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jn0r
Joined: 21 Jan 2021
Last visit: 05 Apr 2024
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Location: Uzbekistan
Concentration: Strategy, Operations
Schools: Yale '25 (A$) Kellogg '25 (A$) McCombs '25 (A$) Cornell '25 (A$$) Fuqua '25 (M$$$) Tuck '25 (A$$$) Haas '25 (WL) Darden '25 (WL) Anderson '25 (A$$)
GMAT 1: 730 Q50 V40
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Schools: Yale '25 (A$) Kellogg '25 (A$) McCombs '25 (A$) Cornell '25 (A$$) Fuqua '25 (M$$$) Tuck '25 (A$$$) Haas '25 (WL) Darden '25 (WL) Anderson '25 (A$$)
GMAT 1: 730 Q50 V40
Posts: 124
18 Jun 2021, 04:02
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (01:15) correct
22%
(01:36)
wrong
based on 27
sessions
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What fraction of the regular hexagon is shaded?
A) \(\frac{2}{3}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{3}\)
D) \(\frac{2}{5}\)
E) \(\frac{3}{7}\)
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Screenshot_6.png [ 17.98 KiB | Viewed 5065 times ]
18 Jun 2021, 09:47
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jn0r
Many questions in Geometry can be done merely by analysing the figure and approximating area.
Make all diagonals and draw perpendiculars as shown.
Now each part is equal and shaded portion is 3 out of 12 equal parts.
Thus, fraction of the shaded region is \(\frac{1}{4}\).
B
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20 Jun 2021, 13:10
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jn0r
First let us focus on the bigger right triangle in the graph, since the non-shaded area inside that triangle has half height and half base compared to the bigger right triangle, that non-shaded area is only \(\frac{1}{2}*\frac{1}{2} = \frac{1}{4}\) of the full triangle. Then the shaded area is \(\frac{3}{4}\) of the bigger triangle.
Then let us compare the big triangle to the hexagon, the hexagon consists of 6 equilateral triangles. The big triangle has the same base as the equilateral triangle, but twice the height. Thus the big triangle = 2 equilateral triangles, and the hexagon = 6 equilateral triangles. We can confirm the big trianlge is \(\frac{1}{3}\) of the hexagon.
Finally the shaded area would be \(\frac{3}{4}*\frac{1}{3} = \frac{1}{4}\) of the whole hexagon.
Ans: B
