A) 10 cm
B) 17 cm
C) 24 cm
D) None of these
Correct Answer: B
Solution :
From O draw \[OL\bot AB\]and \[OM\bot CD\] join OA and OC |
\[AL=\frac{1}{2},\]\[AB=5\,\,cm,\]\[OA=13\,\,cm\] |
\[O{{L}^{2}}=O{{A}^{2}}-A{{L}^{2}}={{(13)}^{2}}-{{(5)}^{2}}\] |
\[=(169-25)=144\] |
\[\Rightarrow \]\[OL=\sqrt{144}=12\,\,cm\] |
Now, \[CM=\frac{1}{2}\times CD=12\,\,cm\] and \[OC=13\,\,cm\] |
\[\therefore \] \[O{{M}^{2}}=O{{C}^{2}}-C{{M}^{2}}={{(13)}^{2}}-{{(12)}^{2}}\] |
\[=(169-144)=25\] |
\[\Rightarrow \] \[OM=\sqrt{25}=5\,\,cm\] |
\[\Rightarrow \] \[ML=OM+OL=(5+12)\,\,cm=17\,\,cm\] |
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