The sides of $\triangle ABC$ are 10 cm, 10.5 cm and 14.5 cm. What is the radius of its circumcircle?

7.25 cm

As we know,

\( {(14.5) }^{2 } \) = \( {(10.5) }^{2 } \) + \( {(10) }^{2 } \)

210.25 = 110.25 + 100

210.25 = 210.25

Now, we can sat triangle is a right angled triangle, then

Circum radius of triangle = \(\frac{Hypotenuse}{2}\) = \(\frac{14.5}{2}\) = 7.25 cm.