Brent@GMATPrepNow wrote:If ∆ABC is an isosceles triangle, what is the perimeter of ∆ABC?
(1) AB = 10√3
(2) AC = 8
Target question: What is the perimeter of ∆ABC?
Given: ∆ABC is an isosceles triangle
Statement 1: AB = 10√3
We only know the measurement of ONE side of the triangle.
So, there's no way to determine the triangle's perimeter.
Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: AC = 8
We only know the measurement of ONE side of the triangle.
So, there's no way to determine the triangle's perimeter.
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that AB = 10√3
Statement 2 tells us that AC = 8
We also know that ∆ABC is an isosceles triangle, so
two sides have the same length.
This means there are two possible cases: the sides have length 10√3, 10√3 and 8 OR the sides have length 10√3, 8 and 8
Since we have two different sets of lengths, we might incorrectly conclude that the statements COMBINED are not sufficient.
However, there's an important rule about the side lengths of a triangle:
If two sides of a triangle have lengths A and B, then . . .
DIFFERENCE between A and B < length of third side < SUM of A and B
In other words, the length of ONE side of a triangle cannot be longer than the sum of the other TWO sides.
For the GMAT, all students should have the following approximations memorized:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2
So, 10√3 ≈ (10)(1.7) ≈ 17
Let's rewrite our two possible cases: the sides have length 17, 17 and 8 OR the sides have length 17, 8 and 8
Notice that, in the second case, the side with length 17 is longer than the sum of the other two sides (8 and 8)
This means the 17, 8, 8 triangle CANNOT EXIST
Since the first case (17, 17, 8) does not break any rules, we can be certain that ∆ABC has lengths 10√3, 10√3 and 8, which means we can definitely determine the perimeter.
Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
