Heron's Formula
Heron's formula was first given by Heron of Alexandria. It is used to find the area of different types of triangles like equilateral, isosceles, and scalene triangles or quadrilaterals. We can use heron's formula to find the area of triangles when the sides of the triangle are given. We use the semiperimeter of the triangle and the side lengths to find the area of the triangle using heron's formula.
In this lesson, we will find how to determine the value of the area of triangles or quadrilaterals using Heron's formula with the help of solved examples for a better understanding of the application of the formula.
1.  What is Heron's Formula? 
2.  Heron's Formula for Area of Triangle Proof 
3.  Area of Triangle By Heron's Formula 
4.  Applications of Heron's Formula 
5.  FAQs on Heron's Formula 
What is Heron's Formula?
Heron's formula is used to determine the area of triangles when lengths of all their sides are given or to find the area of quadrilaterals. We also know it as Hero's formula. This formula for finding the area does not depend on the angles of a triangle. It solely depends on the lengths of all sides of triangles. It contains the term "s" which is known as semiperimeter, which is obtained by halving the perimeter of a triangle. Similarly, this concept of finding the area is further extended to determine the area of quadrilaterals as well.
History of Heron's Formula
Heron's formula was written in 60 CE by Heron of Alexandria. He was a Greek Engineer and Mathematician who determined the value of the area of the triangle using only the lengths of its sides and further extended it to calculate areas of quadrilaterals. He used this formula to prove the trigonometric laws such as Laws of cosines or Laws of cotangents.
Heron's Formula Definition
As per Heron's formula, the value of the area of any triangle having lengths, a, b, c, perimeter of the triangle, P, and semiperimeter of the triangle as 's' is determined using the belowgiven formula:
Area of triangle ABC = √s(sa)(sb)(sc), where s = Perimeter/2 = (a + b + c)/2
Example: Find the area of a triangle whose lengths are 5 units, 6 units, and 9 units respectively.
Solution: As we know, a = 5 units, b = 6 units and c = 9 units
Thus, Semiperimeter, s = (a + b + c)/2 = (5 + 6 + 9)/2 = 10 units
Area of triangle = √(s(sa)(sb)(sc)) = √(10(105)(106)(109))
⇒ Area of triangle = √(10 × 5 × 4 × 1) = √200 = 14.142 unit^{2}
∴ The area of the triangle is 14.142 unit^{2}
Heron's Formula for Area of Triangle Proof
We will use some Pythagoras theorem, area of a triangle formula, and algebraic identities to derive Heron's formula. Let us take a triangle having lengths of sides, a, b, and c. Let the semiperimeter of the triangle ABC be "s", the perimeter of the triangle ABC is "P" and the area of triangle ABC is "A". Let us assume the side length b is divided into two parts p and q as a perpendicular(h) falls from the vertex B on the side AC at point M. Consider the triangle below:
As we know, the area of a triangle = (1/2) b × h where b is the base and h is the height of the triangle. Let us begin to calculate the value of h.
