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# In the figure above, if the area of the smaller square region is 1/2

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Joined: 02 Sep 2009
Posts: 43335

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In the figure above, if the area of the smaller square region is 1/2 [#permalink]

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21 Sep 2017, 23:35
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In the figure above, if the area of the smaller square region is 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) √2 – 1
(B) 1/2
(C) √2/2
(D) (√2 + 1)/2
(E) √2

[Reveal] Spoiler:
Attachment:

2017-09-20_1023_001.png [ 2.51 KiB | Viewed 628 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 139519 [1], given: 12794

Intern
Joined: 28 Aug 2017
Posts: 35

Kudos [?]: 3 [0], given: 57

Re: In the figure above, if the area of the smaller square region is 1/2 [#permalink]

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22 Sep 2017, 03:52
Bunuel wrote:

In the figure above, if the area of the smaller square region is 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) √2 – 1
(B) 1/2
(C) √2/2
(D) (√2 + 1)/2
(E) √2

[Reveal] Spoiler:
Attachment:
2017-09-20_1023_001.png

Side of the bigger square = $$a$$
Area of the bigger square = $$a^2$$
Diagonal of the bigger square = $$a\sqrt{2}$$

Side of the smaller square = $$b$$
Area of the smaller square = $$b^2$$
Diagonal of the smaller square = $$b\sqrt{2}$$

Given: $$b^2 = a^2/2$$
Since a = 1$$, a^2$$= 1, therefore $$b^2$$= $$a^2/2$$ = 1/2
So, $$b = \frac{1}{\sqrt{2}}$$

$$a\sqrt{2} - b\sqrt{2} = \sqrt{2}(a-b) = \sqrt{2}(1-1/\sqrt{2}$$) = $$\sqrt{2} - 1$$

Kudos [?]: 3 [0], given: 57

VP
Joined: 22 May 2016
Posts: 1252

Kudos [?]: 463 [0], given: 683

In the figure above, if the area of the smaller square region is 1/2 [#permalink]

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22 Sep 2017, 07:02
Bunuel wrote:

In the figure above, if the area of the smaller square region is 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) √2 – 1
(B) 1/2
(C) √2/2
(D) (√2 + 1)/2
(E) √2

[Reveal] Spoiler:
Attachment:
2017-09-20_1023_001.png

I used the one length given: the large square's side = 1

Find the area of the small square, find its side length, find length of both diagonals, and subtract.

1) Area of small square

Large square area, s = 1 is (1 * 1) = 1

The area of smaller square is half of 1 = $$\frac{1}{2}$$

2) Side of small square

From area, where $$s$$ = $$\sqrt{area}$$:
Side of the small square is $$\sqrt{\frac{1}{2}}$$

3) Diagonal length for both

Length of a diagonal** of a square with side $$s$$ is $$s\sqrt{2}$$

Large square diagonal = $$1\sqrt{2}$$ = $$\sqrt{2}$$

Small square diagonal:

$$(\sqrt{\frac{1}{2}}$$ * $$\sqrt{2})$$ = $$\sqrt{\frac{2}{2}}$$ = $$\sqrt{1} = 1$$

4) Difference: Large square's diagonal is how many inches longer than small square's diagonal?

(Diagonal$$_{large}$$) - (Diagonal$$_{small}$$):

$$\sqrt{2} - 1$$

**Or find diagonals with Pythagorean theorem, where hypotenuse = diagonal = d, e.g., for large square:

$$1^2 + 1^2 = d^2$$
$$2 = d^2$$
$$d = \sqrt{2}$$
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Kudos [?]: 463 [0], given: 683

In the figure above, if the area of the smaller square region is 1/2   [#permalink] 22 Sep 2017, 07:02
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