Question

The vertices of an equilateral triangle are on a circle.
 

Column A	Column B
The length of a side of the triangle	The diameter of the circle

 

Hint:

Visualize the figure. The diameter is the longest possible straight line that can be drawn inside a circle. Since the side of the inscribed triangle does not pass through the center of the circle, its length must be less than the diameter. This intuitive check confirms the mathematical result.

Answer 1

Step 1: Understanding the Concept:
This is a geometry problem that compares a dimension of an inscribed equilateral triangle to a dimension of the circle that circumscribes it.
Step 2: Key Formula or Approach:
There is a specific relationship between the side length (\(s\)) of an inscribed equilateral triangle and the radius (\(r\)) of the circumscribing circle: \(s = r\sqrt{3}\). We need to compare this to the diameter (\(d\)), where \(d = 2r\).
Step 3: Detailed Explanation:
Let \(s\) be the length of a side of the triangle and \(d\) be the diameter of the circle. Let the radius of the circle be \(r\).
Column A is \(s = r\sqrt{3}\).
Column B is \(d = 2r\).
We are comparing \(r\sqrt{3}\) and \(2r\). Since the radius \(r\) must be a positive number, we can divide both quantities by \(r\) without changing the inequality.
This simplifies the comparison to \(\sqrt{3}\) versus 2.
We know that \((\sqrt{3})^2 = 3\) and \(2^2 = 4\).
Since \(3<4\), it follows that \(\sqrt{3}<2\).
Therefore, \(r\sqrt{3}<2r\), which means \(s<d\).
Step 4: Final Answer:
The length of the side of the triangle is less than the diameter of the circle. The quantity in Column B is greater.