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18 Nov 2022, 08:57
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
39% (01:43) correct
61%
(02:02)
wrong
based on 28
sessions
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Not Attempted Yet
The figure above represents a rectangular tile. The eight hexagons in the design are equilateral and equiangular. What is the area of the tile?
(1) w = 5 inches
(2) q = \(3\sqrt{3}\) inches
Attachment:
2022-11-18_19-55-13.png [ 47.95 KiB | Viewed 1563 times ]
28 Nov 2022, 09:23
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Bunuel
Attachment:
1.6 Gmat.jpg [ 25.37 KiB | Viewed 815 times ]
Please refer to the attached fig.
A hexagon can be divided into \(6\) equilateral \(\Delta's \)
If the whole figure can be decomposed into equilateral \(\Delta's \) and if we can find a side or height of one equilateral \(\Delta\) then certainly we can find the area of the whole fig.
From the fig. can see \(w = 5s\) and \(q =6h \) ( \(s =\) side of equilateral \(\Delta\) and \(h =\) height of equilateral \(\Delta \))
(1) w = 5 inches
Which means \(5s = 5 \) and \(s=1 \)
Now we know the side we can find the height using \(h = \frac{\sqrt3}{2}*s \)
\(h=\frac{\sqrt3}{2}*1 \)
The horizontal measure \( q \) of the figure \(= 6*h = 6 * \frac{\sqrt3}{2}= 3\sqrt3\)
Area of the fig. \(= w*q = 5*3\sqrt3\)
SUFF.
(2)q = \(3\sqrt{3}\) inches
Again from the fig. we can see \(q= 6*h\) so \(6*h = 3\sqrt3 \) solving further we get \(h = \frac{\sqrt3}{2}\)
Now that we know \(h\) , again using \( \frac{\sqrt3}{2}*s = h \) we can get side
\(\frac{\sqrt3}{2}*s = \frac{\sqrt3}{2}\) and so \(s=1 \)
from the fig. we know vertical measure \(w = 5s \), hence \(w= 5 \)
Area \(w*q = 5 * 3\sqrt3\)
SUFF.
Ans D
Hope it's clear
