Question

What is the area of a triangle (with vertices at FCE) that is inscribed in a hexagon with vertices at ABCDEF?
(1) The hexagon is regular and BE = 14.
(2) EC = 7√3.

• A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
• B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
• C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
• D. EACH statement ALONE is sufficient.
• E. Statements (1) and (2) TOGETHER are NOT sufficient.

Hint:

In geometry Data Sufficiency problems, be careful about assumptions. Don't assume a shape is regular unless it's explicitly stated. Statement (1) provides the crucial "regular" property, while statement (2) does not.

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We need to find the area of a triangle FCE, where F, C, and E are vertices of a hexagon ABCDEF. To find the area, we need to know the dimensions and properties of the hexagon and the resulting triangle.
Step 2: Analysis of Statement (1)
Statement (1) states that the hexagon is regular and the length of the long diagonal BE is 14.
In a regular hexagon, all sides are equal, and all interior angles are equal. A regular hexagon can be divided into six equilateral triangles with their common vertex at the center of the hexagon.
The long diagonal (e.g., BE) passes through the center and is equal to twice the side length of the hexagon (s). So, \(BE = 2s = 14\), which means the side length is \(s = 7\).
The vertices of triangle FCE form a specific shape within the regular hexagon. Let's consider the geometry. The distance from the center to any vertex is s=7. The triangle FCE is an isosceles triangle. The side FC is a short diagonal, EC is a short diagonal, and FE is a long diagonal. The length of a short diagonal is \(s\sqrt{3}\). So, \(FC = EC = 7\sqrt{3}\). The side FE is a long diagonal with length \(2s=14\).
We can find the area of triangle FCE. It is a right-angled triangle, since the angle at C (angle FCE) subtends the diameter of the circumscribed circle. The base and height can be considered FC and EC, which are short diagonals. Wait, FCE is not right-angled. Let's use coordinates or another method.
Let's place vertex E on the x-axis at (-7, 0) and F at (7,0), with the center at (0,0). Then vertex C would be at \((-7\cos(60^\circ), 7\sin(60^\circ)) = (-3.5, 7\sqrt{3}/2)\). No, this is incorrect.
A simpler way: The triangle FCE is an isosceles triangle with base FE = 14 and equal sides FC and EC. The height of this triangle is the perpendicular distance from C to the line segment FE. This height is equal to 1.5 times the side length of the equilateral triangles that form the hexagon, which is \(1.5 \times (s\sqrt{3}/2) = 1.5 \times (7\sqrt{3}/2) = 21\sqrt{3}/4\). This is getting complicated.
Let's use a simpler decomposition. The area of the regular hexagon is \( \frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(7^2) = \frac{147\sqrt{3}}{2} \). The triangle FCE covers a specific fraction of this area. Triangle FCE consists of three smaller triangles: F-Center-E, F-Center-C, C-Center-E. These are not easy to calculate.
Let's reconsider the triangle shape. Vertices F, C, E. Let's use the shoelace formula or place E at origin. Let's use the property that the area of triangle FCE is composed of triangle FDE and triangle CDE. This is also not simple.
Let's go back to the simplest observation: The line segment FE is a diagonal that passes through the center. C is another vertex. The triangle FCE is inscribed in the hexagon's circumcircle. The base is the diameter FE=14. The height is the perpendicular distance from C to the diameter FE. The height is the apothem of the hexagon, which is \(s\sqrt{3}/2 = 7\sqrt{3}/2\). No, the height is the y-coordinate of C if FE is on the x-axis. This would be \(s\sqrt{3}/2\). Ah, no, the vertices are F, C, E. The triangle is made of two small equilateral triangles (area \(s^2\sqrt{3}/4\)) and one isoceles triangle made of two radii and a 120 degree angle.
Area = Area(F-Center-E) + Area(C-Center-F). This is not correct.
Let's use coordinates: Center at (0,0), s=7. A=(7,0). B=(3.5, 3.5\(\sqrt{3}\)). C=(-3.5, 3.5\(\sqrt{3}\)). D=(-7,0).
E=(-3.5, -3.5\(\sqrt{3}\)). F=(3.5, -3.5\(\sqrt{3}\)).
Vertices are F(3.5, -3.5\(\sqrt{3}\)), C(-3.5, 3.5\(\sqrt{3}\)), E(-3.5, -3.5\(\sqrt{3}\)).
This is a right-angled triangle with the right angle at E.
Base is CE on the vertical line x=-3.5. Length of CE = \(3.5\sqrt{3} - (-3.5\sqrt{3}) = 7\sqrt{3}\).
Height is the horizontal distance from F to the line x=-3.5. Length of height = \(3.5 - (-3.5) = 7\).
Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7\sqrt{3} \times 7 = \frac{49\sqrt{3}}{2} \).
Since we found a unique value for the area, statement (1) is sufficient.
Step 3: Analysis of Statement (2)
Statement (2) gives \(EC = 7\sqrt{3}\). EC is a short diagonal of the hexagon. If we assume the hexagon is regular, then the length of a short diagonal is \(s\sqrt{3}\). So \(s\sqrt{3} = 7\sqrt{3}\), which implies \(s = 7\). This would make it sufficient. However, the statement does not say the hexagon is regular. An irregular hexagon could have a diagonal of length \(7\sqrt{3}\) without all its sides being 7. Without knowing the hexagon is regular, we do not know the coordinates or positions of the other vertices (like F), so we cannot calculate the area of triangle FCE.
Therefore, Statement (2) ALONE is not sufficient.
Step 4: Final Answer
Statement (1) alone is sufficient, but Statement (2) alone is not. The correct answer is (A).