CAT 2003 Slot 2 — QA Question 21

ABCDEF is a regular hexagon, join centre O with vertices A, C and E.

In quadrilateral AFEO, diagonal EA divides it into two equal area triangles.

i.e. Area (∆AFE) = Area (∆AOE)

Similarly, Area (∆ABC) = Area (∆AOC)

And, Area (∆CDE) = Area (∆COE)

$\therefore \frac{Area(∆ACE)}{Area(Hexagon ABCDEF)}=\frac{3}{6}=\frac{1}{2}$

Hence, option (b).