There is no answer to this question. That triangle cannot exist.
Keep in mind the rule that triangles must follow - any two of the sides must always add up to a number greater than the length of the third side. For example, if you had a triangle with sides of lengths 3, 4 and 5: 3 and 5 added is 8, which is greater than 4; 3 and 4 added is 7, which is greater than 5, etc.
In this question, we have a triangle with a perimeter of 4 and one side is 2 feet shorter than another. Let's make up side lengths in order to prove that this is not a real triangle; I'll say that one of the sides is 2.1 ft, and the one that is 2 feet shorter will be 0.1 ft.
Those two sides add up to a total length of 2.2 ft. With a perimeter of 4 ft, that means that the remaining side must be 1.8 ft. Now let's test it. If we add the shorter sides (0.1 ft and 1.8 ft), we will have a length of 1.9 ft...which is not greater than 2.1 ft. Therefore, this triangle is not a real triangle.
2 years ago
Last edited at 2:27AM on 9/21/2011
You can't if you consider any side to be 2 feet less then any other. This way you will obtain one side to be zero. Try considering the longest side (i think it's name is hypotenuse - i mean the one that oppose the right angle, not the two of them that are the sides of the right angle) to be 2 feet longer than any other.