What is the area of rectangle ABGH?

[BTGmoderatorLU]

In the cube above, AH and BG are diagonals and the surface area of side ABFE is 16. What is the area of rectangle ABGH?

$$A.\ \ 4\sqrt{2}$$
$$B.\ \ 16\sqrt{2}$$
$$C.\ \ 16+\sqrt{2}$$
$$D.\ \ 16$$
$$E.\ \ 15\sqrt{3}$$

The OA is B.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

[DavidG@VeritasPrep]

In the cube above, AH and BG are diagonals and the surface area of side ABFE is 16. What is the area of rectangle ABGH?

$$A.\ \ 4\sqrt{2}$$
$$B.\ \ 16\sqrt{2}$$
$$C.\ \ 16+\sqrt{2}$$
$$D.\ \ 16$$
$$E.\ \ 15\sqrt{3}$$

The OA is B.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

If the figure is a cube, we know that all the sides are equal, (and each face is a square) so if the surface area of ABFE is 16, each side must be 4.
We also know that the diagonal of a square splits the square into two 45:45:90 triangles. If triangle ADH is a 45:45:90 triangle, and AD = 4, then we know that AH, the hypotenuse, is equal to 4*√2, as the ratio of the sides of the 45:45:9 triangle is x: x: x√2.

If AD = 4 and AH = 4*√2, then rectangle ABGH has an area of 4 * 4*√2 = 16*√2. The answer is B

[Scott@TargetTestPrep]

In the cube above, AH and BG are diagonals and the surface area of side ABFE is 16. What is the area of rectangle ABGH?

$$A.\ \ 4\sqrt{2}$$
$$B.\ \ 16\sqrt{2}$$
$$C.\ \ 16+\sqrt{2}$$
$$D.\ \ 16$$
$$E.\ \ 15\sqrt{3}$$

The OA is B.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

Since the surface area of ABFE is 16, each side is 4, and thus, each of diagonals AH and BG is 4√2. Thus, the area of ABGH is 4√2 x 4 = 16√2.

Answer: B