Question

The figure below is a regular hexagon with side '2a' cm. If AG = FG and ED || GH, then what is the area of the shaded region?

• A. \((3\sqrt3)a^2\) cm²
• B. \(\left(\frac{3\sqrt3}{2}\right)a^2\) cm²
• C. \(\left(\frac{\sqrt3}{2}\right)a^2\) cm²
• D. \((6\sqrt3)a^2\) cm²

Hint:

Split the hexagon into a central rectangle and two triangular end caps, then use AG = FG to cut one cap exactly in half with the median from E.

Answer 1

The Correct Option is B

Step 1: Split the hexagon into a rectangle and two triangular caps.
In regular hexagon ABCDEF (side 2a), sides AB and DE are opposite and parallel, so A, B, D, E form a rectangle, leaving triangle BCD as the top cap and triangle FAE as the bottom cap.
Step 2: Find the size of each cap.
In triangle BCD, BC = CD = 2a with an included angle of \(120^\circ\) at C. Base \(BD = 2(2a)\sin60^\circ = 2a\sqrt3\), and the height from C to BD is \(2a\cos60^\circ = a\).
Area of triangle BCD = \(\frac{1}{2}(2a\sqrt3)(a) = \sqrt3a^2\). By symmetry, triangle FAE also has area \(\sqrt3a^2\).
Step 3: Use the condition AG = FG.
Since AG = FG, G is the midpoint of side FA. Drawing GH parallel to ED (as given) marks the boundary of the shaded portion, with the median EG in triangle FAE splitting it into two equal halves, triangle FEG and triangle AEG, each of area \(\frac{\sqrt3a^2}{2}\).
Step 4: Add up the shaded pieces.
The shaded region is made up of the whole top cap BCD together with the half of the bottom cap nearer to E (triangle FEG):
Shaded area = \(\sqrt3a^2 + \frac{\sqrt3a^2}{2} = \frac{3\sqrt3}{2}a^2\).\[\boxed{Shaded\ area = \left(\frac{3\sqrt3}{2}\right)a^2\ cm^2}\]