Question

Area of a regular octagon inscribed in a circle of radius 1 unit is

• A. \( 2\sqrt{2} \)
• B. \( \frac{9}{2\sqrt{2}} \)
• C. \( \sqrt{10} \)
• D. \( 2 + \sqrt{2} \)

Hint:

The general formula for the area of a regular n-gon inscribed in a circle of radius r is \( \frac{1}{2}nr^2\sin(\frac{360^\circ}{n}) \). You can use this as a direct formula for such problems. For an octagon, n=8 and r=1, giving \( \frac{1}{2}(8)(1)^2\sin(45^\circ) = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A regular octagon inscribed in a circle can be divided into 8 congruent isosceles triangles, with their common vertex at the center of the circle. The area of the octagon is the sum of the areas of these triangles.
Step 2: Key Formula or Approach:
1. The angle at the center of the circle subtended by each side of the octagon is \( \frac{360^\circ}{8} = 45^\circ \). 2. The area of a triangle can be calculated using the formula \( \text{Area} = \frac{1}{2}ab\sin(C) \), where a and b are two sides and C is the included angle. 3. Total Area = 8 \( \times \) Area of one triangle.
Step 3: Detailed Explanation:
Let the circle have its center at the origin O and radius r = 1. The regular octagon can be divided into 8 identical isosceles triangles, such as \( \triangle OAB \), where A and B are adjacent vertices of the octagon. The sides OA and OB are radii of the circle, so \( OA = OB = r = 1 \). The angle between these two sides, \( \angle AOB \), is one-eighth of the total angle around the center: \[ \angle AOB = \frac{360^\circ}{8} = 45^\circ \] Now, we can find the area of one of these triangles using the formula \( \text{Area} = \frac{1}{2}ab\sin(C) \): \[ \text{Area of } \triangle OAB = \frac{1}{2} \times OA \times OB \times \sin(\angle AOB) \] \[ \text{Area of } \triangle OAB = \frac{1}{2} \times 1 \times 1 \times \sin(45^\circ) \] We know that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \). \[ \text{Area of } \triangle OAB = \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} \] The total area of the octagon is 8 times the area of one triangle. \[ \text{Area of Octagon} = 8 \times (\text{Area of } \triangle OAB) = 8 \times \frac{\sqrt{2}}{4} \] \[ \text{Area of Octagon} = 2\sqrt{2} \] Step 4: Final Answer:
The area of the regular octagon is \( 2\sqrt{2} \) square units.