 It is currently 18 Jun 2019, 17:38 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. Figure shown below consists of two squares having  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Intern Joined: 19 Feb 2018
Posts: 5
Followers: 0

Kudos [?]: 0 , given: 3

Figure shown below consists of two squares having [#permalink] 00:00

Question Stats: 72% (01:19) correct 27% (01:49) wrong based on 29 sessions
Figure shown below consists of two squares having the same centers. Sides of the inner square, each of length 4 units, are parallel to the sides of the outer square each having length of 12 units.

 Quantity A Quantity B The perimeter of the shaded region. Area of the shaded region.

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Attachments IMG_20180220_142335.jpg [ 852.37 KiB | Viewed 2212 times ]

Last edited by Carcass on 20 Feb 2018, 11:39, edited 2 times in total.
Edited by Carcass Manager  Joined: 15 Jan 2018
Posts: 147
GMAT 1: Q V
Followers: 3

Kudos [?]: 185  , given: 0

Re: Figure shown below consists of two squares having [#permalink]
2
KUDOS
For Quantity A, the vertical sides are obviously 4 and 12. Less obviously, the shaded region can be broken up into a square with a right isosceles triangle on the top and bottom. (The picture isn't drawn to scale at all.) The diagonals of the right triangles would thus be 4√2. Thus, the perimeter is 16 + 8√2.

For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.

Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.

Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.

(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
_________________

-
-
-
-
-

Need help with GRE math? Check out our ground-breaking books and app.

Manager  Joined: 26 Jun 2017
Posts: 104
Followers: 0

Kudos [?]: 41 , given: 38

Re: Figure shown below consists of two squares having [#permalink]
SherpaPrep wrote:
For Quantity A, the vertical sides are obviously 4 and 12. Less obviously, the shaded region can be broken up into a square with a right isosceles triangle on the top and bottom. (The picture isn't drawn to scale at all.) The diagonals of the right triangles would thus be 4√2. Thus, the perimeter is 16 + 8√2.

For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.

Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.

Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.

(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
excellent
_________________

What you think, you become.

Intern Joined: 24 Feb 2018
Posts: 15
Followers: 0

Kudos [?]: 1 , given: 0

Re: Figure shown below consists of two squares having [#permalink]
B

The outer square's side will get divided in 3 sections of length 4 each. This will make a right triangle at the corners, with base and perpendicular sides of length 4 each, so the hypotenuse will be 4 root 2.
Quantity A: - Perimeter = 4 + 4 root 2 + 12 + 4 root 2 = 16 + 8 root 2 = 27.3
Quantity B:- Area (this becomes a trapezium) = height * (average of sum of lengths of parallel sides) = 4 * {(12 + 4)/ 2} = 4 * 8 = 32
Director Joined: 09 Nov 2018
Posts: 508
Followers: 0

Kudos [?]: 27 , given: 1

Re: Figure shown below consists of two squares having [#permalink]
B but did with diagonal. Re: Figure shown below consists of two squares having   [#permalink] 19 Jan 2019, 18:15
Display posts from previous: Sort by

Figure shown below consists of two squares having  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.