Thus, as per the image, b = p + q
⇒ q = b  p ....(1)
On squaring both sides we get,
⇒ q^{2} = b^{2} + p^{2}  2bp ....(2)
Adding h^{2} on both sides we get,
q^{2} + h^{2 }= b^{2} + p^{2}  2bp + h^{2} ....(3)
Applying Pythagoras Theorem in the triangle BCM we get,
h^{2} + q^{2} = a^{2} ....(4)
Applying Pythagoras Theorem in the triangle MBA we get,
p^{2} + h^{2} = c^{2} ....(5)
Substituting the value of (4) and (5) in (3) we get,
q^{2} + h^{2 }= b^{2} + p^{2}  2bp + h^{2}
⇒ a^{2} = b^{2} + c^{2}  2bp
⇒ p = (b^{2} + c^{2}  a^{2})/2b ....(6)
From (5)
p^{2} + h^{2} = c^{2}
⇒ h^{2} = c^{2}  p^{2} = (c + p) (c  p) ....(7) (As a^{2}  b^{2} = (a+b)(ab))
Substituting (6) in (7) we get,
h^{2} = (c + p) (c  p)
⇒ h^{2} = (c + (b^{2} + c^{2}  a^{2})/2b) (c  (b^{2} + c^{2}  a^{2})/2b)
⇒ h^{2} = ((2bc + b^{2} + c^{2}  a^{2})/2b) ((2bc  b^{2}  c^{2} + a^{2})/2b)
⇒ h^{2} = ((b + c)^{2}  a^{2})/2b) ((a^{2}  (b  c)^{2})/2b)
⇒ h^{2} = ((b + c + a)(b + c  a)(a + b  c)(a  b + c))/4b^{2}) ....(8) (As a^{2}  b^{2} = (a+b)(ab))
As perimeter of triangle is P = a + b + c and P = 2s. (Here s = semiperimeter and s = P/2)
∴ 2s = a + b + c ....(9)
Substituting (9) in (8) we get,
h^{2} = ((b + c + a)(b + c  a)(a + b  c)(a  b + c))/4b^{2})
⇒ h^{2} = (2s × (2s  2a) × (2s  2b) × (2s  2c))/4b^{2})
⇒ h^{2} = (2s × 2(s  a) × 2(s  b) × 2(s  c))/4b^{2})
⇒ h^{2} = 16s(s  a)(s  b)(s  c)/4b^{2}
⇒ h = √(4s(s  a)(s  b)(s  c)/b^{2})
⇒ h = 2√(s(s  a)(s  b)(s  c))/b ....(10)
Area of triangle ABC, A = (1/2) × base × height
⇒ A = (1/2) × b × h
⇒ A = (1/2) × b × 2√(s(s  a)(s  b)(s  c))/b (From (10))
⇒ A = √(s(s  a)(s  b)(s  c))
∴ Area of the triangle ABC = √(s(s  a)(s  b)(s  c)) unit^{2}
Area of Triangle By Heron's Formula?
The steps to determine the area using Heron's formula are:
 Step 1: Find the perimeter of the given triangle.
 Step 2: Find the semiperimeter by halving the perimeter.
 Step 3: Find the area of the triangle using Heron's formula √(s(s  a)(s  b)(s  c)).
 Step 4: Once the value is determined, write the unit at the end (For example, m^{2}, cm^{2}, or in^{2}).
Heron's Formula for Equilateral Triangle
An equilateral triangle has all sides of the same length. Thus, in this case, the lengths of all sides are equal. Let us assume the length of all sides is "a", semiperimeter is "s" and the area of the equilateral triangle is "A". Thus, the semiperimeter of the triangle is s = (a + a + a)/2 = 3a/2
Area of the equilateral triangle, A = √(s(sa)(sb)(sc)) = √((3a/2)((3a/2)a)((3a/2)a)((3a/2)a))
⇒ A = √((3a/2)(3a/2a)^{3}) = √((3a/2)(a/2)^{3})
⇒ A = √((3a × a^{3})/2 × 8) = √(3a^{4}/16) = (√3 × a^{2})/4
Heron's Formula for Scalene Triangle
A scalene triangle has all lengths of different sides. Let us assume the length of sides is a, b, c, semiperimeter is "s" and the area of the scalene triangle is "A". Area of scalene triangle = √s(sa)(sb)(sc), where s = (a + b + c)/2
Heron's Formula for Isosceles Triangle
An isosceles triangle has two sides of equal length. Let us assume the length of the two sides is a and one side is b, semiperimeter is "s" and the area of the isosceles triangle is "A". Thus, the semiperimeter of the triangle is s = (a + a + b)/2 = (a+b/2)
Area of the isosceles triangle, A = √(s(sa)(sb)(sc)) = √(a+b/2)((a+b/2)a)((a+b/2)a)((a+b/2)b))
⇒ A = √(a+b/2)((b/2)(b/2)(ab/2))= √((a^{2}  (b/2)^{2})(b/2)^{2})
⇒ A = √(a^{2}  (b^{2}/4))× (b^{2}/4)) = (b/4)√(4a^{2} b^{2})
Heron's Formula for Area of Quadrilateral
We can use Heron's formula to determine the formula for the area of the quadrilateral by dividing it into two triangles. Let us say we have a quadrilateral ABCD with the length of its sides measuring a, b, c, and d. Let us say A and B are joined to show the diagonal of the quadrilateral having length e.
