
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

Chapter Notes – Heron’s Formula
(1) Area of triangle: Generally, the area of triangle is given by the following formula:
Area of a triangle = ½ x base x height
(2) Area of a triangle by Heron’s Formula: The formula given by Heron about the area of triangle is given below:
Area of a triangle =
Where a, b and c are the sides of the triangle and s is semiperimeter i.e. half the perimeter of the triangle = (a + b + c)/2.
For Example: Find the area of triangle having three sides as 4 cm, 6 cm and 8 cm using Heron’s formula.
Let a = 4 cm, b = 6 cm and c = 8 cm.
Therefore, s = (a + b + c)/2 = (4 + 6 + 8)/2 = 9 cm.
Using Heron’s formula, we have,
Area of triangle =
=
=
= cm^{2} = 11.62 cm^{2}
For Example: An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Let the given perimeter p = 30 cm and the two equal sides a = b = 12 cm.
Let the third side be denoted by c.
Now, we know that, perimeter of triangle = sum of all sides
So, p = a + b + c
30 = 12 + 12 + c
Hence, c = 6 cm.
Now, s = Perimeter of triangle/2 = 30/2 = 15 cm.
Now, using Heron’s formula,
Area of triangle =
=
=
= cm^{2} = 11.62 cm^{2}
(3) Application of Heron’s Formula to Find Area of Quadrilaterals:
Let us take examples to understand this:
For Example: Kavita has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?Let the piece of land be denoted by ABCD as shown in figure.
Given, perimeter of land is 400 m and diagonal is of 160 m.
Since, given land is in shape of rhombus, therefore each side will be 400/4 = 100 m.
Let diagonal BD = 160 m.
Now, considering ∆ ABD, its semiperimeter will be
s = (100 + 100 + 160)/2 = 180 m.
Now, area of ∆ ABD =
=
= 4800 m^{2}.
Hence, each one of them will get 4800 m^{2} of area.
For Example: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The nonparallel sides are 14 m and 13 m. Find the area of the field.Now, for the given figure draw line BE parallel to AD and a perpendicular BF on CD as shown in the figure below:
It can be seen from the figure that ABED is a parallelogram, where,
BE = AD = 13 m; ED = AB = 10 m and EC = 25 – ED = 25 – 10 = 15 m.
Now, for ∆BEC, semi – perimeter s = (13 + 14 +15)/2 = 21 m.
Therefore, area of ∆BEC =
=
=
= 84 m^{2
}We know that, area of ∆BEC = ½ x base x height
= ½ x CE x BF
84 = ½ x 15 x BF
On solving,
BF = 11.2 m
Now, area of quadrilateral ABED = base x height
= DE x BF
= 10 x 11.2 = 112 m^{2}.
Now, total area of field = area of ∆BEC + area of quadrilateral ABED
= 84 + 112 = 196 m^{2}.
Thus, area of field is 196 m^{2}.