Question:
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

1. 12√3
2. 8π
3. 18√2
4. 24√3
5. 48

Correct Answer: D
Solution and Explanation:

Approach Solution 1:
In a regular hexagon inscribed in a circle, its side is equal to the radius.
We can divide the hexagon in 6 triangles each with the base of 4. The height will equal √4^2−2^2= √12= 2√3. To obtain this just use Pythagoras, the hypotenuse of each triangle it's the radius, and the bases it's 4/2= 2.
Now we have the height of each triangle, so At= (4∗2√3)/2=4√3.
Ah= 6*At= 6*4√3= 24√3.

Approach Solution 2:
1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120.
2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides.
3) Find the area of one of the triangles:
- Base = 4
- find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3
- one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3
4) find the area of the hexagon by multiplying the one triangle's area by 6:
- 6 x 4√3 = 24√3

Approach Solution 3:
We may recall that a regular hexagon can be divided into 6 equilateral triangles, and thus the area of a regular hexagon is:
6[(s^2√3)/4], where s = side of the equilateral triangle and (s^2√3)/4 = the area of the equilateral triangle.
Since the radius of 4 also represents one side of the equilateral triangle, we can now determine the area of the hexagon.
Area = 6[(4^2√3)/4]
Area = 6[(16√3)/4]
Area = 6(4√3)
Area = 24√3

“If a regular hexagon is inscribed in a circle with a radius of 4, the”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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