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27 Mar 2020, 05:42
Kudos
Bookmarks
D
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:
67% (02:11) correct
33%
(01:57)
wrong
based on 60
sessions
History
Date
Time
Result
Not Attempted Yet
Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is
A. 5: 6
B. 4: 5
C. 3: 4
D. 2: 3
E. 2:5
A. 5: 6
B. 4: 5
C. 3: 4
D. 2: 3
E. 2:5
ArunSharma12
Joined: 25 Oct 2015
Last visit: 20 Jul 2022
Posts: 512
Given Kudos: 74
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
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27 Mar 2020, 10:15
Kudos
Bookmarks
Bunuel
three triangles cut at the corner will be equilateral triangles, external angle of regular hexagon is 60 degrees making the corner triangles equilateral.
lets consider the side of regular hexagon H be a, then the side of triangle T will be 3*a.
Area of H= \(\frac{3*\sqrt{3}*a^2}{2}\)
Area of T = \(\frac{\sqrt{3}*(3*a)^2}{4}\)
\(\frac{Area of H}{Area of T}\) = \(\frac{2}{3}\)
Ans: D
28 Mar 2020, 08:49
Kudos
Bookmarks
Bunuel
Please see the solution in the image
All 9 triangle are of the same area including corners cut off from an equilateral triangle T to form regular hexagon H.
H: T = 6:9 = 2:3
IMO D
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