Question:hard

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?

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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.
Updated On: Jul 20, 2026
  • 1 : 2
  • 1 : 3
  • 1 : 4
  • 2 : 3
  • 2 : 5
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The Correct Option is A

Solution and Explanation

Work with fractions of the hexagon's width instead of raw coordinates.

A regular hexagon of side $2a$ splits into a central rectangle of width $2\sqrt3 a$ and height $2a$ (using the vertical edges AB and DE), plus two end triangles (cap at C, cap at F), each of area $\sqrt3 a^2$. The central rectangle has area $2\sqrt3a \times 2a = 4\sqrt3 a^2$, and together with the two caps ($\sqrt3a^2$ each) the total is $4\sqrt3a^2+\sqrt3a^2+\sqrt3a^2=6\sqrt3a^2$, matching the standard hexagon area formula $\frac{3\sqrt3}{2}(2a)^2$.

Placing A at the origin, F sits at $(\sqrt3a,-a)$ measuring along the bottom cap, and setting up $AG=FG$ for $G=(g,0)$ on line AE gives, after squaring and simplifying:
$g^2=(g-\sqrt3a)^2+a^2 \Rightarrow 0=-2\sqrt3ag+4a^2 \Rightarrow g=\dfrac{2a}{\sqrt3}$

This is exactly $\dfrac13$ of the full base $AE=2\sqrt3a$, so G splits AE in the ratio $1:2$ (AG : GE $= \tfrac{2a}{\sqrt3} : \tfrac{4a}{\sqrt3} = 1:2$).

Because the central rectangle splits into ABHG and GHDE in the same width ratio $1:2$, ABHG $=\tfrac13(4\sqrt3a^2)=\tfrac{4\sqrt3}{3}a^2$ and GHDE $=\tfrac23(4\sqrt3a^2)=\tfrac{8\sqrt3}{3}a^2$.

The bottom cap ($\sqrt3a^2$) splits at G in the same $1:2$ ratio (same base ratio, same apex F), so triangle AGF $=\tfrac13\sqrt3a^2=\tfrac{\sqrt3}{3}a^2$.

Shaded $=$ GHDE $+$ AGF $=\tfrac{8\sqrt3}{3}a^2+\tfrac{\sqrt3}{3}a^2=3\sqrt3a^2$, and hexagon $=6\sqrt3a^2$.

\[\boxed{Shaded : Hexagon = 1:2}\]

This matches the coordinate-geometry method exactly; the provided answer key (1:4) could not be reproduced.
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