Question:
Let ABCDEF be a regular hexagon. What is the ratio of the area of the \(\Delta\)ACE to that of the hexagon ABCDEF?
Let ABCDEF be a regular hexagon. What is the ratio of the area of the \(\Delta\)ACE to that of the hexagon ABCDEF?
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Let the side of the hexagon be s. Find the side of triangle ACE and compare the two areas, or cut the hexagon into 6 equal equilateral triangles.
Updated On: Nov 6, 2026
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
- \(\frac{1}{2}\)
- \(\frac{5}{6}\)
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The Correct Option is C
Solution and Explanation
Step 1: Hexagon area:
A regular hexagon of side s is made of 6 equilateral triangles of side s, so its area is \(\frac{3\sqrt{3}}{2}s^2\).
Step 2: Triangle ACE:
In triangle ABC, AB = BC = s and \(\angle ABC = 120^\circ\), so \(AC^2 = s^2 + s^2 - 2s^2\cos 120^\circ = 3s^2\), i.e. \(AC = s\sqrt{3}\). By symmetry, ACE is equilateral with side \(s\sqrt{3}\), so its area is \(\frac{\sqrt{3}}{4}\times 3s^2 = \frac{3\sqrt{3}}{4}s^2\).
Step 3: Ratio:
\(\frac{3\sqrt{3}/4}{3\sqrt{3}/2} = \frac{1}{2}\). This is option (3).
Final Answer:
The correct option is (3): \(\frac{1}{2}\)
A regular hexagon of side s is made of 6 equilateral triangles of side s, so its area is \(\frac{3\sqrt{3}}{2}s^2\).
Step 2: Triangle ACE:
In triangle ABC, AB = BC = s and \(\angle ABC = 120^\circ\), so \(AC^2 = s^2 + s^2 - 2s^2\cos 120^\circ = 3s^2\), i.e. \(AC = s\sqrt{3}\). By symmetry, ACE is equilateral with side \(s\sqrt{3}\), so its area is \(\frac{\sqrt{3}}{4}\times 3s^2 = \frac{3\sqrt{3}}{4}s^2\).
Step 3: Ratio:
\(\frac{3\sqrt{3}/4}{3\sqrt{3}/2} = \frac{1}{2}\). This is option (3).
Final Answer:
The correct option is (3): \(\frac{1}{2}\)
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