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# Geometry

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Points $A$, $B$, and $C$ are on a circle such that $AB = 8$, $BC = 15$, and $AC = 12$. Find the radius of the circle.

Mar 28, 2024

#1
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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,

$$R = \frac {2520} {5\sqrt{1463}}$$

$$R = \frac {504\sqrt{1463}} {1463}$$

Mar 28, 2024

#1
+14
+1

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,

$$R = \frac {2520} {5\sqrt{1463}}$$

$$R = \frac {504\sqrt{1463}} {1463}$$

WillBill Mar 28, 2024