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26 Aug 2020, 03:52
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E
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:
50% (02:17) correct
50%
(01:50)
wrong
based on 34
sessions
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Greedy John is celebrating his birthday with 6 of his friends. His mother baked him a birthday cake in the shape of a regular hexagon. He makes cuts linking the midpoints of every 2 adjacent sides so that he can keep most of the cake, and distributes these 6 slices to his friends. What proportion of the cake does he leave for himself?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
26 Aug 2020, 04:39
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nick1816
When you join the midpoints of each side you will get another regular hexagon, as shown in the figure.
Now the area of Hexagon is nothing but the sum of area of 6 equilateral triangles whose side is same as side of the hexagon, say y. => \(6*\frac{\sqrt{3}*a^2}{4}=\frac{3*\sqrt{3}*a^2}{2}\)
Thus the area of bigger hexagon with side 2x =\(\frac{3*\sqrt{3}*(2x)^2}{2}=6*(\sqrt{3}*x^2)\)
To calculate the side of inner hexagon in terms of smaller hexagon -
take \(\triangle\) AGH, which is 30-60-90 \(\triangle\)
\(GH = x*\sqrt{3}/2\), so side = \(x*\sqrt{3}\)
Area = \(\frac{3*\sqrt{3}*(\sqrt{3}*x)^2}{2}=\frac{9*\sqrt{3}*x^2}{2}=4.5*(\sqrt{3}*x^2)\)
Proportion = \(\frac{4.5}{6}=\frac{3}{4}\)
Also, the area of two regular hexagon will be in the ratio of square of their sides...
so \((\frac{\sqrt{3}x}{2x})^2=\frac{3}{4}\)
E
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images.png [ 8.44 KiB | Viewed 1924 times ]
ShankSouljaBoi
Joined: 21 Jun 2017
Last visit: 12 Aug 2026
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Location: India
Concentration: Finance, Economics
Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 1: 660 Q49 V31
GMAT 2: 620 Q47 V30
GMAT 3: 650 Q48 V31
GPA: 3.1
WE:Corporate Finance (Non-Profit and Government)
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Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 3: 650 Q48 V31
Posts: 593
26 Aug 2020, 05:47
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Hexagon can be divided into 4 big triangles. Property --- When we join the mid points we actually segregate the triangle in half as it is a regular hexagon.
Let x be the area of one such triangle that he leaves for himself. So, he has 6 triangles for himself i.e. 6x is the amount of cake left for him.
Now, the hexagon has 4 big triangles and each is double the area of a small triangle denoted as x. In all, the area of hexagon is basically just 8x.
Required proportion = 6x / 8x = 3/4 ----> E
Edit : OMG chetan2u your explanation and approach are just amazing !!! Is this question doable in two minutes ?
Let x be the area of one such triangle that he leaves for himself. So, he has 6 triangles for himself i.e. 6x is the amount of cake left for him.
Now, the hexagon has 4 big triangles and each is double the area of a small triangle denoted as x. In all, the area of hexagon is basically just 8x.
Required proportion = 6x / 8x = 3/4 ----> E
Edit : OMG chetan2u your explanation and approach are just amazing !!! Is this question doable in two minutes ?
26 Aug 2020, 09:08
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Another way is-
Just look at the below picture, you'll get the answer (most of the regular polygon problems can be solved by dividing the polygon in smaller similar triangles; it wont take more than 1-1.5 min tbh)
cgm2wzqAG8-1.png [ 171.47 KiB | Viewed 1835 times ]
Just look at the below picture, you'll get the answer (most of the regular polygon problems can be solved by dividing the polygon in smaller similar triangles; it wont take more than 1-1.5 min tbh)
Attachment:
cgm2wzqAG8-1.png [ 171.47 KiB | Viewed 1835 times ]
ShankSouljaBoi
