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# In the figure above, if the area of the smaller square regio

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In the figure above, if the area of the smaller square regio [#permalink]  06 Jul 2018, 08:36
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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. $$\sqrt{3}$$
[Reveal] Spoiler: OA

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Re: In the figure above, if the area of the smaller square regio [#permalink]  07 Jul 2018, 07:17
Carcass wrote:
Attachment:
square.jpg

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. $$\sqrt{3}$$

Here,

Side of the Larger square = 1 inch

Therefore Area Large square = $$1^2 =1$$

But we know Area of a square =$$\frac{diagonal^2}{2}$$

So 1 = $$\frac{diagonal^2}{2}$$

or diagonal = $$\sqrt{2}$$

And we know
smaller square region is $$\frac{2}{3}$$ the area of the larger square region

So it can be written as $$\frac{(Diagonal smaller square)^2}{2}= \frac{2}{3} * 1 = \frac{2}{3}$$ (Since the area of the Larger square = 1)

or $$(diagonal of smaller square)^2 = \frac{4}{3}$$

or diagonal of smaller square = $$\frac{\sqrt4}{\sqrt3} = \frac{2}{(\sqrt3)} = \frac{2}{\sqrt3} * \frac{\sqrt3}{\sqrt3} = \frac{2\sqrt3}{3}$$

Now,
Diagonal of the larger square is longer than the diagonal of the smaller square = $$\sqrt2 - \frac{2\sqrt3}{3}$$
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Re: In the figure above, if the area of the smaller square regio   [#permalink] 07 Jul 2018, 07:17
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