# Math Help - Geometry help, please

I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
Here's what I got:
The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
I'm I correct?

2. Originally Posted by JoanneMac
I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
Here's what I got:
The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
I'm I correct?

There must be a minor mistake (the factor $\sqrt{3}$ is missing)

The area of a regular hexagon consists of 6 equilateral triangles. The height in an equilateral triangle is (according to Pythagorean theorem)

$h = \frac12 s\cdot \sqrt{3}$

The area of the hexagon is:

$a = 6 \cdot \frac12 \cdot s \cdot \frac12 s\cdot \sqrt{3}$

With $s = 2\sqrt{3}$ you'll get:

$a= 6 \cdot \frac12 \cdot 2\sqrt{3} \cdot \frac12 \cdot 2\sqrt{3} \cdot \sqrt{3} = \boxed{18\sqrt{3}}$

3. Originally Posted by JoanneMac
I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
Here's what I got:
The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
I'm I correct?

$A=\frac{1}{2}ap$
The perimeter is $12\sqrt{3}$ and the apothem is 3.
$A=\frac{1}{2}(3)(12\sqrt{3})=18\sqrt{3}$