GRE geometry question: Determining whether one area is greater or equal to another

Question

This problem was in my GRE practice textbook

Their answer was that the two quantities are equal, but I keep on getting a different answer. Let $\overline{BC} = y $ and $\overline{AE}=x$. Then it follows that the diagonal of the square $BCDF$ is equal to $x$. Therefore, we can use the pythagorean theorem to see that $y^2 + y^2 = x^2 \rightarrow x = \sqrt{2}y$. Therefore, the area of $ABDE$ is $x^2=2y^2$ and the area of $BCDF$ is $y^2$. Multiply $ABDE$ by two to get $4y^2$ and multiply the area of $BCDF$ by three to get $3y^2$. Therefore it should be that twice the area of $ABDE$ is greater than three times the area of $BCDF$, but this is not correct. What error am I making here?

Answer 1

Your error is in comparing the areas of the squares, but what the question is asking for is comparing the area of the shaded region (Quantity A) against the area of the white square.

The simpler way to solve this question is to observe that drawing the diagonal $BD$ of square $BCDF$, and continuing the sides $DF$ to $A$ and $BF$ to $E$, partitions the figure into five congruent isosceles triangles. Two of these triangles (in white) comprise the square $BCDF$, and three of these (shaded) comprise the total area of the shaded figure. The comparison then becomes trivial.

Answer 2

Let the edge length of the square (bottom) be $x$.

So, the height and base length of the triangle $BDF$ are $\frac{x}{2}$ and $x$ respectively.

Area of shaded region, $$A_1 = x^2 - \frac{1}{2}.\frac{x^2}{2} = \frac{3x^2}{4}$$

Now, for the square $CDFB$, $BD$ acts as a diagonal. So, $$CD = \ \frac{x}{\sqrt2} $$

Area of upper square, $$A_2 = (\frac{x}{\sqrt2})^2 = \frac{x^2}{2}$$

So, $$2A_1 = \frac{3x^2}{2} = 3A_2$$