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31 Oct 2018, 02:04
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:29) correct
28%
(02:05)
wrong
based on 102
sessions
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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
Attachment:
phd03.png [ 9.57 KiB | Viewed 8144 times ]
31 Oct 2018, 02:31
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Bunuel
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
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
31 Oct 2018, 02:32
Kudos
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Let the side of the larger square be 4 units.
image.jpg [ 51.38 KiB | Viewed 7204 times ]
The ratio of area of RSTU to outer square is \(\frac{(\sqrt 10)^2}{16}\)
\(\frac{10}{16}\)
\(\frac{5}{8}\)
OPTION : E
Attachment:
image.jpg [ 51.38 KiB | Viewed 7204 times ]
The ratio of area of RSTU to outer square is \(\frac{(\sqrt 10)^2}{16}\)
\(\frac{10}{16}\)
\(\frac{5}{8}\)
OPTION : E
