Question

O is the center of the circle and the perimeter of \(\Delta AOB\) is 6. The angle \(\angle AOB\) is \(60^{\circ}\).
Column A: The circumference of the circle
Column B: 12

• A. Quantity A is greater
• B. Quantity B is greater
• C. The two quantities are equal
• D. The relationship cannot be determined from the information given

Hint:

Recognizing that a 60° angle between two radii forms an equilateral triangle is a common shortcut in geometry problems. If the central angle of an isosceles triangle is 60°, the triangle must be equilateral.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires understanding the properties of circles and triangles. Specifically, we use the relationship between the center of a circle, its radii, and the properties of an equilateral triangle to find the circle's circumference.
Step 2: Key Formula or Approach:
The circumference of a circle is given by the formula \(C = 2\pi r\), where \(r\) is the radius.
The perimeter of a triangle is the sum of its three sides.
Step 3: Detailed Explanation:
In the given circle with center O, OA and OB are radii. Therefore, \(OA = OB = r\).
This means that \(\Delta AOB\) is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal. So, \(\angle OAB = \angle OBA\).
The sum of angles in a triangle is \(180^{\circ}\).
\[ \angle AOB + \angle OAB + \angle OBA = 180^{\circ} \] We are given \(\angle AOB = 60^{\circ}\).
\[ 60^{\circ} + \angle OAB + \angle OAB = 180^{\circ} \] \[ 2\angle OAB = 180^{\circ} - 60^{\circ} = 120^{\circ} \] \[ \angle OAB = 60^{\circ} \] Since all three angles of \(\Delta AOB\) are \(60^{\circ}\), it is an equilateral triangle.
In an equilateral triangle, all sides are equal. Therefore, \(OA = OB = AB = r\).
The perimeter of \(\Delta AOB\) is given as 6.
\[ \text{Perimeter} = OA + OB + AB = r + r + r = 3r \] \[ 3r = 6 \] \[ r = \frac{6}{3} = 2 \] Now we can calculate the circumference of the circle.
\[ C = 2\pi r = 2\pi(2) = 4\pi \] Step 4: Comparing the Quantities:
Column A: The circumference of the circle = \(4\pi\).
Column B: 12.
To compare \(4\pi\) and 12, we use the approximation \(\pi \approx 3.14159\).
\[ 4\pi \approx 4 \times 3.14159 = 12.56636 \] Since \(12.56636 \textgreater 12\), the quantity in Column A is greater.