Heron’s Formula Class 9 Notes Maths Chapter 10

Students can go through AP 9th Class Maths Notes Chapter 10 Heron’s Formula to understand and remember the concepts easily.

Class 9 Maths Chapter 10 Notes Heron’s Formula

Heron’s formula is used to find the area of triangle whose sides are given.
The proof of it can be found in the book ‘Metrica’.

When base and height are given, we can find its area by using the formula
A = \(\frac{1}{2}\) bh → (1)
We cannot find the area of triangle. If a height of it is not given.

In the following case a, b, c are length of sides of a triangle as shown below.

Then its perimeter (P) = a + b + c
and its semi-perimeter = \(\frac{a+b+c}{2}\)
Semi perimeter is denoted by “s”.
Then the area of a triangle whose sides are ‘a’, ‘b’, c’ and semi perimeter ‘s’ can be found using the following formula.
Δ (or) A = \(\sqrt{s(s-a)(s-b)(s-c)}\)
This formula is known as Heron’s formula.

Proof : Let the sides ‘BC’ = a, CA = b and ‘AB’ =.c units as shown below.

Let AD ⊥ BC and ‘D’ is at a distance of ‘d’ units from ‘B’.
Then BD = d,
DC = a – d
Let the height ’AD’ is ‘h’ units, then its area = \(\frac{\text { base } \times \text { height }}{2}\) = \(\frac{\mathrm{ah}}{2}\)
Δ = \(\frac{\mathrm{ah}}{2}\) → (1)
Now, consider ΔABD
here ∠D = 90°, BD = d, AD = h, AB = c
Using Pythagoras theorem,
AB² = AD² + BD²
c² = h² + d² → (2)
⇒ h² = c² – d² → (3)
Now, consider ΔADC
here ∠D = 90°, AD = h, DC = a – d, AC = b from Pythagoras theorem,
AC² = AD² + DC²
b² = h² + (a – d)² → (4)
Now, put the value of h² from (3) in (4) we get
b² + c² – d² + (a – d)²

Now put the value of d² in (3) we get
h² = c² – d²


Hence proved.

This heron’s formula is useful to find area of a triangle where its height is difficult to find.
Ex : Consider a right angled triangle whose sides are 32 m, 24 m, 40 m.
Here the largest is hypotenuse = 40 m Then if base is 32 m, A then height will be 24 m (or) vice-versa

then its area = Δ = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 32 × 24
= 384 m² → (1)
Let us, we are not informed that it is a right-triangle.
Then let a = 40 m
b = 32 m
c = 24 m

Then its semi perimeter
s = \(\frac{40+32+24}{2}\) = \(\frac{1}{2}\) = 48 m
s = 48 m
s – a = 48 – 40 = 8m
s – b = 48 – 32 = 16m
s – c = 48 – 24 = 24 m
then from Heron’s formula
Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{48 \times 8 \times 16 \times 24}\)
= \(\sqrt{8 \times 6 \times 8 \times 4 \times 4 \times 6 \times 2 \times 2}\)
= 8 × 6 × 2 × 4 = 384 m² → (2)
(1) = (2)
It is clear that in both ways we get same result.
Ex :
i) Find the area of equilateral triangle whose side is 10 cm.
Solution:
Given that it is equilateral triangle. Hence all sides are equal.
So a = b = c = 10 cm
then semi-perimeter
s = \(\frac{a+b+c}{2}\) = \(\frac{10+10+10}{2}\) = \(\frac{30}{2}\) = 15 cm
Then area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15(15-10)(15-10)(15-10)}\)
= \(\sqrt{15 \times 5 \times 5 \times 5}\) = 25√3 cm²

ii) Find the area of Isosceles triangle whose sides are 8 cm, 5 cm, 5 cm.
Solution:
Here a = 8 cm, b = 5 cm, c = 5 cm
then s = \(\frac{a+b+c}{2}\) = \(\frac{8+5+5}{2}\) = \(\frac{18}{2}\) = 9 cm
then the area of given triangle
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{9(9-8)(9-5)(9-5)}\)
= \(\sqrt{9 \times 1 \times 4 \times 4}\) = \(\sqrt{3 \times 3 \times 1 \times 4 \times 4}\)
= 3 × 4 =12 cm²
∴ Area =12 cm²