Question:hard

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?

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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.
Updated On: Jul 21, 2026
  • \((3\sqrt3)a^2\) cm2
  • \(\left(\frac{3\sqrt3}{2}\right)a^2\) cm2
  • \(\left(\frac{\sqrt3}{2}\right)a^2\) cm2
  • \((6\sqrt3)a^2\) cm2
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The Correct Option is B

Solution and Explanation

Step 1: Set up coordinates.
Place the hexagon's centre at the origin with circumradius 2a (equal to the side length). Going around: C=(0,2a), D=(a√3,a), E=(a√3,-a), F=(0,-2a), A=(-a√3,-a), B=(-a√3,a).
Step 2: Locate G using AG = FG.
Since AG = FG, G is the midpoint of FA: \(G=\left(\frac{0+(-a\sqrt3)}{2},\frac{-2a+(-a)}{2}\right)=\left(-\frac{a\sqrt3}{2},-\frac{3a}{2}\right)\).
Step 3: Find the area of triangle BCD by the shoelace formula.
With B(-a√3,a), C(0,2a), D(a√3,a):
Area = \(\frac{1}{2}\left|x_B(y_C-y_D)+x_C(y_D-y_B)+x_D(y_B-y_C)\right| = \frac{1}{2}\left|(-a\sqrt3)(a)+0+(a\sqrt3)(-a)\right| = \sqrt3a^2\).
Step 4: Find the area of triangle FEG by the shoelace formula.
With F(0,-2a), E(a√3,-a), G(-a√3/2,-3a/2):
Area = \(\frac{1}{2}\left|0+a\sqrt3\left(\frac{a}{2}\right)+\left(-\frac{a\sqrt3}{2}\right)(-a)\right| = \frac{1}{2}\left(\frac{\sqrt3a^2}{2}+\frac{\sqrt3a^2}{2}\right) = \frac{\sqrt3a^2}{2}\).
Step 5: Add the two shaded pieces.
Total shaded area = \(\sqrt3a^2 + \frac{\sqrt3a^2}{2} = \frac{3\sqrt3}{2}a^2\), confirming the same result found by the decomposition method.\[\boxed{Shaded\ area = \left(\frac{3\sqrt3}{2}\right)a^2\ cm^2}\]
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