The perimeter of a triangle consists of its edges; its length is the sum of
the lengths of the edges. Half of this length shows up in some properties of
triangles, so this page is dedicated to the discussion of those properties. It
builds on the discussion of the various centres
of a triangle.
The constructions of the circumcentre
and the centroid of a triangle both
overtly involve bisecting the edges of the triangle, hence the half-lengths of
the edges, whose sum is the semi-perimeter
. Slightly more
surprisingly, as we'll see below, the semi-perimeter also shows up in the
construction of the incentre of a triangle
– and is actually more relevant there than where it's more obviously
implicated.
Note that a straight line segment is the shortest path between its two end-points. The route along two edges of a triangle – from one end of the third, via the other vertex, to the other end of the third – is manifestly a path from one end of the third to the other hence at least as long as that third edge. Thus the sum of the lengths of any two sides of a triangle is greater than the length of the third side of the triangle. As a result, the semi-perimeter of a triangle is greater than the length of any of its sides. In particular, the point half way round the perimeter from each vertex – where a pair of travellers would meet, starting there and walking along the sides around the triangle at equal speed – is on the opposite edge, not on either of the edges out of the vertex.
The area of a triangle can be directly shown to be equal to half the product of the length of one side and the perpendicular distance from that side to the opposite vertex. Let us now see how that can be expressed in terms of the sides of the triangle.
As in the determination of the area of a triangle, construct the perpendicular, from a vertex with its widest internal angle, to the opposite edge, which it meets within the edge. This perpendicular sub-divides that side into two pieces; let their lengths be a and c, with the perpendicular being of length b. The perimeter of the triangle then consists of a, c and the hypotenuses of the right triangles with a or c as one side and b as the other; let's name the three whole sides A, B, C with (thanks to Pythagoras for two of them):
and refer to half their sum, the semi-perimeter, as s, with 2.s = A +B +C, so
The triangle's area is b.(a +c)/2 = b.B/2. So now let's look at the product of s with the results of subtracting each side from it (which we'll meet again, below, in the cotangent rule; and each is positive, as shown above, because the semi-perimeter is greater than the length of each side), using square = power(2) = (: x.x ←x :):
in which A.A = c.c +b.b, C.C = a.a +b.b and B.B = a.a +2.a.c +c.c, so
i.e. the square of the area of our triangle.
This gives rise
to Hero's formula
(Hero was a not uncommon name in classical Greece),
which gives the area of a triangle as the square root of the product of the
semi-circumference and the three results of subtracting one edge's length from
it.
Notice that the points where the incircle touches the edges split each edge in two parts; each part forms, with a radius of the incircle and the angle bisector of the vertex at its end away from the incircle, a right-angle triangle that's the mirror image for the matching triangle on that vertex's part of the other edge into it. Thus the two edge-parts that end in each vertex are equal.
Label the edges by their lengths a, b, c; let A be the length of the edge-parts into the vertex opposite a, B the length of the edge-parts into the vertex opposite b and C the length of the edge-parts into the vertex opposite c. Then a = B+C, b = C+A and c = A+B; the triangle's perimeter is then a+b+c = 2.(A+B+C); the semi-perimeter, half of this, is thus s = A+B+C, from which we can recover A = s−a, B = s−b and C = s−c; so we can now forget we had names for the part-lengths and just deal with the edge-lengths a, b, c and the semi-perimeter s = (a+b+c)/2.
Let r be the radius of the incircle. The area of the pair of right-angle triangles opposite side a is just r.(s−a) and similar for the other pairs gives, thanks to a+b+c = 2.s, a total area of r.(3.s −2.s) = r.s, so the area of a triangle is its semi-perimeter times the radius of its incircle.
Now, dividing the radius r of the incircle by each of the edge-part lengths, we get the tangent of half the triangle's angle at the vertex associated with the given edge-parts. Let the angle opposite the edge of length a be 2.α, that opposite b be 2.β and that opposite c be 2.γ. Thus
This law is more usually stated in terms of cotangents, with r.coTan(α) = s−a, &c; the cotangent of an angle is just one divided by the tangent. I'll use that form of it below, but I find it clearer in the present form.
Now the sines, cosines and thus tangents of sums of angles can be expressed in terms of those of the angles summed; and α +β +γ = turn/4, half the sum of the interior angles of a triangle, so
which we can divide through by Sin(α).Sin(β).Sin(γ), which is necessarily non-zero, to obtain
but, equally,
whence, scaling by r.r.r.s,
from which we can infer the area, r.s, as the square root of the right hand side. This is an alternative to the derivation of the same conclusion, above (which I prefer, because it doesn't need trigonometry).
Written by Eddy.