Problem Solving

[Math Revolution GMAT math practice question]

Which of the following could be the area of a regular hexagon in which the length of the equal sides is an integer?

A. √3/6
B. √3/5
C. √3/4
D. 6√3
E. 3√3/4

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Which of the following could be the area of a regular hexagon in which the length of the equal sides is an integer?

A. √3/6
B. √3/5
C. √3/4
D. 6√3
E. 3√3/4

$$?\,\,\,:\,\,\,{\rm{possible}}\,\,S\, = \,{\rm{regular}}\,\,{\rm{hexagon}}\,\,{\rm{area}}\,\,\,\,\left[ {{\rm{L}}\,\,{\rm{integer}}\,\,{\rm{side}}\,\,\left( * \right)} \right]\,\,$$
$$S = 6 \cdot {S_{\Delta {\rm{equil,}}\,{\rm{side}}\,{\rm{L}}}} = 6\left( {{{{L^2}\sqrt 3 } \over 4}} \right) = {{3\sqrt 3 } \over 2}{L^2}$$
$$\left. \matrix{
\left( {\rm{A}} \right)\,\,\,{{\sqrt 3 } \over 6}\,\,\,\,\, \Rightarrow \,\,\,\,\,{L^2} = {2 \over 3}\left( {{1 \over 6}} \right)\,\,\,\,{\rm{impossible}}\,\,\,\left( * \right)\,\,\, \hfill \cr
\left( {\rm{B}} \right)\,\,\,{{\sqrt 3 } \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,\,{L^2} = {2 \over 3}\left( {{1 \over 5}} \right)\,\,\,\,{\rm{impossible}}\,\,\,\left( * \right) \hfill \cr
\left( {\rm{C}} \right)\,\,\,{{\sqrt 3 } \over 4}\,\,\,\,\, \Rightarrow \,\,\,\,\,{L^2} = {2 \over 3}\left( {{1 \over 4}} \right)\,\,\,\,{\rm{impossible}}\,\,\,\left( * \right) \hfill \cr
\left( {\rm{D}} \right)\,\,\,6\sqrt 3 \,\,\,\,\, \Rightarrow \,\,\,\,\,{L^2} = {2 \over 3}\left( 6 \right)\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,L = 2\,\,\,\,{\rm{possible!}}\,\,\, \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

for a hexagon of equal sides, the area can be calculated using formula
sqrt{3} * x^2/2
this has been derived from equilateral triangle formula (sqrt{3}*x^2/4) * 6

since given that x is an integer so only at x=2 we get option D 6 sqrt{3} IMOD

=>

Let n be the side-length of a regular hexagon. Then the area of the shaded equilateral triangle is (√3/4)n^2 and the area of the regular hexagon is (6√3/4)n^2 = (3√3/2)n^2 .

If n = 2, then (3√3/2)n^2 = (3√3/2)2^2 =6√3. None of the other values are possible for integer values of n.

Therefore, the answer is D.
Answer: D