Question

The radius of the circle passing through the vertices of triangle ABC is
AB=12,AC=12 and BC=6

Answer 1

Please see the attachment

Answer 2

Answer: The radius is $\dfrac{8}{5} \sqrt{15} unit$

Step-by-step explanation:

In Δ ABC      AB= 12,   AC= 12,   BC= 6

By Herons Formula

$s=\dfrac{a+b+c}{2} \\\\\text {ar of } \triangle ABC } =\sqrt{s(s-a)(s-b)(s-c)}$

$s= \dfrac{12+12+6}{2} = 15\\\\\text {ar of } \triangle ABC =\sqrt{15(15-12)(15-12)(15-6)} = 9\sqrt{15} \text{ sq unit }$

Now The radius of the circle passing through the vertices is given by

$r=\dfrac{AB\times AC \times BC}{4\times\text {ar of} \triangle ABC} \\\\=\dfrac{12\times 12\times 6 }{4\times 9\sqrt{15} } \\\\=\dfrac{24}{\sqrt{15} } = \dfrac{8}{5} \sqrt{15} unit$

Hence, the radius is $\dfrac{8}{5} \sqrt{15} unit$