 The area of triangle ADC = √(s(s  a)(s  d)(s  e)), where s = (a + d + e)/2
 The area of triangle ABC = √(s'(s'  b)(s'  c)(s'  e)), where s' = (b + c + e)/2
Thus, area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
⇒ Area of quadrilateral ABCD = √(s(s  a)(s  d)(s  e)) + √(s'(s'  b)(s'  c)(s'  e))
Applications of Heron's Formula
Heron's formula has numerous applications. They are:
 It can be used to determine the area of different types of triangles if the lengths of their different sides are given.
 It can be used to find the area of the quadrilateral if the lengths of all its sides are given.
Application of Heron's Formula in Finding Area of Quadrilateral
The application of Heron's formula in finding the area of the quadrilateral is that it can be used to determine the area of any irregular quadrilateral by converting the quadrilateral into triangles.
Important Notes on Heron's Formula
 Heron's formula is used to find the area of a triangle when all its sides are given.
 We can use heron's formula to find the area of the quadrilateral by dividing it into two triangles.
 The formula uses the semiperimeter and side of lengths of a triangle.
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Heron's Formula Examples

Example 1: If the length of the sides of a triangle ABC are 4 in, 3 in, and 5 in. Calculate its area using the heron's formula.
Solution: To find: Area of the triangle ABC.
Given that, AB = 4 in, BC = 3 in, AC = 5 in (let)
Using Heron's Formula,
A = √(s(sa)(sb)(sc))
As, s = (a+b+c)/2
s = (4+3+5)/2
s = 6 unitsPut the values,
A = √(6(64)(63)(65))
⇒ A = √(6(2)(3)(1))
⇒ A = √(36) = 6 in^{2}Answer: ∴ The area of the triangle is 6 in^{2}.

Example 2: The area of an equilateral triangle is √3 squared units. Find the length of the sides of the triangle.
Solution: To find: The length of the sides of the triangle.
Area = √3 unit^{2 }(given)
Let the length of the triangle is "a" unit.Using Heron's Formula,
Area of equilateral triangle = (√3 × a^{2})/4
√3 = (√3 × a^{2})/4
⇒ a^{2} = 4
⇒ a = √4 = 2 unitsAnswer: ∴ The length of the sides of the equilateral triangle is 2 units.

Example 3: Calculate the area of an isosceles triangle using Heron's formula if the lengths of its sides are 4 units, 8 units, and 8 units.
Solution: To find: Area of triangle
Given the lengths of sides are 4 units, 8 units, and 8 units.Using Heron's Formula,
A = √(s(sa)(sb)(sc))As, s = (4 + 8 + 8)/2 = 10 units
A = √(10(104)(108)(108))
⇒ A = √(10 × 6 × 2 × 2)
⇒ A = 15.491 unit^{2}Answer: ∴ The area of the isosceles triangle is 15.491 unit^{2}.

Example 4: What is the area of the quadrilateral ABCD shown in the figure below using Heron's formula?
