+1234 Points $A$, $B$, and $C$ are on a circle such that $AB = 8$, $BC = 15$, and $AC = 12$. Find the radius of the circle.
+14 According to the Triangle Inequality, ABC is indeed a triangle, since 8 + 12 > 15.
Thus, the radius of the circle is the same as the circumradius of ABC, which can be found by:
\(R = \frac {abc} {4S}\)
Where S is the area of the triangle.
Using Heron's formula, we get:
\(S = \sqrt { \frac {35} {2} \left( \frac {35} {2} - 8 \right) \left( \frac {35} {2} - 15 \right) \left( \frac {35} {2} - 12 \right)}\)
\(S = \sqrt {\frac {35} {2} \left( \frac {19} {2} \right)\left( \frac {5} {2} \right)\left( \frac {11} {2} \right)}\)
\(S = \frac {5} {4} \sqrt {1463}\)
Therefore,
\(R = \frac {2520} {5\sqrt{1463}}\)
\(R = \frac {504\sqrt{1463}} {1463}\)
+14 According to the Triangle Inequality, ABC is indeed a triangle, since 8 + 12 > 15.
Thus, the radius of the circle is the same as the circumradius of ABC, which can be found by:
\(R = \frac {abc} {4S}\)
Where S is the area of the triangle.
Using Heron's formula, we get:
\(S = \sqrt { \frac {35} {2} \left( \frac {35} {2} - 8 \right) \left( \frac {35} {2} - 15 \right) \left( \frac {35} {2} - 12 \right)}\)
\(S = \sqrt {\frac {35} {2} \left( \frac {19} {2} \right)\left( \frac {5} {2} \right)\left( \frac {11} {2} \right)}\)
\(S = \frac {5} {4} \sqrt {1463}\)
Therefore,
\(R = \frac {2520} {5\sqrt{1463}}\)
\(R = \frac {504\sqrt{1463}} {1463}\)
