Area of a Triangle using Heron’s Formula

Heron’s formula provides a method to calculate the area of a triangle when the lengths of all three sides are known. This is particularly useful when the height of the triangle isn’t readily available. Instead of relying on the base and height, Heron’s formula uses the side lengths to directly compute the area.

Formulae

The area of a triangle using Heron’s formula is given by:

Let $a$, $b$, and $c$ be the lengths of the sides of a triangle. First, calculate the semi-perimeter, $s$, using:

$s = \frac{a + b + c}{2}$

Then, the area, $A$, is:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

Examples

Example-1:

Consider a triangle with sides of length $a = 5$ cm, $b = 6$ cm, and $c = 7$ cm.

First, calculate the semi-perimeter:

$s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$ cm

Next, apply Heron’s formula:

$A = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \times 4 \times 3 \times 2} = \sqrt{216} \approx 14.7$ cm$^2$

Example-2:

Calculate the area of a triangle with sides $a=13$, $b=14$, and $c=15$.

Find the semi-perimeter, $s = (13+14+15)/2 = 21$.

Then, $A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21*8*7*6} = \sqrt{7056} = 84$

Theorem with Proof

Theorem: Given a triangle with sides $a$, $b$, and $c$, and semi-perimeter $s = \frac{a+b+c}{2}$, the area $A$ of the triangle is given by $A = \sqrt{s(s-a)(s-b)(s-c)}$.

Proof:

Let’s consider a triangle ABC, where the side lengths are a, b, and c, with the angle opposite of ‘c’ as C. Using the Law of Cosines, we know that: $c^2 = a^2 + b^2 – 2ab\cos(C)$ So, $\cos(C) = \frac{a^2 + b^2 – c^2}{2ab}$

The area of the triangle can also be expressed as:

$Area = \frac{1}{2}ab\sin(C)$. Squaring this equation, we have: $Area^2 = \frac{1}{4} a^2 b^2 \sin^2(C)$ Since $\sin^2(C) = 1 – \cos^2(C)$, we can rewrite it as: $Area^2 = \frac{1}{4} a^2 b^2 (1 – \cos^2(C))$. Replace $\cos(C)$ with what we got before: $Area^2 = \frac{1}{4} a^2 b^2 \left(1 – \left(\frac{a^2 + b^2 – c^2}{2ab}\right)^2\right)$ Simplifying the equation, $Area^2 = \frac{1}{16} \left[4a^2b^2 – (a^2 + b^2 – c^2)^2 \right] $. Using the difference of squares, $Area^2 = \frac{1}{16} \left[ (2ab – (a^2 + b^2 – c^2))(2ab + a^2 + b^2 – c^2) \right]$ $Area^2 = \frac{1}{16} \left[ (c^2 – (a – b)^2)((a+b)^2 – c^2) \right] $ $Area^2 = \frac{1}{16} \left[ (c – a + b)(c + a – b)(a + b + c)(a + b – c)\right]$ We know that, $s = \frac{a+b+c}{2}$, so $2s = a+b+c$, $2s-2a = b+c-a$, $2s-2b = a+c-b$, and $2s-2c = a+b-c$. Replacing the terms in the bracket, we get $Area^2 = \frac{1}{16} \left[ (2s-2a)(2s-2b)(2s)(2s-2c) \right]$ $Area^2 = \frac{16s(s-a)(s-b)(s-c)}{16}$ Therefore, $Area = \sqrt{s(s-a)(s-b)(s-c)}$.

Common mistakes by students

Incorrect Calculation of Semi-perimeter: A common error is miscalculating the semi-perimeter ($s$). Students might forget to divide the sum of the sides by 2.
Incorrect Substitution: Students might incorrectly substitute the side lengths into the formula $(s-a)$, $(s-b)$, and $(s-c)$.
Ignoring Units: Forgetting to include the correct units (e.g., cm², m²) in the final answer.
Using the Formula Incorrectly: Trying to use Heron’s formula for right-angled triangle without being aware of that.

Real Life Application

Heron’s formula is practical in various real-life situations, such as:

Land Surveying: Calculating the area of irregularly shaped plots of land when only the side lengths can be measured.
Construction: Determining the amount of material needed for triangular structures, like roofs or gardens.
Navigation: Calculating the area of a triangular region defined by the positions of three points.

Fun Fact

Heron of Alexandria, a Greek mathematician and engineer, is credited with discovering and documenting this formula in his book “Metrica” around 60 AD. He also made significant contributions to the fields of mechanics and optics.