[tex]\huge\underline\color{skyblue}{Answer ☘}[/tex]
( Note~ refer the attachment for naming of certain points )
Now given that ,
AB = 12 cm
BC = 8 cm
We know that , angle in a semicircle = 90°
[tex]\purple{Therefore \:, triangle\: ABC\: is \:a \:right \:- \:angled\: traingle.}[/tex]
Now in right angled triangle ABC ,
[tex]\purple{by \:Pythagoras \:theorem}[/tex]
[tex]H {}^{2} = P {}^{2} + B {}^{2} \\ AC {}^{2} = AB {}^{2} + BC {}^{2} \\ AC {}^{2} = 12 {}^{2} + 8 {}^{2} \\ AC {}^{2} = 144 + 64 \\ AC {}^{2} = 208 \\ AC = \sqrt{208} = 14.42cm (approx.)[/tex]
Now , AC is the diameter of the circle and also the hypotenuse of the given right angled triangle.
And ,
[tex]\pink{AC = diameter\: of \:circle = 14.42 cm}[/tex]
[tex] \blue{Therefore \: , radius = \frac{diameter}{2} }\\[/tex]
[tex]\pink{radius = 7.21 cm}[/tex] ( approx. )
hope helpful~
