There is a formula $R = \dfrac{abc}{4\Delta}$, where $a$, $b$, and $c$ are the triangle's side lengths and $\Delta$ is its area. Focus next on finding the area of the triangle from the three side lengths.

Use Heron's formula: first compute the semiperimeter $s = \dfrac{a+b+c}{2}$, then compute $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ for $a=13$, $b=14$, and $c=15$.

After you find the area, substitute $a=13$, $b=14$, $c=15$, and your area value into $R = \dfrac{abc}{4\Delta}$, then simplify the resulting fraction carefully to match one of the answer choices.

Type s = (13 + 14 + 15)/2 and note the value of s that Desmos shows. This is the semiperimeter of the triangle.

Enter A = sqrt(s*(s-13)*(s-14)*(s-15)). Desmos will display the area of the triangle as the value of A.

Now type R = 13*14*15/(4*A). The value Desmos gives for R is the circumradius. Compare this decimal value to the answer choices by converting each fraction choice to a decimal and matching it.

Triangle $ABC$ is inscribed in a circle, so the circle is the circumcircle of the triangle. For a triangle with side lengths $a$, $b$, $c$ and area $\Delta$, the radius $R$ of its circumcircle is

Here, $a=13$, $b=14$, $c=15$, so once we find the area $\Delta$, we can compute $R$.

To find the area using Heron's formula, first compute the semiperimeter $s$:

Heron's formula for the area $\Delta$ of a triangle with sides $a,b,c$ and semiperimeter $s$ is

Substitute $s=21$, $a=13$, $b=14$, $c=15$:

So the area of triangle $ABC$ is $84$ square units.

Now use $a=13$, $b=14$, $c=15$, and $\Delta=84$ in

We get

This is the radius, but we should simplify this fraction.

Simplify $\dfrac{2730}{336}$ step by step:

• Divide numerator and denominator by $6$:

• Now divide numerator and denominator by $7$:

So the radius of the circle is $\dfrac{65}{8}$, which corresponds to choice A.