HERON’S FORMULA-Notes
Maths - Notes
Heron’s Formula
Heron’s Formula is a special method used to find the area of a triangle when the lengths of all three
sides are known.
It is extremely useful because we don’t need to know the height of the triangle.
Why We Need Heron’s Formula
In many triangles, especially scalene triangles, drawing or measuring the height is difficult. For example, a triangle where none of the sides are equal or perpendicular. In such cases, Heron’s Formula makes the calculation simple.
Semi-Perimeter (s)
Before using the formula, we must find the semi-perimeter of the triangle.
If the sides are \(a,\;b,\;c\) then the semi-perimeter is \[s=\frac{a+b+c}{2}\]
It is simply half of the total perimeter of the triangle.
Heron’s Formula for Area
Once we get the semi-perimeter, the area is calculated by:
Area of Triangle
\[
\Delta= \sqrt{s(s-a)(s-b)(s-c)}
\]
This formula can be used for any triangle:
- Scalene
- Isosceles
- Obtuse
- Acute
- Even right triangles (though in right triangles, normal \(\frac{1}{2}base\times height\) is easier)
Important Points to Remember
- All three sides must be known.
- The triangle must be valid (the sum of any two sides should be greater than the third side).
- The formula works for both small and large triangles.
- We use square root at the final step, so calculations should be neat.
Heron’s Formula for Quadrilaterals
A quadrilateral can be divided into two triangles.
So we can apply Heron’s Formula to each triangle separately and then add both areas.
Example-1
Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm
Solution:
Given:
The triangle has two sides of lengths \(8\text{ cm, }11\text{ cm and } \text{the perimeter is } 32\text{ cm}\)-
Third side
$$ \begin{aligned} \text{Third side} &= 32 - (8 + 11) \\ &= 32 - 19 \\ &= 13~\text{cm} \end{aligned} $$
-
Side lengths
$$ \begin{aligned} a &= 8\text{ cm, } \\ b &= 11 \text{ cm, } \\ c &= 13\text{ cm } \end{aligned} $$
-
Calculate the semi-perimeter
$$ s = \dfrac{a + b + c}{2} = \dfrac{32}{2} = 16~\text{cm} $$
-
Find the area using Heron's formula
$$\small\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $$
Substitute the values:
$$\begin{aligned} \text{Area} &= \tiny\sqrt{16 \times (16-8) \times (16-11) \times (16-13)} \\ &= \sqrt{16 \times 8 \times 5 \times 3} \\ &= \sqrt{64 \times 30} \\ &= 8\sqrt{30}~\text{cm}^2 \end{aligned} $$
The area of the triangle is $$8\sqrt{30}~\text{cm}^2$$
Example-2
A triangular park ABC has sides 120m, 80m and 50m. 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
Given: Sides of triangle \(ABC\) are \(a=120\ \text{m},\; b=80\ \text{m},\; c=50\ \text{m}.\)
1. Area (Heron's formula)
Semi-perimeter \(s\):
$$\begin{aligned}s&=\frac{a+b+c}{2}\\&=\frac{120+80+50}{2}\\&=\frac{250}{2}\\&=125\ \text{m}.\end{aligned}$$
Compute the three terms \(s-a,\; s-b,\; s-c\):
$$\begin{aligned} s-a&=125-120=5,\\ s-b&=125-80=45,\\ s-c&=125-50=75\end{aligned}$$
Heron's formula gives the area \( \Delta \):
$$\begin{aligned}\Delta&=\sqrt{s(s-a)(s-b)(s-c)}\\\\&=\sqrt{125\cdot 5\cdot 45\cdot 75}\end{aligned}$$
Simplify the product under the square root by prime/factor grouping:
\[\begin{aligned} 125\cdot 5\cdot 45\cdot 75 &=5^{3+1+1+2}\cdot 3^{2+1}\\ &=5^7\cdot 3^3. \end{aligned}\]
Therefore
$$\begin{aligned} \Delta=\sqrt{5^7\cdot 3^3} &=5^{3}\cdot 3\cdot \sqrt{5\cdot 3}\\ &=125\cdot 3\cdot \sqrt{15}\\ &=375\sqrt{15}\ \text{m}^2. \end{aligned}$$
Numeric value (to two decimal places):
$$\Delta\approx 375\sqrt{15}\approx 1452.37\ \text{m}^2.$$
2. Length of wire required and cost of fencing
Perimeter of the park:
$$\begin{aligned}P&=a+b+c\\&=120+80+50\\&=250\ \text{m}\end{aligned}$$
A gate of width \(3\ \text{m}\) is left open, so the length of wire to be used is
$$\begin{aligned}\text{Wire length}&=P-\text{gate width}\\&=250-3\\&=247\ \text{m}\end{aligned}$$
Rate of barbed wire = ₹20 per metre. Thus the cost is
$$\text{Cost}=20\times 247=\text{₹ }4940$$
Answers
- Area to be planted: \(\boxed{375\sqrt{15}\ \text{m}^2\approx 1452.37\ \text{m}^2}\)
- Length of wire needed: \(\boxed{247\ \text{m}}\)
- Cost of fencing: \(\boxed{₹\,4940}\)
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
Given: The sides of the triangular plot are in the ratio \(3:5:7\) and the perimeter is \(300\ \text{m}.\)
1. Find the sides
Sum of ratio parts = \(3+5+7=15\).
Each part equals \(\dfrac{300}{15}=20\ \text{m}.\)
So the sides are
\[\begin{aligned} a=3\times 20&=60\ \text{m},\\ b=5\times 20&=100\ \text{m},\\ c=7\times 20&=140\ \text{m} \end{aligned}\]
2. Area by Heron's formula
Semi-perimeter \(s\):
$$\begin{aligned}s&=\frac{a+b+c}{2}\\\\&=\frac{300}{2}\\\\&=150\ \text{m}\end{aligned}$$
Compute the three differences:
$$\begin{aligned}s-a&=150-60=90,\\ s-b&=150-100=50,\\ s-c&=150-140=10\end{aligned}$$
Using Heron:
$$\begin{aligned} \Delta&=\sqrt{s(s-a)(s-b)(s-c)}\\&=\sqrt{150\cdot 90\cdot 50\cdot 10}. \end{aligned}$$
Simplify by prime-factor grouping:
\[150=2\cdot3\cdot5^2,\\\ 90=2\cdot3^2\cdot5,\\\ 50=2\cdot5^2,\\\ 10=2\cdot5.\]Hence \[ 150\cdot90\cdot50\cdot10=2^{4}\cdot 3^{3}\cdot 5^{6}. \] Taking the square root, \[\begin{aligned} \Delta&=\sqrt{2^{4}\cdot 3^{3}\cdot 5^{6} }\\&=2^{2}\cdot 3^{1}\cdot 5^{3}\cdot\sqrt{3}\\&=4\cdot 3\cdot125\sqrt{3}\\&=1500\sqrt{3}\ \text{m}^2. \end{aligned}\]
Numeric value (to two decimals):
$$\begin{aligned}\Delta&\approx 1500\sqrt{3}\\&\approx 2598.08\ \text{m}^2\end{aligned}$$
Final answer
- Area of the triangular plot \[\begin{aligned}&=\boxed{1500\sqrt{3}\ \text{m}^2}\\&\approx\boxed{2598.08\ \text{m}^2}\end{aligned}\]
