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Updated on: 30 Jul 2014, 18:43
Originally posted by PareshGmat on 05 Mar 2014, 19:23.
Last edited by PareshGmat on 30 Jul 2014, 18:43, edited 2 times in total.
Last edited by PareshGmat on 30 Jul 2014, 18:43, edited 2 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:09) correct
29%
(02:47)
wrong
based on 296
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Attachment:
sq.jpg [ 25.89 KiB | Viewed 80953 times ]
A: \(\frac{1}{12}\)
B: \(\frac{\sqrt{2}}{8}\)
C: \(\frac{1}{16}\)
D: \(\frac{1}{8}\)
E: \(\frac{\sqrt{2}}{16}\)
05 Mar 2014, 22:36
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PareshGmat
There are various ways of approaching it.
Method 1:
Area of ABCD is 1.
Area of PQRS = (1/2)*(diagonal1)*(diagonal2)
Both diagonals of PQRS are 1 each since they are the same lengths as sides of ABCD
Area of PQRS = (1/2)*1*1 = 1/2
Area of leftover region = 1 - 1/2 = 1/2
The leftover region after you cut out PQRS is split into 8 equal areas and the red region is one of those 8.
Hence area of red region is (1/8)*(1/2) = 1/16
Method 2:
The perimeter of ABCD is 4 so each side is 1. So each half side is 1/2.
In triangle APQ, AP and AQ are 1/2 each so \(PQ = 1/\sqrt{2}\) (using Pythegorean theorem)
So half of PQ is \(1/2\sqrt{2}\)
Area of red triangle = \((1/2)*(1/2\sqrt{2})*(1/2\sqrt{2}) = 1/16\)
General Discussion
05 Mar 2014, 22:38
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Side of outer square = 1
Half of side of outer square = 0.5
Area of outer square = 1
Area of inner square = \((side of inner square)^2\)
\((side of inner square)^2= 0.5^2 + 0.5^2 = 0.25 + 0.25 = 0.5\)
Area of outer square - area of inner square = 1 - 0.5 = 0.5
Shaded area = \(\frac{1}{8}(0.5) = \frac{1}{16}\)
Answer is C
Half of side of outer square = 0.5
Area of outer square = 1
Area of inner square = \((side of inner square)^2\)
\((side of inner square)^2= 0.5^2 + 0.5^2 = 0.25 + 0.25 = 0.5\)
Area of outer square - area of inner square = 1 - 0.5 = 0.5
Shaded area = \(\frac{1}{8}(0.5) = \frac{1}{16}\)
Answer is C
