Question

The radius of the circle passing through the vertices of the triangle ABC, is

• A. $\frac{8\sqrt{15}}{5}$
• B. $\frac{3\sqrt{15}}{5}$
• C. $3\sqrt{5}$
• D. $3\sqrt{2}$

Answer 1

Answer: (C)

$3\sqrt{5}$

Answer 2

The question asks for the radius of the circumcircle of triangle ABC. Assuming the triangle ABC is a right-angled triangle at C, with sides AC = 6 and BC = 12, we can find the hypotenuse AB using the Pythagorean theorem: $AB^2 = AC^2 + BC^2$ $AB^2 = 6^2 + 12^2$ $AB^2 = 36 + 144$ $AB^2 = 180$ $AB = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$

For a right-angled triangle, the radius of the circumcircle is half the length of the hypotenuse. $R = \frac{AB}{2} = \frac{6\sqrt{5}}{2} = 3\sqrt{5}$