nick1816
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 left for himself?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
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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