The triangle inequality, Heron's formula and the test for acute, right or obtuse using side lengths.

You probably already learned in high school that the interior angles of a triangle always add up to 180°.  Just as importantly, the Triangle Inequality tells us that any two side lengths of a triangle must add up to more than the third side length.  Another way to say this is that every side length must be less than half the total perimeter.

The idea of half the perimeter or semi perimeter shows up in Heron's Formula, a way to find the area of a triangle if you are given the three side lengths.

We also have a test using side lengths to see if a triangle is acute, right or obtuse.  If we have three side u, v and w and we declare that w is the long side, then we can use the sum of the squares of the short side to see how a triangle is classified.

u² + v² < w² : The triangle is obtuse


u² + v² = w² : The triangle is right

u² + v² > w² : The triangle is acute

Here are some practice problems for area and classification.

For each of these triples of numbers:
1) Determine if the triangle is equilateral, isosceles or scalene.
2) Determine if the triangle is acute, right or obtuse.
3) Find the area as a simplified square root
4) Find the area rounded to the nearest thousandth

a) 1, 1, 1
b) 1, 2, 2
c) 1, 3, 3
d) 1, 4, 4
e) 2, 2, 2
f) 2, 2, 3
g) 2, 3, 3
h) 2, 3, 4
i) 2, 4, 4
j) 3, 3, 3
k) 3, 3, 4
L) 3, 4, 4
m) 4, 4, 4

Answers in the comments.

The triangle inequality and Heron's Formula with LOTS of practice problems.

In class, we discussed the Triangle Inequality, a way to check if three given numbers u, v and w, could possibly the lengths of the sides of some triangle. If the numbers are given in any old order, we have two inequalities to check.

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.

If a triangle is defined for us by the lengths of three sides, the "easiest" way to find the area is Heron's Formula, named for the ancient Greek mathematician Heron, sometimes known as Hero. We add up the three sides to get the perimeter p, then we take ½p and call it s, the semi-perimeter. The area is the square root of the product

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.