what is the area of a regular hexagon with the apothem 5 root 3?

Answer 1

You can break your hexagon down into six equilateral triangles. 5 root 3 would be the height and 10 would be the base of each triangle. (This comes from the ratios of 2 : 1 : sqrt(3)) for a 30-60-90 triangle, or 10 : 5 : 5 sqrt(3) for your case. So the sides are 10).

b = 10
h = 5 sqrt(3)

Putting it together:
A = 1/2(b*h) * 6
A = 1/2(10*5 sqrt(3)) *6
A = 150 sqrt(3)
A ≈ 259.81 sq. units

Answer 2

A regular hexagon breaks down nicely into 6 equilateral triangles that all meet at the center.

The apothem a, is the height of the triangle.

For an equilateral triangle, the base b = (2/√3)a.

a = 5√3
b = (2/√3)a = (2/√3)(5√3) = 10

So the area K of the hexagon is:

K = 6(½ab) = 3ab = 3(5√3)(10) = 150√3 ≈ 259.80762

Answer 3

Divide the hexagon into 6 equilateral triangles. One side of an equilateral triangle is 5\/3. The area of an equilateral triangle is \/3 a^2 all over 4 where 'a' is a side. \/3(5\/3)(5\/3) / 4 = 75\/3 all over 4. Multiply 75\/3 / 4 by 6 because there are 6 triangles. The answer is 225\/3 / 2

Answer 4

You treat the hexagon as 12 similar triangles.
The area of each triangle is tan((360/(6*2)))*5sqr3* 5sqr3*.5 = 21.6506 approximately.

So, total area is 21.6506*12 = 259.8076 approxiamtely.

Answer 5

259.8