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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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08 Jul 2018, 21:42
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Difficulty:

45% (medium)

Question Stats:

66% (01:53) correct 34% (02:25) wrong based on 29 sessions

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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 [ 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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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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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.

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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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