Heron's Formula: Area of a Triangle

Heron's Formula

Heron's formula is named after Hero of Alexandria, a Greek mathematician who lived around 60 A.D.. It provides a method for finding the area of any triangle, given its three sides. Here's how it works:

Let (a), (b), and (c) represent the lengths of the triangle's sides opposite angles (\angle A), (\angle B), and (\angle C), respectively. Then, the semiperimeter of the triangle can be calculated using the formula[S = \frac{a + b + c}{2}]. In general, the area of a triangle with side lengths (a), (b), and (c) can be found by calculating half the product of all three sides and one semiside, i.e., [A = \sqrt{\left(S - \frac{a}{2}\right)\left(S - \frac{b}{2}\right)\left(S - \frac{c}{2}\right)}]. This leads to the following expression for the area when we substitute our value of (S):

[ A = \sqrt{(ab + bc + ac)(ac + bd + cd)} - ab + ac + bd ]

This expression may seem complex, but once you plug in values from your diagram, it becomes much simpler! For example, let's say (AB = 7,\ BC=8,\ AC=9). Plugging these into the formula gives us [A = \sqrt{(7 * 8 + 8 * 9 + 7 * 9)(7 * 9 + 8 * 6 + 7 * 6)} - 7 * 8 + 7 * 9 + 8 * 6\ A = \sqrt{(176 + 72 + 63)(162 + 48 + 42)} - 136 + 63 + 48\ A = \sqrt{311}(100)\ A = 56.07]

So, the area of this triangle is approximately 56 square units. Now that we have the concept down, let's move on to proving why this formula actually works.

Proof of Heron's Formula

The proof of Heron's formula relies on some basic concepts of geometry and algebra. We will consider each step separately:

1. First, recall that the area of every triangle can be expressed two ways: either through the base times height or through the sum of products of base and corresponding heights divided by 2.

2. Next, take two triangles which share their bases and vertices. Imagine placing them together so that they form a quadrilateral that has equal areas because they were both constructed using the same formula.

3. Since these triangles have equal areas, they must also have equal perimeters. Therefore, if the measures of AC and AD are interchanged while keeping AB constant, the medians from point D measure only half the distance between points C and A, leaving considerable space above point D uncovered.

4. Finally, imagine drawing lines parallel to the base connecting points A and D, forming a trapezoid with area (x.) Because we started with triangles sharing common boundaries, these lines extend symmetrically from those boundaries. If we subtract (x) from the total area of the original set of quadrilaterals, we obtain Triangle ABC remaining intact with new areas being assigned to points a, b, and c along edges AB, BC, and CA, respectively.

Now, let's go back in time to prove that Heron's formula holds true for all possible triangles regardless of shape. Consider a triangle with side lengths (a), (b), and (c), and let (h_a), (h_b), and (h_c) denote the distances from vertices A, B, and C to the median line passing through point O. By symmetry, we know that (h_a = h_b = h_c = x). Let S denote the semiperimeter of the triangle. Applying the formula for the area of a triangle twice yields:

[ A = \frac{ah_a+bh_b+ch_c}{2} \ A = \frac{ax+ay+az}{2} \ 2A = ax+ay+az ]

Expanding this equation further, including all six terms involving the other side lengths yields:

[ 2A+ab+bc+ca=ax+ay+az+bx+by+cz ]

Since each term appears exactly twice, this simplifies to:

[ 2A=(a+b+c)(x)+ab+bc+ca ]

By dividing by 2, we arrive at the final version of Heron's formula proving validity across all possible configurations of length ratios within a triangle:

[ A=\frac{(a+b+c)}{2} * x+\frac{ab+bc+ca}{2} ]