Question

The figure is a regular hexagon ABCDEF (in order) with side '2a' cm, drawn with C at the top vertex, F at the bottom vertex, B and A forming the left vertical edge, and D and E forming the right vertical edge. A rectangle is drawn using side AB as its left edge, with its bottom edge AG horizontal (G lying on line AE) and its right edge GH vertical (parallel to ED, with H directly above G). Given AG = FG and ED || GH, what is the ratio of the area of the shaded region (the triangle below AG down to F, plus the strip to the right of GH up to the hexagon's right edge DE) to the area of the hexagon?

• A. 1 : 2
• B. 1 : 3
• C. 1 : 4
• D. 2 : 3
• E. 2 : 5

Hint:

Place the hexagon on coordinates with side 2a; use AG = FG to pin down where G falls on line y = -a, then split the shaded area into a triangle and a rectangle.

Answer 1

The Correct Option is A

Step 1: Set up coordinates.
Place the regular hexagon with side \(s=2a\) and circumradius \(R=2a\) (for a regular hexagon, side = circumradius) centred at the origin, vertex-up:
\(C=(0,2a)\), \(B=(-\sqrt{3}a,a)\), \(A=(-\sqrt{3}a,-a)\), \(F=(0,-2a)\), \(E=(\sqrt{3}a,-a)\), \(D=(\sqrt{3}a,a)\).

Step 2: Locate G using AG = FG.
Since AG is horizontal (the rectangle's bottom edge from A), G lies on the line \(y=-a\), so \(G=(g,-a)\).
\(AG = g+\sqrt{3}a\) and \(FG=\sqrt{g^2+a^2}\).
Setting \(AG=FG\): \((g+\sqrt{3}a)^2 = g^2+a^2 \Rightarrow 2\sqrt{3}ag+3a^2=a^2 \Rightarrow g=-\dfrac{a}{\sqrt{3}}\).
So \(AG = -\dfrac{a}{\sqrt3}+\sqrt3a = \dfrac{2a}{\sqrt3}\) (this is exactly \(\tfrac13\) of the full width AE, since \(AE=2\sqrt3a\)).

Step 3: Locate H.
Since \(GH \parallel ED\) (vertical), H is directly above G: \(H=\left(-\dfrac{a}{\sqrt3},\,a\right)\), completing rectangle ABHG.

Step 4: Compute the shaded pieces.
Triangle AGF: base \(AG=\dfrac{2a}{\sqrt3}\), height = perpendicular distance from F to line AG \(=a\).
Area \(=\dfrac12\times\dfrac{2a}{\sqrt3}\times a=\dfrac{a^2}{\sqrt3}\).
Rectangle GHDE: width \(=\sqrt3a-g=\sqrt3a+\dfrac{a}{\sqrt3}=\dfrac{4a}{\sqrt3}\), height \(=2a\).
Area \(=\dfrac{4a}{\sqrt3}\times 2a=\dfrac{8a^2}{\sqrt3}\).
Total shaded \(=\dfrac{a^2}{\sqrt3}+\dfrac{8a^2}{\sqrt3}=\dfrac{9a^2}{\sqrt3}=3\sqrt3a^2\).

Step 5: Compute the hexagon's area and the ratio.
Area of regular hexagon \(=\dfrac{3\sqrt3}{2}s^2=\dfrac{3\sqrt3}{2}(2a)^2=6\sqrt3a^2\).
Ratio \(=\dfrac{3\sqrt3a^2}{6\sqrt3a^2}=\dfrac12\), i.e. 1 : 2.

Note on the answer key: This solution was independently re-derived and numerically cross-checked (including with an origin shifted to A) and consistently gives 1 : 2 (option a) from the literal conditions AG = FG and GH || ED. The listed answer key for this question is (c) 1 : 4, which could not be reproduced from the stated conditions even after repeated rechecking; the discrepancy is flagged here rather than silently matching the key.