Brief Theory & Notes on Heron’s Formula

Let us give you a brief overview of Heron’s formula and its significance.

Area of a Triangle

In general, the area of a triangle is calculated as half the product of the height and base of the triangle.

Therefore,

Area of a Triangle =  1/2(base x height)

What is Heron’s Formula?

Heron’s formula also known as Hero’s formula is named after Hero of Alexandria a Greek Engineer and Mathematician in 10 – 70 AD. It is applied to find out the area of a triangle by requiring no arbitrary choice of side as base or vertex as origin, as compared to the other formulas widely applied to find out the area of a triangle, such as half the base times the height or half the norm of a cross product of two sides. Thus, one can use this formula to find the area of a triangle using the 3 side lengths.

In order to find out the area of a triangle using Heron’s formula, we will first have to find out the semi-perimeter of the triangle, which is denoted by ‘s’

The formula for calculating the perimeter of a triangle is-

P=  a+b+c

Where a,b,c represents the three respective sides of a triangle

Therefore,

Semi-perimeter of a triangle (s) = (a+b+c)/2

Therefore,

Area of a Triangle using Heron’s Formula is-

A= √s (s-a) (s-b) (s-c)

Example:

• Find the area of a triangle where each side is 10 long

Solution:

A=b=c=10

Therefore, semi-perimeter (s)= (10+10+10)/2 =15

Hence using Heron’s Formula, the Area of the Triangle is

A = √15(15-10) (15-10) (15-10)

=√15(5) (5) (5)

=43.30 square units (approx.)

Proofs of Heron’s Formula

Below the proofs of the validity of Heron’s formula to find out the area of triangles-

• Trigonometric proof using the law of cosines

This proof basically uses the cosine rule.

In a triangle, the altitude to a side is equal to the product of the sine of the angle subtending the altitude and aside from the angle to the vertex of the triangle.

For Triangle in the above picture, the altitude to side c is   b sin A or  a sin B

Here the area of the triangle will be

Area= ½(bc sinA)

Now we look for a substitution for sin A in terms of a, b, and c. It is readily (if messy) available from the Law of Cosines

cos A= (b²+c²-a²)/2bc

and substitution in the identity

sin A= √1-(cos A)²

sin A= √1-{(b²+c²-a²)/2bc}²

Factor (easier than multiplying it out) to get

sin A= √{1- (b²+c²-a²)/2bc} {1+ (b²+c²-a²)/2bc)}

On solving the variables we get,

sin A= √[{2bc-(b²+c²-a²)}/2bc] [{2bc + (b²+c²-a²)}/2bc}

sin A= (1/2bc) √{2bc-(b²+c²-a²)} {2bc + (b²+c²-a²)}

sin A= (1/2bc) √{a²– (b²-2bc+c²)} {(b²+2bc+c²)-a²}

Factor

sin A= (1/2bc) √{a²– (b-c)²} {(b⁺c)²-a²}

Factor and rearrange

sin A= (1/2bc) √(a+b+c) (-a+b+c) (a-b+c) (a+b-c)

Now where the semi-perimeter s is defined by

s= (a+b+c)/2

the four expressions under the radical are 2s, 2(s – a), 2(s – b), and 2(s – c).   So,

sin A= 2/bc √s (s-a) (s-b) (s-c)

Since,

Area = ½(bc sin A)

we have.

A= √s (s-a) (s-b) (s-c)

• The Algebraic Proof

Let us consider the below triangle given in the image

Let a, b, c be the lengths of the sides of the above triangle and h be the height to the side of the lenghth c.

We know,

Semi-perimeter of a triangle (s) = (a+b+c)/2

So,

2s = a + b + c
2(s – a) = – a + b + c
2(s – b) = a – b + c
2(s – c) = a + b – c

There is at least one side of our triangle for which the altitude lies “inside” the triangle.             For convenience make that the side of length c.

We need to express h in terms of a, b, c and then substitute h in the formula of area of the triangle, which is A= 1/2 (ch)

Let p + q = c as indicated. Then,

h² + p²= a²

and

h² + q² = b²

since,

q= c – p

so, q² = (c – p)²

and

q² = c² – 2cp + p²

Adding h² to each side, we get

h² + q² = h² + c² – 2cp + p²

Substituting the values, we get

b² = a² – 2cp + c²

To solve p, we get

p = (a² + c² – b²) 2c

 

Now as we know, h² = a² – p² substituting the values for p we get the entire expression in terms of a, b, c.

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