
Fig. 12.5
Page 5


CHAPTER 12
HERON’S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles  and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
 
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
 Area of a triangle =
1
2
 × base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
area of ? ABC is given by
1
2
 × base × height = 
1
2
 × 12 × 5 cm
2
, i.e., 30 cm
2
Note that we could also take 5 cm as the base and 12 cm as height.
Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height of
this triangle?
Let us recall how we find its height when we
know its sides. This is possible in an equilateral
triangle. Take the mid-point of QR as M and join it to
P. We know that PMQ is a right triangle. Therefore,
by using Pythagoras Theorem, we can find the length
PM as shown below:
PQ
2
 =PM
2
 + QM
2
i.e., (10)
2
 =PM
2
 + (5)
2
, since QM = MR.
Therefore, we have PM
2
 = 75
i.e., PM = 75 cm = 53 cm.
Then area of ? PQR = 
1
2
 × base × height = 
2
1
10 5 3 cm 25 3
2
×× = cm
2
.
Let us see now whether we can calculate the area of an isosceles triangle also
with the help of this formula. For example, we take a triangle XYZ with two equal
sides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we draw
a perpendicular XP to side YZ. You can see that this perpendicular XP divides the
base YZ of the triangle in two equal parts.
Therefore, YP = PZ = 
1
2
 YZ = 4 cm
Then, by using Pythagoras theorem, we get
XP
2
 =XY
2
 – YP
2
=5
2
 – 4
2
 = 25 – 16 = 9
So, XP = 3 cm
Now, area of ? XYZ = 
1
2
 × base YZ × height XP
 = 
1
2
 × 8 × 3 cm
2
 = 12 cm
2
.
Fig. 12.2
Fig. 12.3
HERON’S FORMULA 199
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Now suppose that we know the lengths of the sides of a scalene triangle and not
the height. Can you still find its area? For instance, you have a triangular park whose
sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want
to apply the formula, you will have to calculate its height. But we do not have a clue to
calculate the height. Try doing so. If you are not able to get it, then go to the next
section.
12.2 Area of a Triangle — by Heron’s Formula
Heron was born in about 10AD possibly in Alexandria in
Egypt. He worked in applied mathematics. His works on
mathematical and physical subjects are so numerous and
varied that he is considered to be an encyclopedic writer
in these fields. His geometrical works deal largely with
problems on mensuration written in three books. Book I
deals with the area of squares, rectangles, triangles,
trapezoids (trapezia), various other specialised
quadrilaterals, the regular polygons, circles, surfaces of
cylinders, cones, spheres etc. In this book, Heron has
derived the famous formula for the area of a triangle in
terms of its three sides.
The formula given by Heron about the area of a triangle, is also known as Hero’ s
formula. It is stated as:
Area of a triangle = ()( )( ) ss a sb sc - -- (II)
where a, b and c are the sides of the triangle, and s = semi-perimeter i.e. half the
perimeter of the triangle = 
2
ab c + +
,
This formula is helpful where it is not possible to find the height of the triangle
easily. Let us apply it to calculate the area of the triangular park ABC, mentioned
above (see Fig. 12.5).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s = 
40 24 32
2
++
 m = 48 m.
Fig. 12.4
Heron (10AD – 75 AD)
200 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
s – a = (48 – 40) m = 8 m,
s – b = (48 – 24) m = 24 m,
s – c = (48 – 32) m = 16 m.
Therefore, area of the park ABC
= ()( )( ) -- - ssa s b s c
= 
22
48 8 24 16 m 384m ×× × =
We see that 32
2
 + 24
2
 = 1024 + 576 = 1600 = 40
2
. Therefore, the sides of the park
make a right triangle. The largest side i.e. BC which is 40 m will be the hypotenuse
and the angle between the sides AB and AC will be 90°.
By using Formula I, we can check that the area of the park is 
1
2
 × 32 × 24 m
2
= 384 m
2
.
We find that the area we have got is the same as we found by using Heron’s
formula.
Now using Heron’s formula, you verify this fact by finding the areas of other
triangles discussed earlier viz;
(i) equilateral triangle with side 10 cm.
(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm.
You will see that
For (i), we have s = 
10 10 10
2
+ +
 cm = 15 cm.
Area of triangle  = 15(15 10) (15 10) (15 10) -- - cm
2
                         =
22
15 5 5 5 cm 25 3 cm ×× × =
For (ii), we have s = 
85 5
cm 9 cm
2
++
=
.
Area of triangle = 9(98)(95)(9 5) -- - cm
2 
= 
22
9 1 4 4 cm 12 cm . ×× × =
Let us now solve some more examples:
Fig. 12.5
HERON’S FORMULA 201
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Example 1 : Find the area of a triangle, two sides of which are 8 cm and 11 cm and
the perimeter is 32 cm (see Fig. 12.6).
Solution : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cm
So, 2s = 32 i.e. s = 16 cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11) cm = 5 cm,
s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = ()( )( ) ss a sb sc - --
=
22
16 8 5 3 cm 8 30 cm ×× × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 12.7). A
gardener Dhania has to put a fence all around it and also plant grass inside. How
much area does she need to plant? Find the cost of fencing it with barbed wire at the
rate of Rs 20 per metre leaving a space 3m wide for a gate on one side.
Solution : For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.
i.e., s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,
s – c = (125 – 50) m = 75 m.
Therefore, area of the park = ()( )( ) ss a sb sc - --
= 125 5 45 75 ×× × m
2
=
2
375 15 m
Also, perimeter of the park =  AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs 20 × 247 = Rs 4940
Fig. 12.6
Fig. 12.7
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NCERT Textbook- Heron's Formula Class 9 Notes | EduRev

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NCERT Textbook- Heron's Formula Class 9 Notes | EduRev

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video lectures

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Summary

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Viva Questions

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mock tests for examination

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Previous Year Questions with Solutions

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