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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]  26 Aug 2018, 04:38
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In the figure above, if the area of the smaller square region is $$\frac{1}{2}$$ 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. $$\sqrt{2} - 1$$

B. $$\frac{1}{2}$$

C. $$\frac{\sqrt{2}}{2}$$

D. $$\frac{\sqrt{2} + 1}{2}$$

E. $$\sqrt{2}$$
[Reveal] Spoiler: OA

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Re: In the figure above, if the area of the smaller square regio [#permalink]  26 Aug 2018, 06:28
Carcass wrote:
Attachment:
square.jpg

In the figure above, if the area of the smaller square region is $$\frac{1}{2}$$ 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. $$\sqrt{2} - 1$$

B. $$\frac{1}{2}$$

C. $$\frac{\sqrt{2}}{2}$$

D. $$\frac{\sqrt{2} + 1}{2}$$

E. $$\sqrt{2}$$

$$\frac{{Diagonal^2}}{2} = Area of square$$

So for large square of side 1 inch

$$\frac{{Diagonal^2}}{2} = 1$$ ( area of the large square with side 1 inch)

or $$Diagonal = \sqrt2$$

Now for small square

$$\frac{{Diagonal^2}}{2} = \frac{1}{2}$$ ( area of the small square = $$\frac{1}{2}$$ the area of larger square)

or $$Diagonal = 1$$

So the diagonal of the larger square is longer than the diagonal of the smaller square = $$\sqrt2 - 1$$
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Re: In the figure above, if the area of the smaller square regio   [#permalink] 26 Aug 2018, 06:28
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