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# Question

How can the area of the triangle be 2? When I attempted to solve this problem, I thought about the third side rule- the unknown side must be less than the sum of the other two sides and greater than the difference... so I came up with a range of 3-13. I thought if 3 was the smallest possible side, then 6 would be the smallest possible area.

For any triangle, the sum of any two sides must be greater than the sum of the third side. So if you are given two sides, the third side must be greater than the positive difference of those two sides, but less than the sum of those sides For this triangle, we are given sides of 6 and 8. We know the third side must be less than 6 + 8 = 14, and the third side must also be greater than 2 so that, when added to 6, we have a result greater than 8. So, the third side must be more than 2 and less than 14. Note: the third side doesn't have to be of integer length. So it could be as small as 2.0000000000001 (or smaller) or as large as 13.9999999999999... (or larger) as long as it's greater than 2 exactly and less than 14 exactly.

However, in this problem, even though our third side must be greater than 2, we can still make the area of the triangle infinitely close to zero:

Please refer again to the 2nd diagram under the text explanation.

Imagine we have a line segment of length 8 with a point "A" that divides the segment into lengths of 6 and 2. That's not a triangle of course—it's a line segment. But if we raise point "A" an infinitely small amount so it's perpendicular to its original place on the line segment, we now have a triangle with a base of 8 and a height infinitely close to zero. You will notice the third side is greater than 2—albeit by an infinitely small amount, but greater than 2 nonetheless.

Since we can make an area infinitely close to zero, we know we can make a triangle of any area greater than 0 and less than or equal to the maximum area, which is 6 * 8 /2 = 24.