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# In the figure shown, two identical squares are inscribed in

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In the figure shown, two identical squares are inscribed in  [#permalink]

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Updated on: 18 Sep 2012, 00:54
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In the figure shown, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
Attachment:

Rectangle.png [ 19.46 KiB | Viewed 19345 times ]

A. 8√2
B. 12
C. 12√2
D. 16
E. 18

Originally posted by udaymathapati on 27 Aug 2010, 23:01.
Last edited by Bunuel on 18 Sep 2012, 00:54, edited 1 time in total.
Edited the question.
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Re: Geometry-Square within Rectangle  [#permalink]

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28 Aug 2010, 07:45
12
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udaymathapati wrote:
In the figure attached (refer file), two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
A. 8√2
B. 12
C. 12√2
D. 16
E. 18

The rectangle's $$width=d$$ and $$length=2d$$, where $$d$$ is the diagonal of each square.

$$P_{rectangle}=2(d+2d)=18\sqrt{2}$$ --> $$d=3\sqrt{2}$$.

Now, $$d^2=s^2+s^2$$, where $$s$$ is the side of a square --> $$d^2=(3\sqrt{2})^2=18=2s^2$$ --> $$s=3$$ --> $$P_{square}=4s=12$$.

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Re: Geometry-Square within Rectangle  [#permalink]

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28 Aug 2010, 07:45
2
Hello

Let's name:

A width of the rectangle (the biggest line)
B height of the rectangle (the smallest line)
C width of the square

We know that 2 (A + B) = 18√2, so A + B = 9√2

We can also infer that A = 2B since A = 2 diagonal of the square and B = 1 diagonal of the square (see it on the figure to understand it more easily)

A = 3√2 and B = 6√2

From Pythagor, we have C² + C² = B²
<=> 2c² = (3√2)²
<=> 2c² = 9 * 2
<=> C = 3

So the perimeter of each square is 4 * 3 = 12
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Re: Geometry-Square within Rectangle  [#permalink]

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28 Aug 2010, 07:49
let each square is with side a & diagonal b. hence a = 1/\sqrt{2}b.
b is breadth of the bigger rectangle & 2b is the length of the rectangle.

perimeter of the rectangle is 2X(2b+b) = 6b = 18\sqrt{2}
b = 3\sqrt{2}

=> a = 3.
perimeter of each square = 12.

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Square within Rectangle.docx [17.79 KiB]

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Re: Geometry-Square within Rectangle  [#permalink]

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19 Apr 2011, 18:09
l+b = 9root(2) (l - length of rectange, b - breadth of rectangle)

Also, 2d + d = 9root(2) (d = Diagonal of square)

d = 3root(2)

Side of square = 3, so permieter = 4 * 3 = 12

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Re: Geometry question  [#permalink]

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17 Sep 2012, 22:15
2
1
dineesha wrote:
In the figure shown, two identical squares are inscribed in the rectangle. If the perimeter of the
rectangle is 18\sqrt{2}, then what is the perimeter of each square?

A. 8\sqrt{2}
B. 12
C. 12\sqrt{2}
D. 16
E. 18

Please see figure in the attached file.

PERIMETER=2(A+B) WHERE A AND B ARE TWO SIDES OF THE RECTANGLE.....
A --> THE LENGTH
B-- > THE BREADTH

AS THE TWO SQUARES ARE IDENTICAL THE DIAGONALS ARE EQUAL TO B . THEREFORE A=2B ..

ON EQUATING WE WILL GET THE ANSWER
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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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26 Sep 2012, 01:40
Interesting questions and i like such questions.
Since diagonal of the square is equal to side of the square*sqrt2 then we have one side of the reqtangle is equal to two diagonal of the square and another side of the rectangle is equal to one diagonal. All the sides (perimiter) are equal to 6 diagonals. So the side of the square is equal to 18\sqrt{2}/6\sqrt{2}=3. Then perimiter of the square 3*4=12
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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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26 Sep 2012, 02:25
2
1
Answer is B. See Solution.
Attachments

solution mixture.jpg [ 31.89 KiB | Viewed 18452 times ]

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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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21 Nov 2013, 14:45
udaymathapati wrote:
In the figure shown, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
Attachment:
Rectangle.png

A. 8√2
B. 12
C. 12√2
D. 16
E. 18

If y'all take a look you can tell that the length + width is equal to 3 diagonals of the square.
Therefore, Since 2(x+y) = 18 sqrt (2) then x+y = 9 sqrt (2)
Now as stated before we have 3s sqrt (2) = 9 sqrt (2)
s = 3, 's' stands for side of the square.
Perimeter = 12

Hope it helps
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Re: Geometry-Square within Rectangle  [#permalink]

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03 Jan 2014, 07:23
Bunuel wrote:
udaymathapati wrote:
In the figure attached (refer file), two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
A. 8√2
B. 12
C. 12√2
D. 16
E. 18

The rectangle's $$width=d$$ and $$length=2d$$, where $$d$$ is the diagonal of each square.

$$P_{rectangle}=2(d+2d)=18\sqrt{2}$$ --> $$d=3\sqrt{2}$$.

