What is the area of rectangle ABGH?

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BTGmoderatorLU: Moderator; Posts: 2554; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Tue Nov 28, 2017 7:47 am

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.

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Source: Beat The GMAT — Problem Solving | Tue Nov 28, 2017 7:47 am

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Tue Nov 28, 2017 8:19 am

LUANDATO wrote:

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*âˆš2, as the ratio of the sides of the 45:45:9 triangle is x: x: xâˆš2.

If AD = 4 and AH = 4*âˆš2, then rectangle ABGH has an area of 4 * 4*âˆš2 = 16*âˆš2. The answer is B

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8152; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: What is the area of rectangle ABGH?

by Scott@TargetTestPrep » Thu Jan 16, 2020 6:53 am

BTGmoderatorLU wrote: ↑
Tue Nov 28, 2017 7:47 am

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

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