Sides of a triangle are 12 cm, 9 cm and 9 cm. What is the radius of the circumcircle of this triangle ?

$ (27\sqrt{5})/10$

Area of triangle = \(\sqrt {[s(s\; - \;a)(s \;- \;b)(s\; - \;c)] }\) where s = \(\frac{a\;+\;b\;+\;c}{2}\)

Side of triangle are, a = 12 cm, b = 12 cm and c = 9 cm, then

s = \(\frac{(12\; + \;9 \;+ \;9)}{2}\) = \(\frac{(30)}{2}\) =  15 cm

Area of the triangle = \(\sqrt {[15(15\; - \;12)(15 \;- \;9)(15\; - \;9)] }\) = 18\(\sqrt {5 }\) \( {cm }^{2 } \)

As we know,

Circumference - radius = \(\frac{abc}{4\; ×\; area of triangle}\) = \(\frac{12 \;×\; 9\;×\;9}{4\;×\;18√5}\) = \(\frac{27√5}{10}\)

Therefore, answer is \(\frac{27√5}{10}\)