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The perimeter of square S is 40. Square T is inscribed in square S.

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The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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Updated on: 09 Feb 2014, 00:08
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The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?

A. 45
B. 48
C. 49
D. 50
E. 52

Originally posted by Fabino26 on 08 Feb 2014, 05:55.
Last edited by Fabino26 on 09 Feb 2014, 00:08, edited 1 time in total.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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05 Mar 2014, 02:42
5
1
Perimeter of square = 40
So length of each side = 10

Area of the square = 100

Area of an (inscribed square) in a square = Half the area of the square

So 100/2 = 50 = Answer D
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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08 Feb 2014, 11:34
1
Given that the perimeter of Square 40 , gives you the information that each side is 10.
The case of a square in a square , you can see as square T as a ''diamond'' inside the square S. The midpoint of this square become 5 (in the middle ). The side of this new square can be obtained by the use of Pythagorean 5-5-5*sqrt(2). This gives you the information that the side of area = 5*sqrt(2).

5*sqrt(2)*5*sqrt(2)=50.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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08 Feb 2014, 12:30
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Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square AS. What is the least possible value of the area of square T ?

A. 45
B. 48
C. 49
D. 50
E. 52

Consider this picture:

The perimeter of square S is 40 implies each side of S is 10, which also means that diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (45-45-90) 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).
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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09 Feb 2014, 00:11
1
Abdul29 wrote:
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square AS. What is the least possible value of the area of square T ?

A. 45
B. 48
C. 49
D. 50
E. 52

Consider this picture:

The perimeter of square S is 40 implies each side of S is 10, which also means that diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (45-45-90) 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).

Thanks for the correction! T is inscribed in square S of course. I edited the question.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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18 Jun 2014, 05:35
1
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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18 Jun 2014, 05:42
2
maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.

Square T is inscribed in square S means that the vertices of square T are on the sides of square S. You cannot inscribe a square with sides of 7 into a square with sides of 10.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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18 Jun 2014, 06:09
Thanks much Bunuel
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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18 Jun 2014, 20:26
1
maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.

Also note that area of inscribed square is always half than that of the original square
As Bunuel pointed out, if it goes less than 50, it means some of the vertex is not touching side of the original square.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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01 Jul 2015, 18:04
If $$x^{2}$$ is area of square, then find x, one side of the square. If square is inscribed, then diagonal is the length of larger square and therefore the diagonal is $$10$$. To determine the side, the formula also includes the area of the square, $$x^{2}$$. So, if $$2x^{2} = 100$$ then $$x^{2}=50$$

D.

Thanks,
A
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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31 Jul 2016, 20:43
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?

A. 45
B. 48
C. 49
D. 50
E. 52

Perimeter of S= 40 ; side = 10
Now square T area can only be minimum if this 10 is its diagonal
Therefore area of square t = diagonal^2/2 = 100/2 = 50
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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22 Sep 2017, 08:51
first, test takers have to understand that inscribe means the square T touches the square S.
Secondly, using feeling and imagination and experience in geometry, 4 points of T can move on sides of square S. The minimum is if all 4 points are mid-points of sides of square S.

If that is the case, then the area of square T is half of that of square S.
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Re: The perimeter of square S is 40. Square T is inscribed in square S.  [#permalink]

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06 Oct 2018, 04:47
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Re: The perimeter of square S is 40. Square T is inscribed in square S.   [#permalink] 06 Oct 2018, 04:47
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