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03 Jul 2022, 16:00
Kudos
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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:
55%
(hard)
Question Stats:
46% (01:45) correct
54%
(02:08)
wrong
based on 24
sessions
History
Date
Time
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Not Attempted Yet
What is the area of a triangle (with vertices at FCE) that is inscribed in a hexagon with vertices at ABCDE?
(1) The hexagon is regular and BE = 14.
(2) \(EC = 7\sqrt{3}\).
Are You Up For the Challenge: 700 Level Questions
(1) The hexagon is regular and BE = 14.
(2) \(EC = 7\sqrt{3}\).
Are You Up For the Challenge: 700 Level Questions
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
KabirKanha
08 Jul 2022, 21:40
Kudos
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Attachment:
hex.jpg [ 24.21 KiB | Viewed 1304 times ]
(1) We know that a regular hexagon can be seen to be made up of 6 equilateral triangles, as shown in the figure above.
It is clear from the image that \(BE = \) Twice the length of a side of the equilateral triangle. Hence, \(EF = 7\) and \(CF = 14\).
Now that we have the hypotenuse and the base, we can apply Pythagoras' Theorem to find the height too.
(Notice that the triangle \(FCE\) is right-angled and this can be proved.)
\(CE^2 + EF^2 = CF^2\)
\(\Rightarrow CE^2 + 49 = 196\)
\(\Rightarrow CE^2 = 147\)
\(\Rightarrow CE = 7 \sqrt{3}\)
Hence,
Area of Triangle \(FCE \)= \(\frac{1}{2}(EF)(CE)\)
\(= \frac{1}{2}(7)(7 \sqrt{3})\)
\(= \frac{49}{2}\sqrt{3}\)
=> Sufficient
(2) Just given \(EC = 7 \sqrt{3}\) does not tell us anything about the nature of the hexagon and we cannot deduce the other sides of the triangle, and hence cannot find the area.
=> Insufficient
Ans: Option A
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