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.

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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 YES

b) 1, 1, 2 STRAIGHT

c) 1, 1, 3 NO

d) 1, 1, 4 NO

e) 1, 2, 2 YES

f) 1, 2, 3 STRAIGHT

g) 1, 2, 4 NO

h) 1, 3, 3 YES

i) 1, 3, 4 STRAIGHT

j) 1, 4, 4 YES

k) 2, 2, 2 YES

l) 2, 2, 3 YES

m) 2, 2, 4 STRAIGHT

n) 2, 3, 3 YES

o) 2, 3, 4 YES

p) 2, 4, 4 YES

q) 3, 3, 3 YES

r) 3, 3, 4 YES

s) 3, 4, 4 YES

t) 4, 4, 4 YES

Part 2.

For every triplet of numbers that is a triangle, find the area as a square root and rounded to the nearest thousandth.

a) 1, 1, 1 YES

p = 3, s = 1½

Area = sqrt(3)/4 ~= 0.433e) 1, 2, 2 YES

p = 5, s = 2½

Area = sqrt(15)/4 ~= 0.986h) 1, 3, 3 YES

p = 7, s = 3½

Area = sqrt(35)/4 ~= 1.479j) 1, 4, 4 YES

p = 9, s = 4½

Area = 3*sqrt(7)/4 ~= 1.984k) 2, 2, 2 YES

p = 6, s = 3

Area = sqrt(3) ~= 1.732l) 2, 2, 3 YES

p = 7, s = 3½

Area = 3*sqrt(7)/4 ~= 1.984n) 2, 3, 3 YES

p = 8, s = 4

Area = 2*sqrt(2) ~= 2.828o) 2, 3, 4 YES

p = 9, s = 4½

Area = 3*sqrt(15)/4 ~= 2.905p) 2, 4, 4 YES

p = 10, s = 5

Area = sqrt(15) ~= 3.873q) 3, 3, 3 YES

p = 9, s = 4½

Area = 9*sqrt(3)/4 ~= 3.897r) 3, 3, 4 YES

p = 10, s = 5

Area = 2*sqrt(5) ~= 4.472s) 3, 4, 4 YES

p = 11, s = 5½

Area = 3*sqrt(55)/4 ~= 5.562t) 4, 4, 4 YES

p = 12, s = 6

Area = 4*sqrt(3) ~= 6.928Post a Comment