Bunuel

A rectangle is inscribed in a hexagon that has all sides of equal length and all angles of equal measure, as shown in the figure above. If the total area of the hexagon is x, then the sum of the areas of the two shaded regions is
A. x/6
B. x/4
C. x/3
D. x/2
E. 2x/3
PS20418
KarishmaBhow about an alternative solution where you subtract an area of rectangle from an area of hexagon
i did this but smth went wrong

\(\frac{3*x^2 \sqrt{3} }{2 }- 2x^2\sqrt{3}\)
area of rectangle i got this way: each shaded region represents isosceles triangle, so if i draw perpendicular line i get 30:60:90 trangle
hence length of rectangle is \(2x\sqrt{3}\) and width x
I would much rather use the symmetry of a regular hexagon and the triangles I can carve out of it as done above.
But if you wanted to use geometry as suggested by you, note that x is the area of the hexagon, not a side.
Since a hexagon is made up of 6 equilateral triangles, its area will be \(6 * \frac{\sqrt{3} * a^2}{4} = x\) ... (I)
If side of the hexagon is a, in the 30-60-90 shaded triangle (half of each shaded triangle), the side opposite to 60 is \(\frac{\sqrt{3}a}{2}\) which means the width of the rectangle is \(\sqrt{3}a\).
Area of rectangle is \(a*\sqrt{3}a = \sqrt{3}a^2 = \frac{2x}{3}\) (from (I) above)