A regular hexagon is inscribed in a circle of radius 6 cm. Find its area enclosed by the hexagon:
A18√3
B36√3
C54√3
D72√3
Answer:
C. 54√3
Read Explanation:
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