Sides of a triangle are 9 cm, 6 cm and 5 cm. What is the value of circumradius of this triangle ?

$\frac{27\sqrt{2}}{8}$ cm

Let be assume a, b and c is the sides of the triangle.

= Area of triangle = \(\sqrt {s(s\;-\;a)(s\;-\;b)(s\;-\;c) }\)

= s = \(\frac{a\;+\;b\;+\;c}{2}\) = \(\frac{20}{2}\) = 10 cm

= \(\sqrt {10(10\;-\;9)(10\;-\;6)(10\;-\;5) }\)

= A = \(\sqrt {200 }\)

= A = 10\(\sqrt {2 }\)

= R = \(\frac{abc}{4A}\)

= R = \(\frac{9 \;×\; 6\;×\;5 }{4\;×\;10√2}\)

= R = \(\frac{270}{40√2}\)

= R = \(\frac{27√2}{8}\) cm

Therefore, R is \(\frac{27√2}{8}\) cm.