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In the figure above, if the area of the smaller square region is

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In the figure above, if the area of the smaller square region is  [#permalink]

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New post 08 Jul 2018, 21:42
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A
B
C
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E

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In the figure above, if the area of the smaller square region is \(\frac{2}{3}\) the area of the larger square region, then the diagonal of the larger square is how many inches longer than the diagonal of the smaller square?


A. \({\sqrt{2} - \frac{2\sqrt3}{3}\)

B. \(\frac{2}{3}\)

C. \(\frac{2\sqrt3}{3}\)

D. \(\frac{\sqrt2 - 2}{3}\)

E. \(\sqrt{3}\)


Attachment:
square (2).jpg
square (2).jpg [ 7.19 KiB | Viewed 256 times ]

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Re: In the figure above, if the area of the smaller square region is  [#permalink]

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New post 08 Jul 2018, 22:40
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The diagonal of the larger square is \(\sqrt{1+1}\) inches = \(\sqrt{2}\) inches

The area of the larger square is 1 square inches

Let the side of smaller square be a

The area of the smaller square is \(a^2\) square inches = \(\frac{2}{3}\) * area of larger square = \(\frac{2}{3} * 1\)

=> a = \(\sqrt{\frac{2}{3}}\)

length of diagonal of smaller square is \(\sqrt{a^2 + a^2}\) = \(\sqrt{2a^2}\) = \(\sqrt{\frac{4}{3}}\)

Length of diagonal of larger square - Length of diagonal of smaller square = \(\sqrt{2} - \sqrt{\frac{4}{3}}\) = \({\sqrt{2} - \frac{2\sqrt3}{3}\)

Hence option A
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Re: In the figure above, if the area of the smaller square region is  [#permalink]

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New post 08 Jul 2018, 22:49

Solution



Given:
    • The area of the smaller square region is \(\frac{2}{3}\) the area of the larger square region

To find:
    • The diagonal of the larger square is how many inches longer than the diagonal of the smaller square

Approach and Working:
Let us consider each of the squares individually.

For larger square:
    • Side length = 1
    • Area = 1
    • Diagonal = \(\sqrt{2}\)

For smaller square:
    • Area = \(\frac{2}{3} * 1 = \frac{2}{3}\)
    • Side length = \(\sqrt{\frac{2}{3}}\)
    • Diagonal = \(\sqrt{2} * \sqrt{\frac{2}{3}} = \frac{2}{\sqrt{3}} = 2\sqrt{3}/3\)

Therefore, the difference in the length of diagonals = \(\sqrt{2} – 2\sqrt{3}/3\)

Hence, the correct answer is option A.

Answer: A
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Re: In the figure above, if the area of the smaller square region is &nbs [#permalink] 08 Jul 2018, 22:49
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