Find circumradius of an equilateral triangle of side 7$\text{cm}$ 
I know that each length is 7 cm but how would I use that to work out the radius.
Thank you and your help is appreciated.
 A: 
As shown above, let $O$ be the center of the circle, $r$ be the circle radius and $D$ be the midpoint of $AB$. Draw lines from $O$ to $A$, $B$ and $D$. Note $OA$ bisects $\angle BAC$, so $\angle OAB = 30^{\circ}$. Also, Since $OA = OB = r$, $\triangle AOB$ is an isosceles triangle, so $DO$ is perpendicular to $AB$, which means $\angle ODA = 90^{\circ}$. By the sum of angles in a triangle being $180^{\circ}$, this means $\angle AOD = 60^{\circ}$.
Note the length of $AD$ is half that of $AB$, so it's $\frac{7}{2}$ cm. Also, since $30^{\circ}$- $60^{\circ}$- $90^{\circ}$ triangles have side lengths in a ratio of $1:\sqrt{3}:2$ (e.g., see the 30°–60°–90° triangle section of Wikipedia's "Special right triangle" article), this means
$$\frac{AO}{AD} = \frac{2}{\sqrt{3}} \implies r = (AD)\frac{2}{\sqrt{3}} = \left(\frac{7}{2}\right)\left(\frac{2}{\sqrt{3}}\right) = \frac{7}{\sqrt{3}} \text{ cm} \tag{1}\label{eq1A}$$
A: Use the triangle sine rule,
$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2r}$$
where $r$ is the circumradius. For the equilateral triangle of side 7,
$$r = \frac{7}{2\sin60} = \frac{7}{\sqrt3}$$
