Page 1
CHAPTER 10
HERON’S FORMULA
10.1 Area of a Triangle — by Heron’s Formula
We know that the area of triangle when its height is given, is
1
2
× base × height. 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.
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 = ( ) ( ) ( ) s s a s b s c - - - (I)
Heron (10 C.E. – 75 C.E.)
Fig. 10.1
Rationalised 2023-24
Page 2
CHAPTER 10
HERON’S FORMULA
10.1 Area of a Triangle — by Heron’s Formula
We know that the area of triangle when its height is given, is
1
2
× base × height. 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.
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 = ( ) ( ) ( ) s s a s b s c - - - (I)
Heron (10 C.E. – 75 C.E.)
Fig. 10.1
Rationalised 2023-24
132 MATHEMA TICS
where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the
perimeter of the triangle =
2
a b 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. 10.2).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s =
40 24 32
2
+ +
m = 48 m.
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
= ( ) ( ) ( ) - - - s s a s b s c
=
2 2
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°.
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.
Fig. 10.2
Rationalised 2023-24
Page 3
CHAPTER 10
HERON’S FORMULA
10.1 Area of a Triangle — by Heron’s Formula
We know that the area of triangle when its height is given, is
1
2
× base × height. 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.
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 = ( ) ( ) ( ) s s a s b s c - - - (I)
Heron (10 C.E. – 75 C.E.)
Fig. 10.1
Rationalised 2023-24
132 MATHEMA TICS
where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the
perimeter of the triangle =
2
a b 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. 10.2).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s =
40 24 32
2
+ +
m = 48 m.
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
= ( ) ( ) ( ) - - - s s a s b s c
=
2 2
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°.
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.
Fig. 10.2
Rationalised 2023-24
HERON’S FORMULA 133
Area of triangle = 15(15 10) (15 10) (15 10) - - - cm
2
=
2 2
15 5 5 5 cm 25 3 cm × × × =
For (ii), we have s =
8 5 5
cm 9 cm
2
+ +
=
Area of triangle = 9(9 8) (9 5) (9 5) - - - cm
2
=
2 2
9 1 4 4 cm 12 cm . × × × =
Let us now solve some more examples:
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. 10.3).
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 = ( ) ( ) ( ) s s a s b s c - - -
=
2 2
16 8 5 3 cm 8 30 cm × × × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). 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 `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.
Fig. 10.3
Fig. 10.4
Rationalised 2023-24
Page 4
CHAPTER 10
HERON’S FORMULA
10.1 Area of a Triangle — by Heron’s Formula
We know that the area of triangle when its height is given, is
1
2
× base × height. 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.
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 = ( ) ( ) ( ) s s a s b s c - - - (I)
Heron (10 C.E. – 75 C.E.)
Fig. 10.1
Rationalised 2023-24
132 MATHEMA TICS
where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the
perimeter of the triangle =
2
a b 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. 10.2).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s =
40 24 32
2
+ +
m = 48 m.
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
= ( ) ( ) ( ) - - - s s a s b s c
=
2 2
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°.
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.
Fig. 10.2
Rationalised 2023-24
HERON’S FORMULA 133
Area of triangle = 15(15 10) (15 10) (15 10) - - - cm
2
=
2 2
15 5 5 5 cm 25 3 cm × × × =
For (ii), we have s =
8 5 5
cm 9 cm
2
+ +
=
Area of triangle = 9(9 8) (9 5) (9 5) - - - cm
2
=
2 2
9 1 4 4 cm 12 cm . × × × =
Let us now solve some more examples:
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. 10.3).
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 = ( ) ( ) ( ) s s a s b s c - - -
=
2 2
16 8 5 3 cm 8 30 cm × × × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). 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 `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.
Fig. 10.3
Fig. 10.4
Rationalised 2023-24
134 MATHEMA TICS
Therefore, area of the park = ( ) ( ) ( ) s s a s b s c - - -
= 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 = `20 × 247 = `4940
Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter
is 300 m. Find its area.
Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 10.5).
Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)
Therefore, 15x = 300, which gives x = 20.
So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m
i.e., 60 m, 100 m and 140 m.
Can you now find the area [Using Heron’s formula]?
We have s =
60 100 140
2
+ +
m = 150 m,
and area will be 150(150 60) (150 100) (150 140) - - - m
2
= 150 90 50 10 × × × m
2
=
2
1500 3 m
EXERCISE 10.1
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with
side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is
180 cm, what will be the area of the signal board?
Fig. 10.5
Rationalised 2023-24
Page 5
CHAPTER 10
HERON’S FORMULA
10.1 Area of a Triangle — by Heron’s Formula
We know that the area of triangle when its height is given, is
1
2
× base × height. 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.
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 = ( ) ( ) ( ) s s a s b s c - - - (I)
Heron (10 C.E. – 75 C.E.)
Fig. 10.1
Rationalised 2023-24
132 MATHEMA TICS
where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the
perimeter of the triangle =
2
a b 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. 10.2).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s =
40 24 32
2
+ +
m = 48 m.
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
= ( ) ( ) ( ) - - - s s a s b s c
=
2 2
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°.
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.
Fig. 10.2
Rationalised 2023-24
HERON’S FORMULA 133
Area of triangle = 15(15 10) (15 10) (15 10) - - - cm
2
=
2 2
15 5 5 5 cm 25 3 cm × × × =
For (ii), we have s =
8 5 5
cm 9 cm
2
+ +
=
Area of triangle = 9(9 8) (9 5) (9 5) - - - cm
2
=
2 2
9 1 4 4 cm 12 cm . × × × =
Let us now solve some more examples:
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. 10.3).
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 = ( ) ( ) ( ) s s a s b s c - - -
=
2 2
16 8 5 3 cm 8 30 cm × × × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). 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 `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.
Fig. 10.3
Fig. 10.4
Rationalised 2023-24
134 MATHEMA TICS
Therefore, area of the park = ( ) ( ) ( ) s s a s b s c - - -
= 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 = `20 × 247 = `4940
Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter
is 300 m. Find its area.
Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 10.5).
Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)
Therefore, 15x = 300, which gives x = 20.
So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m
i.e., 60 m, 100 m and 140 m.
Can you now find the area [Using Heron’s formula]?
We have s =
60 100 140
2
+ +
m = 150 m,
and area will be 150(150 60) (150 100) (150 140) - - - m
2
= 150 90 50 10 × × × m
2
=
2
1500 3 m
EXERCISE 10.1
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with
side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is
180 cm, what will be the area of the signal board?
Fig. 10.5
Rationalised 2023-24
HERON’S FORMULA 135
2. The triangular side walls of a flyover have been used for advertisements. The sides of
the walls are 122 m, 22 m and 120 m (see Fig. 10.6). The advertisements yield an
earning of ` 5000 per m
2
per year. A company hired one of its walls for 3 months. How
much rent did it pay?
Fig. 10.6
3. There is a slide in a park. One of its side walls has been painted in some colour with a
message “KEEP THE P ARK GREEN AND CLEAN” (see Fig. 10.7 ). If the sides of the
wall are 15 m, 11 m and 6 m, find the area painted in colour.
Fig. 10.7
4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is
42cm.
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find
the area of the triangle.
Rationalised 2023-24
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