Hero’s Formula and The Area of Triangles

Let’s say you have a triangle ABC with sides of length a, b and c.

triangle

Imagine the vertical height and the angles of the triangle are unknown. How would you go about finding the area of the triangle? Well, using the simple \frac{1}{2} \times base \times height formula would obviously be a little tricky since it would require calculating the vertical height of the triangle.

It would be a bit simpler to find the area using the \frac{1}{2} \times a \times b \times \sin c method, but then you will need to use something like the cosine rule to find the angle of at least one of the vertexes. If only there was some simple, elegant method that allowed one to calculate the area of a triangle just from the length of its sides without first having to first figure out the vertical height or the angles of the vertexes of the triangle!

Hero’s formula allows one to do exactly that! Here is how you calculate the area, A, of a triangle with sides ab and c using Hero’s formula…

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

where s is the semi-perimeter of the triangle i.e

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

Now, say we have a triangle PQR with sides of length 5, 11 and 12 units respectively…

triangle

Let’s use Hero’s formula to find its area. First let’s calculate its semi-perimeter, s.

s = \frac{a + b + c}{2} \Rightarrow \frac{5 + 11 + 12}{2} \Rightarrow \frac{28}{2} \Rightarrow 14 \: \mathtt{units}

Next use Hero’s formula with s = 14, = 5, = 11 and = 12 …

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

\Rightarrow \sqrt{14(14-5)(14-11)(14-12)}

\Rightarrow \sqrt{14 \times 9 \times 3 \times 2}

\Rightarrow \sqrt{756} \approx 27.5 \: \mathtt{sq. units}

and that’s all there is to it!

Hero’s formula is widely believed to be invented/discovered by the Greek mathematician and engineer, Hero of Alexandria in 60 AD.

Note: For a rigorous proof of Hero’s formula, see this link.

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