Now, $$d^2=s^2+s^2$$, where $$s$$ is the side of a square --> $$d^2=(3\sqrt{2})^2=18=2s^2$$ --> $$s=3$$ --> $$P_{square}=4s=12$$.

Can I please ask why the width is D and length 2D?

Thank You
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Re: Geometry-Square within Rectangle  [#permalink]

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03 Jan 2014, 07:31
theGame001 wrote:
Bunuel wrote:
udaymathapati wrote:
In the figure attached (refer file), two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
A. 8√2
B. 12
C. 12√2
D. 16
E. 18

The rectangle's $$width=d$$ and $$length=2d$$, where $$d$$ is the diagonal of each square.

$$P_{rectangle}=2(d+2d)=18\sqrt{2}$$ --> $$d=3\sqrt{2}$$.

Now, $$d^2=s^2+s^2$$, where $$s$$ is the side of a square --> $$d^2=(3\sqrt{2})^2=18=2s^2$$ --> $$s=3$$ --> $$P_{square}=4s=12$$.

Can I please ask why the width is D and length 2D?

Thank You

The length is twice the width, so if $$width=d$$, then $$length=2d$$.
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Re: Geometry-Square within Rectangle  [#permalink]

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03 Jan 2014, 07:35
Bunuel wrote:

The length is twice the width, so if $$width=d$$, then $$length=2d$$.

This may sound a silly question but where is it stated that Length is twice the width? Is this a property of rectangle?
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Re: Geometry-Square within Rectangle  [#permalink]

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03 Jan 2014, 07:43
1
theGame001 wrote:
Bunuel wrote:

The length is twice the width, so if $$width=d$$, then $$length=2d$$.

This may sound a silly question but where is it stated that Length is twice the width? Is this a property of rectangle?

Not all rectangles have the ratio of width to length as 1 to 2.

From the figure we can see that the width equals to the diagonal of the inscribed square and the length equals to the two diagonals.
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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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02 Sep 2014, 20:48
Perimeter of rectangle$$= 18\sqrt{2}$$

Lets say one side = x

other side $$= 9\sqrt{2} - x$$

When we divide the rectangle (as shown in fig), two squares would be formed

one side = x; other side $$= \frac{9\sqrt{2}}{2} - \frac{x}{2}$$

As square ABCD is formed, both sides should be equal

$$x = \frac{9\sqrt{2}}{2} - \frac{x}{2}$$

$$x = 3\sqrt{2}$$

Area of Square ABCD$$= 3\sqrt{2} * 3\sqrt{2} = 18$$

Area of inscribed square PQRS $$= \frac{1}{2} * 18 = 9$$ (This is a thumb rule/property for inscribed square)

Length of a side of square PQRS $$= \sqrt{9} = 3$$

Perimeter of square PQRS= 3 * 4 = 12

Attachments

Rectangle.png [ 29 KiB | Viewed 16592 times ]

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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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12 Apr 2016, 20:55
Attached is a visual that should help.
Attachments

Screen Shot 2016-04-12 at 8.54.23 PM.png [ 131.08 KiB | Viewed 13185 times ]

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In the figure shown, two identical squares are inscribed in  [#permalink]

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30 Jul 2016, 06:07
2
udaymathapati wrote:
In the figure shown, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18√2, then what is the perimeter of each square?
Attachment:
The attachment Rectangle.png is no longer available

A. 8√2
B. 12
C. 12√2
D. 16
E. 18

Given $$2l+2b=18√2$$
$$l+b=9√2$$ {equation 1}

As seen in the diagram that length of the RECTANGLE is diagonal + diagonal OF SQUARE ; length = $$2d$$
As seen in the diagram that breadth of the RECTANGLE is diagonal of the SQUARE =$$d$$
As seen in the diagram the side of the square is $$x$$

Substituting these values in equation 1 gives us
$$2d+d=9√2$$
$$3d=9√2$$
$$d=3√2$$ so the diagonal of the square is $$3√2$$
now $$side^2 + side^2 = diagonal ^2$$ {simple pythagorus theorum}
$$x^2+x^2= (3√2)^2$$

$$2x^2= 9*2=18$$

$$x^2=\frac{18}{2} = 9$$

$$x=\sqrt{9}$$

$$x= 3$$the side of the square is 3 therefore its perimeter is 3*4=12

Attachments

Rectangle.png [ 101.26 KiB | Viewed 12103 times ]

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Re: In the figure shown, two identical squares are inscribed in  [#permalink]

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02 Jul 2017, 05:35
I solved it in a very easy way.
Lets take side of square is x. You can see from figure, two diagonals of squares = length of rectangle.
And one diagonal of square = width of rectangle.
So, as Length x Width = 36,
we can say (2 * root2x)* (root2x) = 36
x = 3
Perimeter of square = 12

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