a/Kẻ \(OH\perp AC,OK\perp BC.\)
Ta có :\(OA=OB,OH\text{//}BC\)
\(\Rightarrow OH\) là đường trung bình.
\(\Rightarrow BC=2OH=12\left(cm\right)\)
Ttự, ta có:\(AC=2OK=2.8=16\left(cm\right)\)
Ta có:\(AB=\sqrt{AC^2+BC^2}\)
\(=\sqrt{16^2+12^2}=20\left(cm\right).\)