cfarrera wrote:Triangle A has sides:
1. 4x-3
2. 2x+1
3. x+8
What is the perimeter of triangle A?

(1) Triangle A is an isosceles
(2) x+8 is the greatest side


Answer is D, help!! I do not get the answer....

We know that the sum of any two sides of a triangle must be greater than the third. So we know, before even looking at the statements, that

4x - 3 + 2x + 1 > x + 8
x > 2

and

2x + 1+ x + 8 > 4x - 3
12 > x

(the third combination of sides doesn't give us anything useful). That is, 2 < x < 12.

Considering Statement 1, which pair of sides could be equal? 2x + 1 cannot be equal to 4x - 3, because then x = 2, and we know that's impossible from the analysis above. Still, the other pairs could be equal:

4x - 3 = x + 8
x = 11/3

or

2x + 1 = x + 8
x = 7

You get a different perimeter for each x, so the Statement is not sufficient.

Considering Statement 2, we know that x + 8 is greater than either of the other two sides, from which we can produce two inequalities:

x + 8 > 2x + 1
x < 7

and

x+ 8 > 4x - 3
x < 11/3

So we know that x < 11/3, and from above, we know that x > 2. Still, we can't determine the perimeter, since we only have a range of values for x. So the Statement is not sufficient.

The two Statements are clearly contradictory, so it's impossible to consider both of them together, which means there's something wrong with the question. Where is it from?

nasa wrote:Ian your analysis was great. However, a small correction to your conclusion that 1 is insufficient.
x = 11/3 or 7
=> the three sides are (35/3,35/3,25/3) or (15,15,25)
Given the contraint that 2 < x < 12 leaves us with x = 11/3
So (1) is sufficient.

I think you might have confused the value of x with the lengths of the sides. We know that 2 < x < 12, which means that x can be 7, and x can be 11/3. When x = 7, it's true that this produces sides which are longer than 12, but the inequality provides a minimum and maximum for x itself, not for the lengths of the sides of the triangle.