The figure shows a regular hexagon ABCDEF with each side equal to \(2a\) cm. Point G lies on side CD and point H lies on side EF. If \(AG = FG\) and \(ED \parallel GH\), then what is the ratio of the area of the shaded region to the area of the hexagon?
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Place the hexagon on coordinates so AB and DE become vertical sides, then use AG = FG to pin down G and the parallel condition to pin down H before measuring the shaded pieces.
Step 1: Place the hexagon on coordinates. Since AB and DE are drawn as the two vertical sides of the hexagon, put the centre of the hexagon at the origin. With side \(2a\), the vertices work out to \(A(-\sqrt3a,-a)\), \(B(-\sqrt3a,a)\), \(C(0,2a)\), \(D(\sqrt3a,a)\), \(E(\sqrt3a,-a)\), \(F(0,-2a)\). Check: \(AB = 2a\) and \(BC=\sqrt{(\sqrt3a)^2+a^2}=\sqrt{4a^2}=2a\), so every side is \(2a\), matching a regular hexagon.
Step 2: Locate G on CD using AG = FG. Write \(G = C+t(D-C) = (t\sqrt3a,\,(2-t)a)\) for \(t\) between 0 and 1. Setting \(AG^2=FG^2\) and simplifying gives \(4t^2+12 = 4t^2-8t+16\), so \(8t=4\) and \(t=\dfrac12\). So G is exactly the midpoint of CD: \(G=\left(\dfrac{\sqrt3a}{2},\,1.5a\right)\).
Step 3: Locate H on EF using GH parallel to ED. ED joins \(E(\sqrt3a,-a)\) and \(D(\sqrt3a,a)\), which is a vertical line. So GH must also be vertical, meaning H has the same x-coordinate as G, \(\dfrac{\sqrt3a}{2}\). Line EF runs from \(E(\sqrt3a,-a)\) to \(F(0,-2a)\); putting \(x=\dfrac{\sqrt3a}{2}\) shows H is the midpoint of EF too: \(H=\left(\dfrac{\sqrt3a}{2},\,-1.5a\right)\).
Step 4: Identify and measure the shaded pieces. The diagram shades two pieces: the quadrilateral GDEH on the right, and the small triangle at the bottom cut off near F (whose other two corners are H and the point where the hexagon's own left-right symmetry places the mirror of G on side FA). GDEH is a trapezium because \(GH \parallel DE\) (both vertical). \(GH = 1.5a-(-1.5a)=3a\) and \(DE = a-(-a)=2a\), while the gap between these two parallel sides is \(\sqrt3a-\dfrac{\sqrt3a}{2}=\dfrac{\sqrt3a}{2}\). \[ \text{Area(GDEH)} = \frac12(GH+DE)\times \frac{\sqrt3a}{2} = \frac12(3a+2a)\times\frac{\sqrt3a}{2} = \frac{5\sqrt3a^2}{4} \] The bottom triangle has its top side of length \(\sqrt3a\) (by the same mirror symmetry that gave GH) sitting at height \(-1.5a\), and its apex is F at \(-2a\), so its height is \(0.5a\). \[ \text{Area(triangle)} = \frac12\times\sqrt3a\times0.5a = \frac{\sqrt3a^2}{4} \] Adding these, shaded area \(= \dfrac{5\sqrt3a^2}{4}+\dfrac{\sqrt3a^2}{4}=\dfrac{6\sqrt3a^2}{4}=\dfrac{3\sqrt3a^2}{2}\).
Step 5: Divide by the hexagon's area. A regular hexagon of side \(s\) has area \(\dfrac{3\sqrt3}{2}s^2\); here \(s=2a\), so hexagon area \(=\dfrac{3\sqrt3}{2}(2a)^2=6\sqrt3a^2\). \[ \text{Ratio} = \frac{3\sqrt3a^2/2}{6\sqrt3a^2} = \frac{3}{12} = \frac14 \]
Final Answer: The shaded region is one-quarter of the hexagon's area. \[ \boxed{1:4} \]
