
u + v >= w
AND
w >= |u - v|
Example: u = 6, v = 12, w = 4
6 + 12 >= 4 Yes, this is true.
AND
4 >= |6 - 12| No, this is false, so these three lengths cannot be the sides of a triangle.
Note: If you make w the long side no matter which way they are handed to you, all you have to check is if u + v >= w.
Let's do another example. u = 10, v = 12 and w = 6.
If we make the long side of 12 be w and the side of 6 be v, then all we have to check is
10 + 6 >= 12, and that is true. These could be the sides of a triangle.

s(s - u)(s - v)(s - w).
Let's take the example of sides 6, 10 and 12. The perimeter is 28 and the semi-perimeter is 14. Our formula reads as follows
Area = sqrt(14(14-12)(14-10)(14-6)) = sqrt(14*2*4*8).
Factoring we get sqrt(7*2*2*2*2*2*2*2), which means we have three pairs of twins and one 2 that will be left under the square root sign. This gives us 2*2*2*sqrt(7*2) or 8*sqrt(14). If we want to round to the nearest thousandth 8*sqrt(14) ~= 29.933.
Practice problems . Someone in class asked for a LOT of practice problems so here goes. Here is a list of every possible combination of three positive numbers less than 4. The largest number will be listed last. If u + v = w, then it is a straight line segment with a point on the segment as the vertex. If u + v < w, it's not a triangle at all and only if u + v > w do the three sides create an actual triangle with area.
Part 1. For all these examples, determine if the lengths can create a triangle (answer YES), a straight line (answer STRAIGHT) or not a triangle (answer NO).
a) 1, 1, 1
b) 1, 1, 2
c) 1, 1, 3
d) 1, 1, 4
e) 1, 2, 2
f) 1, 2, 3
g) 1, 2, 4
h) 1, 3, 3
i) 1, 3, 4
j) 1, 4, 4
k) 2, 2, 2
l) 2, 2, 3
m) 2, 2, 4
n) 2, 3, 3
o) 2, 3, 4
p) 2, 4, 4
q) 3, 3, 3
r) 3, 3, 4
s) 3, 4, 4
t) 4, 4, 4
Part 2.
For every triplet of numbers that is a triangle, find the area as a square root and rounded to the nearest thousandth.
Answers in the comments.