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Heron’s Formula - Class 9 : Notes


 

(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 semi-perimeter i.e. half the perimeter of the triangle = (a + b + c)/2.

For ExampleFind 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 =
                          =
                          =
                          = cm2 = 11.62 cm2

For ExampleAn 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 =
                          =
                          =
                          = cm2 = 11.62 cm2

(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 semi-perimeter will be
s = (100 + 100 + 160)/2 = 180 m.
Now, area of ∆ ABD =
                                  =
                                  = 4800 m2.
Hence, each one of them will get 4800 m2 of area.

For ExampleA field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel 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 m2
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 m2.
Now, total area of field = area of ∆BEC + area of quadrilateral ABED
                                                    = 84 + 112 = 196 m2.
Thus, area of field is 196 m2.



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