Question:hard

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.
Updated On: Jul 21, 2026
  • 1 : 2
  • 1 : 3
  • 1 : 4
  • 2 : 3
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The Correct Option is C

Solution and Explanation

Step 1: Set up coordinates from corner A instead of the centre.
Take \(A=(0,0)\). Moving around the regular hexagon of side \(2a\) gives \(B=(0,2a)\), \(C=(\sqrt3a,3a)\), \(D=(2\sqrt3a,2a)\), \(E=(2\sqrt3a,0)\), \(F=(\sqrt3a,-a)\). This is just the earlier hexagon shifted, so all the side lengths still come out to \(2a\).

Step 2: Write down G, H and the mirror point on FA.
Since \(AG=FG\) forces G to the midpoint of CD (shown by the same distance argument as before), and \(GH\parallel ED\) forces H to the midpoint of EF, we get: \[ G=(1.5\sqrt3a,\,2.5a),\quad H=(1.5\sqrt3a,\,-0.5a) \] and the third corner of the bottom triangle is the midpoint of FA, \(M=(0.5\sqrt3a,-0.5a)\).

Step 3: Use the shoelace (coordinate determinant) formula on quadrilateral GDEH.
For points \((x_1,y_1),\dots,(x_4,y_4)\), area \(=\dfrac12\left|\sum (x_iy_{i+1}-x_{i+1}y_i)\right|\). Plugging in \(G,D,E,H\):
\(G\to D: 1.5\sqrt3a(2a)-2\sqrt3a(2.5a)=-2\sqrt3a^2\)
\(D\to E: 2\sqrt3a(0)-2\sqrt3a(2a)=-4\sqrt3a^2\)
\(E\to H: 2\sqrt3a(-0.5a)-1.5\sqrt3a(0)=-\sqrt3a^2\)
\(H\to G: 1.5\sqrt3a(2.5a)-1.5\sqrt3a(-0.5a)=4.5\sqrt3a^2\)
Sum \(=-2.5\sqrt3a^2\), so Area(GDEH) \(=\dfrac{2.5\sqrt3a^2}{2}=\dfrac{5\sqrt3a^2}{4}\), matching the trapezium answer.

Step 4: Apply the same formula to triangle M, H, F.
\[ \text{Area} = \frac12\left|x_M(y_H-y_F)+x_H(y_F-y_M)+x_F(y_M-y_H)\right| \] Substituting the coordinates gives \(0.25\sqrt3a^2=\dfrac{\sqrt3a^2}{4}\).
Total shaded area \(=\dfrac{5\sqrt3a^2}{4}+\dfrac{\sqrt3a^2}{4}=\dfrac{3\sqrt3a^2}{2}\).

Step 5: Compare to the hexagon.
Hexagon area \(=6\sqrt3a^2\), so the ratio is \(\dfrac{3\sqrt3a^2/2}{6\sqrt3a^2}=\dfrac14\). This lines up exactly with the trapezium method, confirming the shaded region is a quarter of the hexagon. \[ \boxed{1:4} \]
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