In the figure, is the area of triangular region ABC equal to the area of triangular region DBA?
1) (AC)² = 2(AD)²
2) ∆ABC is Isosceles
Area of a Right Triangle = 1/2(length side 1)(length side 2)
So Area ∆ABC = 1/2(AC)(BC), and Area ∆DBA = 1/2(AD)(AB).
So the question can be rephrased as the following. Is (AC)(BC) = (AD)(AB)?
Statement 1:
(AC)² = 2(AD)² can be simplified to AC = √2(AD)
So we have the relationship between AC and AD, but we also need the relationship between them and AB and BC.
Try different easy values and see if you get different areas.
Let AC = √2, AD = 1, BC = 1.
Using the Pythagorean theorem, (AC)² + (BC)² = (AB)². So if AC = √2 and BC = 1, AB = √3.
AC x BC = √2 and AD x AB = √3. (AC)(BC) ≠(AD)(AB), and the areas are not equal.
I notice that Statement 2 says that ABC is Isosceles. So try AC = BC = √2 and AD = 1.
Using the Pythagorean theorem, AB = 2.
AC x BC = 2 and AD x AB = 2. (AC)(BC) = (AD)(AB), and the areas are equal.
So we can make the areas of ABC and DBA equal or not equal.
Insufficient.
Statement 2:
While, via the Pythagorean theorem, this statement gives us the relationships between AC, BC and AB, AD could be any length between 0 and infinity, and so the area of DBA could be greater than, smaller than or equal to that of ABC.
Insufficient.
Statement 1 and Statement 2 combined:
In our second test for Statement 1, ABC is Isosceles, and the areas are equal.
Given the rules of similar triangles, if we change the length of one side of ABC or DBA, all of the sides of both triangles will increase proportionally.
So if AC = AD, and if ABC is Isosceles, it is always the case that (AC)(BC) = (AD)(AB), and the areas of ABC and DBA will always be equal.
Sufficient.
The correct answer is
C.
