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22 Sep 2017, 00:35
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:45) correct
33%
(02:24)
wrong
based on 80
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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
Attachment:
2017-09-20_1023_001.png [ 2.51 KiB | Viewed 16527 times ]
22 Sep 2017, 04:52
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Bunuel
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, \(b = \frac{1}{\sqrt{2}}\)
\(a\sqrt{2} - b\sqrt{2} = \sqrt{2}(a-b) = \sqrt{2}(1-1/\sqrt{2}\)) = \(\sqrt{2} - 1\)
Therefore, answer is A.
22 Sep 2017, 08:02
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Bunuel
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\(_{small}\)):
\(\sqrt{2} - 1\)
ANSWER A
**Or find diagonals with Pythagorean theorem, where hypotenuse = diagonal = d, e.g., for large square:
\(1^2 + 1^2 = d^2\)
\(2 = d^2\)
\(d = \sqrt{2}\)
