Question

A regular hexagon is inscribed in a circle of radius 6 cm. Find its area enclosed by the hexagon:

• A. 18√3
• B. 36√3
• C. 54√3
• D. 72√3

Answer 1

Answer:

C. 54√3

We know that the angle subtended by the sides of a regular polygon (in this case hexagon) is equal to 2π/n

where n is the number of sides of the regular polygon.

In this case, n=6

Thus, the angle AOB = 2π/6 = 60

Now, we know that in triangle AOB, OA=OB, since both are radii of the same circle. So, we can say that triangle OAB is an equilateral triangle.

Thus, the side AB is of length 6 units.

The formula for area of an equilateral triangle is √3/4 × (side)²

Since the length of the side of regular hexagon is 6 cm, so the area of the equilateral triangle AOB is

√3/4 × (side)² = √3/4 × (6)²

= √3/4 × 36

= 9√3

Since there are six such equilateral triangles, the area of the regular hexagon is

6 × 9√3

= 54√3 sq cm