Solution: Note that we have only been given the lengths of the four sides, but not the length of any diagonal. Fortunately, AB and AD are perpendiculars, which means that the Pythagoras theorem can be used to calculate BD:
In triangle ABD,
Given a = 4 units, b = 5 units and c = 4 units
s = (4 + 5 + 3)/2 = 6 units
A = √(s(sa)(sb)(sc)) = √(6(64)(65)(63)) = 6 unit^{2}In triangle CDB,
Given a = 3 units, b = 5 units and c = 5 units
s = (3 + 5 + 5)/2 = 13/2 units
A = √(s(sa)(sb)(sc)) = √((13/2)((13/2)3)((13/2)5)((13/2)5)) = 7.15 unit^{2}Area of ABCD = Area of triangle ABD + Area of triangle CDB
⇒ Area of ABCD = (6 + 7.15) unit^{2} = 13.15 unit^{2}Answer: ∴ The area of the quadrilateral is 13.15 unit^{2}.
FAQs on Heron's Formula
What is Heron's Formula for Area of Triangle?
Heron's formula is used to find the area of the triangle when the lengths of all triangles are given. It can be used to determine areas of different types of triangles, equilateral, isosceles, or scalene triangles. For a triangle having three sides, a, b, and c, semiperimeter, s, and area of triangle A, the semiperimeter and area of the triangle are given as (a + b + c)/2 and √s(sa)(sb)(sc) respectively.
How Does Heron's Formula Work?
The heron's formula depends only on the semiperimeter of a triangle and the length of its three sides. We first determine the value of the semiperimeter using the lengths of three sides of the triangle. Once the value of the semiperimeter is obtained we can find the area of the shape.
Who Discovered Heron's Formula?
Heron's Formula was discovered by Heron of Alexandria (also known as Hero of Alexandria) who was a Greek Engineer and Mathematician. He found the area of the triangle using only the lengths of its sides which made it possible to apply to any type of triangle be it, equilateral, isosceles, or scalene. This formula was further extended by him to calculate areas of quadrilaterals and proved the trigonometric laws such as Laws of cosines or Laws of cotangents.
What Does S Stand for in Heron's Formula?
S stands for semiperimeter in Heron's formula. It is obtained by halving the value of the perimeter.
Can We Use Heron's Formula In Equilateral Triangle?
Yes, we can use Heron's formula in an equilateral triangle. The area of an equilateral triangle is given as (√3 × a^{2})/4 unit^{2}.
How to Derive Heron's Formula?
We can derive Heron's formula by using the Pythagoras theorem, area of a triangle formula, and algebraic identities. We construct an altitude from the top vertex to the base of the triangle, which divides the triangle into 2 triangles. Thereby applying the Pythagoras theorem, on both triangles and substituting the values obtained we derive Heron's formula.
How to Find Area of Trapezium Using Heron's Formula?
We can find the area of the trapezium by drawing a perpendicular from one of the vertices at the top to the base. Once this is done we obtain one parallelogram and a triangle. In this case, the area of the trapezium can be found by adding the area of a parallelogram and a triangle. Now, the area of the parallelogram can be found by using the formula of the area of a parallelogram or by dividing the parallelogram into two parts by drawing a diagonal. In this case, we obtain three triangles. Thus, we calculate the area of all the 3 triangles using Heron's formula. In either case, area values are calculated individually and then are added to obtain the value of the area of trapezium.
Where is Heron's Formula Used?
Heron's formula is used to determine the value of the area of any type of triangle, area of quadrilaterals, and area of polygons when the lengths of their sides are given.
How to Find Height of a Triangle with Heron's Formula?
As Heron's formula gives the value of the area of the triangle, it is equal to the area of the triangle obtained by the formula (1/2) × base × height. Thus, we can obtain the value of the height of the triangle.
How to Solve Heron's Formula Questions?
We solve Heron's formula questions using the below steps:
 Step 1: First find the perimeter of the given triangle.
 Step 2: Now, divide the perimeter by 2 to obtain the semiperimeter.
 Step 3: Use the semiperimeter to find the area of the triangle using Heron's formula √(s(s  a)(s  b)(s  c)).
 Step 4: Write the unit at the end once the value of the area is obtained(For example, m^{2}, cm^{2}, or in^{2}).
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