Question:
Let ABCDEF be a regular hexagon. What is the ratio of the area of $\triangle ACE$ to that of the hexagon ABCDEF?
Let ABCDEF be a regular hexagon. What is the ratio of the area of $\triangle ACE$ to that of the hexagon ABCDEF?
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In problems involving regular polygons like hexagons, divide the polygon into smaller congruent parts (such as triangles) to find the required area ratio.
Updated On: Mar 23, 2026
- $\frac{1}{3}$
- $\frac{1}{2}$
- $\frac{2}{3}$
- $\frac{5}{6}$
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The Correct Option is C
Solution and Explanation
In a regular hexagon, the area of $\triangle ACE$ is formed by drawing diagonals between alternate vertices. The area of this triangle is a part of the entire hexagon, which can be divided into 6 equilateral triangles. Each triangle has an area that is $\frac{1}{6}$ of the total hexagon area.
Since the triangle $\triangle ACE$ covers two of these smaller triangles, the area of $\triangle ACE$ is $\frac{2}{3}$ of the total hexagon area.
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