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The figure above shows four adjacent small squares, forming one large

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Bunuel
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The figure above shows four adjacent small squares, forming one large [#permalink] New post   31 Oct 2018, 02:04
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The figure above shows four adjacent small squares, forming one large square. The vertices of square RSTU are midpoints of the sides of the small squares. What is the ratio of the area of RSTU to the area of the large outer square?


A. 1/2

B. 5/9

C. 7/12

D. 3/5

E. 5/8

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Re: The figure above shows four adjacent small squares, forming one large [#permalink] New post   31 Oct 2018, 02:31
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Bunuel wrote:


The figure above shows four adjacent small squares, forming one large square. The vertices of square RSTU are midpoints of the sides of the small squares. What is the ratio of the area of RSTU to the area of the large outer square?


A. 1/2

B. 5/9

C. 7/12

D. 3/5

E. 5/8

Show Spoiler
Attachment:
phd03.png


For the sake of dealing with easy numbers assume the bigger side of a square is 8, it creates 2 small squares that are 4 each. since the lines bisect in the middle of the the small 4/2 = 2

Now using pythagorean theorem we can find the side of the square RSTU the width is 4+2 = 6 and the height is 2

so it becomes 36 + 4 = 40

now the area of RSTU = 40

Area of bigger square = 8*8 = 64

\(\frac{40}{64}=\frac{4 * 2 * 5}{4* 2 * 8} = \frac{5}{8}\)

Answer choice E
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Re: The figure above shows four adjacent small squares, forming one large [#permalink] New post   31 Oct 2018, 02:32
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Let the side of the larger square be 4 units.

Attachment:
image.jpg
image.jpg [ 51.38 KiB | Viewed 5575 times ]


The ratio of area of RSTU to outer square is \(\frac{(\sqrt 10)^2}{16}\)

\(\frac{10}{16}\)

\(\frac{5}{8}\)

OPTION : E
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Re: The figure above shows four adjacent small squares, forming one large [#permalink] New post   14 Jan 2021, 05:19
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Bunuel wrote:


The figure above shows four adjacent small squares, forming one large square. The vertices of square RSTU are midpoints of the sides of the small squares. What is the ratio of the area of RSTU to the area of the large outer square?


A. 1/2

B. 5/9

C. 7/12

D. 3/5

E. 5/8

Show Spoiler
Attachment:
phd03.png


Solution:

If we let each side of the small squares be 2, then the side of the large outer square is 4, and thus it has an area of 4^2 = 16.
Let’s label the upper left vertex of the figure as Q, and let’s focus on right triangle RUQ. We know that the length of side QR is 3, and the length of side QU is 1. Using the Pythagorean theorem, we find that the length of the hypotenuse UR is √((3^2 + 1^2) = √10. Since UR is also the length of the side of square RSTU, we see that the area of RSTU = 10.
Thus, the ratio of the area of RSTU to that of the larger square is 10/16 = 5/8.

Answer: E
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Re: The figure above shows four adjacent small squares, forming one large [#permalink]
14 Jan 2021, 05:19
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