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D
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Difficulty:
55%
(hard)
Question Stats:
71% (03:21) correct
29%
(03:10)
wrong
based on 347
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
Triangle.PNG [ 33.97 KiB | Viewed 19148 times ]
A. \(2\sqrt{2}\)
B. \(2\sqrt{7}\)
C. \(\frac{2\sqrt{3}}{3}\)
D. \(\frac{7\sqrt{2}}{2}\)
E. \(\frac{7\sqrt{3}}{3}\)
14 Feb 2012, 08:14
Kudos
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Attachment:
Triangle.PNG [ 33.97 KiB | Viewed 18995 times ]
A. \(2\sqrt{2}\)
B. \(2\sqrt{7}\)
C. \(\frac{2\sqrt{3}}{3}\)
D. \(\frac{7\sqrt{2}}{2}\)
E. \(\frac{7\sqrt{3}}{3}\)
Since the area of unshaded region is equal to the are of shaded region, then the area of the big triangle is twice the area of the little triangle (unshaded region): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)
Next, triangles ABC and MNP are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\). Thus \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}\) --> \(\frac{2}{1}=\frac{7^2}{MP^2}\) --> \(MP^2=\frac{7^2}{2}\) --> \(MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\).
Answer: D.
General Discussion
14 Feb 2012, 08:35
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Neat little rule Bunuel ! Kudos !
