Heron’s Formula

Heron of Alexandria was the first to reveal Heron’s formula. In the years 10–70 AD, he was a Greek engineer and mathematician. He constructed the  Aeolipile, the first known steam engine, among other things, but it was viewed as a toy!

It’s used to calculate the area of various triangles, including equilateral, isosceles, and scalene triangles, as well as quadrilaterals. In this lesson, we’ll learn how to use Heron’s formula to calculate the area of triangles and quadrilaterals.

What is Heron’s Formula?

When the lengths of all of a triangle’s sides are known, Heron’s formula is used to calculate the area of the triangle or the area of quadrilaterals. Hero’s formula is another name for it. This formula for calculating the area of a triangle is independent of the triangle’s angles. It is totally determined by the lengths of all triangle sides. It bears the letter “s,” which stands for semi-perimeter, which is obtained by halves a triangle’s perimeter. Similarly, the principle of determining the area is extended to determine the area of quadrilaterals.

History of heron’s formula

Heron of Alexandria penned Heron’s formula about 60 CE. He was a Greek engineer and mathematician who calculated the area of a triangle using only the lengths of its sides and went on to calculate the areas of quadrilaterals using the same method. This formula was used to prove trigonometric laws such as the Laws of Cosines and the Laws of Cotangents.

Derivation of the Heron’s Formula for Area of the triangle:-

Heron of Alexandria was the first to reveal Heron’s formula. It’s used to calculate the area of various triangles, including equilateral, isosceles, and scalene triangles, as well as quadrilaterals.

To derive Heron’s formula, we’ll use the Pythagoras theorem, the area of a triangle formula, and algebraic identities. Considering a triangle whose sides are a, b and c. Let “s” be the triangle ABC’s semi-perimeter, “P” be the triangle ABC’s perimeter, and “A” be the triangle ABC’s area. Assume that a perpendicular(h) descends from the vertex B on the side AC at point M, dividing the side length b into two portions p and q. Consider the following triangle: 

We know that area of a triangle is:-

A= (1/2) b*h

b = base of the triangle and h = height of the triangle

Now we will calculate the height of h

Thus, as per the image, b = p + q

• Q = b – p ….(i)

On squaring both sides we get,

• Q² = b² + p² – 2bp ….(ii)

Adding h^2on both sides we get,

Q² + h² = b² + p² – 2bp + h² ….(iii)

After applying the Pythagoras theorem in ABM we have

H² + q² = a² ….(iv)

After applying the Pythagoras theorem in ACM we have

P2 + h2 = c2 ….(v)

Substituting the value of (iv) and (v) in (iii) we get,

Q2 + h2 = b2 + p2 – 2bp + h2

A2 = b2 + c2 – 2bp

P = (b2 + c2 – a2)/(2b) ….(vi)

From (v)

P2 + h2 = c2 

H2 = c2 – p2 = (c + p) (c – p) ….(vii) 

(As a2 – b2 = (a+b)(a-b))

Substituting (vi) in (vii) we get,

H² = (c + p) (c – p)

H2 = (c + (b2 + c2 – a2)/(2b)) (c – (b2 + c2 – a2)/(2b))

H2 = ((2bc + b2 + c2 – a2)/(2b)) ((2bc – b2 – c2 + a2)/2b)

H2 = ((b + c)2 – a2)/(2b)) ((a2 – (b – c)2)(/2b))

H2 = ((b + c + a)(b + c – a)(a + b – c)(a – b + c))/(4b2) ….(viii) 

(As a2 – b2 = (a+b)(a-b))

As perimeter of triangle is P = a + b + c and P = 2s. (Here s = semi-perimeter and s = P/2)

s = (a + b + c )/2 ….(ix)

Substituting (ix) in (viii) we get,

H2 = ((b + c + a)(b + c – a)(a + b – c)(a – b + c))/(4b2)

H2 = (2s* (2s – 2a) * (2s – 2b) * (2s – 2c))/(4b2)

H2 = (2s * 2(s – a) * 2(s – b) * 2(s – c))/(4b2)

H2 = (16s(s – a)(s – b)(s – c))/(4b2)

H = sqrt(4s(s – a)(s – b)(s – c)/(4b2))

H = (2sqrt((s(s – a)(s – b)(s – c)))/b ….(x)

Area of triangle ABC, A = (1/2) × b × h

A = (1/2)* b* h

A = (1/2) * b * (2√((s(s – a)(s – b)(s – c)))/b

A = √((s(s – a)(s – b)(s – c))

Therefore, area of the triangle ABC is:

A = sqrt(s(s – a)(s – b)(s – c)) unit2

How to Find the Area Using Heron’s Formula?

The following are the steps to use Heron’s formula to calculate the area:

• Calculate the perimeter of the triangle.
• Halve the perimeter to find the semi-perimeter.
• Using Heron’s formula (s(s – a)(s – b)(s – c)), calculate the triangle’s area.
• After you’ve determined the value, write the unit at the end.

Application of heron’s formula

Heron’s formula can be used in a variety of situations. They are as follows:

(1) the lengths of the different sides of a triangle are known, it can be used to calculate their area.

(2) If the lengths of all of the quadrilateral’s sides are known, it can be used to calculate its area.

Application of the Heron’s Formula in Finding the Area of the Quadrilaterals:-

Rectangles, squares, and trapeziums have typical quadrilateral areas that we can calculate. However, our quadrilaterals may not be in any of these shapes at any one time.

Because the quadrilateral in question does not fit into any of these categories, we can’t utilize our typical formulas. In such instances, Heron’s formula comes in handy. In the diagram above, we can build two triangles by joining any two opposed vertices. The area can then be calculated.