In AB = 5 cm, AC = 12 cm and \[AB\bot AC,\] then the radius of the circumcircle of \[\Delta ABC\] is [SSC (CPO) 2014] |
A) 6.5 cm
B) 6 cm
C) 5 cm
D) 7 cm
Correct Answer: A
Solution :
Given, \[\Delta ABC\] with \[AB=5\,\,cm,\] \[AC=12\,\,cm\] |
and \[AB\bot AC\] |
Since, \[\angle BAC=90{}^\circ \] |
So, BC is the diameter of the circle,[by property of semi-circle] |
\[{{\text{(Hypotentuse)}}^{\text{2}}}\text{+(Perpendicular}{{\text{)}}^{\text{2}}}\text{+(Base}{{\text{)}}^{\text{2}}}\] |
We have, \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] |
\[\Rightarrow \]\[B{{C}^{2}}={{5}^{2}}+{{12}^{2}}=25+144=169\] |
\[\therefore \] \[BC=13\,\,cm\] |
\[\Rightarrow \]Radius \[=\frac{BC}{2}=6.5\,cm\] |
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