Author 
Message 
TAGS:


Moderator
Joined: 18 Apr 2015
Posts: 5130
Followers: 77
Kudos [?]:
1029
[0], given: 4636

The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
05 Mar 2017, 08:59
Question Stats:
55% (00:38) correct
44% (00:37) wrong based on 54 sessions
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible area of square T ? A) 45 B) 48 C) 49 D) 50 E) 52
_________________
Get the 2 FREE GREPrepclub Tests




GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
WE: Business Development (Energy and Utilities)
Followers: 93
Kudos [?]:
1653
[1]
, given: 396

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
11 Mar 2017, 17:27
1
This post received KUDOS
ExplanationAn inscribed figure is one that is drawn inside another with only the sides touching, so square T will be placed inside of square S. Hence the target diagram the question, both with square T as large as possible and with T as small as possible: As we can see, the closer the vertex of square T to a vertex of square A, the larger the area of square T. Square T has its least possible area when its vertex bisects each side of square S. Now use square S to find the length of one side of square T: The triangle that constitutes the space between square T and square S is a 45:45:90 triangle. This helps us determine that the length of one side of square T is \(5 \sqrt{2}\) . Finally, find the area of square T: \(Area = side^2\) \(Area = (5 \sqrt{2} )^2 = 50\). Hence option D is correct.
Attachments
di new1.jpg [ 9.37 KiB  Viewed 14964 times ]
di new.jpg [ 13.19 KiB  Viewed 15202 times ]
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Intern
Joined: 14 Jun 2018
Posts: 36
Followers: 0
Kudos [?]:
7
[0], given: 100

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
28 Jul 2018, 02:01
sandy wrote: ExplanationAn inscribed figure is one that is drawn inside another with only the sides touching, so square T will be placed inside of square S. Hence the target diagram the question, both with square T as large as possible and with T as small as possible: As we can see, the closer the vertex of square T to a vertex of square A, the larger the area of square T. Square T has its least possible area when its vertex bisects each side of square S. Now use square S to find the length of one side of square T: The triangle that constitutes the space between square T and square S is a 45:45:90 triangle. This helps us determine that the length of one side of square T is \(5 \sqrt{2}\) . Finally, find the area of square T: \(Area = side^2\) \(Area = (5 \sqrt{2} )^2 = 50\). Hence option D is correct. Thank you for your efforts. But why it's strange that it must be the bisector! I understand that it it's at circumscribed square corners, then the area would be 100. On the opposite, if it's at the middle of the circumscribed square side, it will be 50. As it approaches the sides of circumscribed vertex at the corners, the area is getting larger and larger. But still the whole thing is strange! Is not the inscribed figure should have a fixed area or it's made of Rubber!



Moderator
Joined: 18 Apr 2015
Posts: 5130
Followers: 77
Kudos [?]:
1029
[1]
, given: 4636

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
29 Jul 2018, 12:45
1
This post received KUDOS
The perimeter of square S is 40 implies each side of S is 10, which also means that the diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (454590) triangles, which, as you know, have length ratios of \(x:x:x\sqrt{2}\). As you can see, the sides of square T represent the hypotenuse of each of the smaller 4 triangles, thus each side of triangle T has a length of \(5\sqrt{2}.\) Since the area of the triangle is \(x^2\), \((5\sqrt{2})^2\) = 25*2 = 50, thus choice (D). The explanation provided by Sandy is correct. Regards
_________________
Get the 2 FREE GREPrepclub Tests



GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
WE: Business Development (Energy and Utilities)
Followers: 93
Kudos [?]:
1653
[1]
, given: 396

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
29 Jul 2018, 13:42
1
This post received KUDOS
Avraheem wrote: Thank you for your efforts. But why it's strange that it must be the bisector! I understand that it it's at circumscribed square corners, then the area would be 100. On the opposite, if it's at the middle of the circumscribed square side, it will be 50. As it approaches the sides of circumscribed vertex at the corners, the area is getting larger and larger. But still the whole thing is strange! Is not the inscribed figure should have a fixed area or it's made of Rubber! You have pointed out absolutely correctly that area would approach 100 inscribed squares vertices approaches the corners of the outer square. Now visualize the inner square (rotating) with vertices near the outer square vertices and start slowly moving away to the center, 'My bisector assumption', and then keep on moving till the corners of the squares again reach corners of the larger square (after crossing the midpint and ). As you may have notices as you move the inscribed squares vertices closer to the mid point of the larger square the area keeps on decreasing and once it crosses the mid point the area starts increasing again! This is because of the symmetry of the square in question. If no information is given how exactly the inner square is you can place it in any orientation as you like such that the area of the inner square is minimum. Hence the inner square is a variable whose exact orientation needs to be defined. So in order to solve this problem the inner square is assumed to be variable. Just like a rubber! (a variable whose values are not fixed!)
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Intern
Joined: 21 Jun 2018
Posts: 35
Followers: 0
Kudos [?]:
6
[0], given: 11

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
07 Aug 2018, 00:09
Carcass wrote: The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible area of square T ? A) 45 B) 48 C) 49 D) 50 E) 52 Just out of curiosity, why can't a square of side 7 be inscribed in a square of side 10? This way the least area would be 49.



Moderator
Joined: 18 Apr 2015
Posts: 5130
Followers: 77
Kudos [?]:
1029
[0], given: 4636

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
07 Aug 2018, 01:12
Please see my explanation above why you have to follow certain parameters. Regards
_________________
Get the 2 FREE GREPrepclub Tests



GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
WE: Business Development (Energy and Utilities)
Followers: 93
Kudos [?]:
1653
[1]
, given: 396

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
07 Aug 2018, 08:21
1
This post received KUDOS
ShubhiNigam29 wrote: Just out of curiosity, why can't a square of side 7 be inscribed in a square of side 10? This way the least area would be 49. A square of side 7 cannot be inscribed as a square of side 7 would not touch all the sides of the larger square. The mininimum size of the square inside is \(5\sqrt{2}\).
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Manager
Joined: 09 Nov 2018
Posts: 73
Followers: 0
Kudos [?]:
5
[1]
, given: 1

Re: The perimeter of square S is 40. Square T is inscribed in sq [#permalink]
18 Nov 2018, 17:45
1
This post received KUDOS
Avraheem wrote: sandy wrote: ExplanationAn inscribed figure is one that is drawn inside another with only the sides touching, so square T will be placed inside of square S. Hence the target diagram the question, both with square T as large as possible and with T as small as possible: As we can see, the closer the vertex of square T to a vertex of square A, the larger the area of square T. Square T has its least possible area when its vertex bisects each side of square S. Now use square S to find the length of one side of square T: The triangle that constitutes the space between square T and square S is a 45:45:90 triangle. This helps us determine that the length of one side of square T is \(5 \sqrt{2}\) . Finally, find the area of square T: \(Area = side^2\) \(Area = (5 \sqrt{2} )^2 = 50\). Hence option D is correct. Thank you for your efforts. But why it's strange that it must be the bisector! I understand that it it's at circumscribed square corners, then the area would be 100. On the opposite, if it's at the middle of the circumscribed square side, it will be 50. As it approaches the sides of circumscribed vertex at the corners, the area is getting larger and larger. But still the whole thing is strange! Is not the inscribed figure should have a fixed area or it's made of Rubber! Yes, by definition, inscribed square in a square act like a rubber. For that reason, the question asks "least possible area " not greater than the least.




Re: The perimeter of square S is 40. Square T is inscribed in sq
[#permalink]
18 Nov 2018, 17:45





