A) 6 cm
B) 7 cm
C) 10 cm
D) 8 cm
Correct Answer: D
Solution :
[d] Let the chord of greater circle which touches the smaller. Circle be AB at M. Given OM = 3 cm, OA = 5 cm In \[\Delta AMO,\]\[{{(OA)}^{2}}={{(AM)}^{2}}+{{(OM)}^{2}}\] \[\Rightarrow \] \[{{(5)}^{2}}={{(AM)}^{2}}+{{(3)}^{2}}\] \[\Rightarrow \] \[25={{(AM)}^{2}}+9\] \[\Rightarrow \] \[{{(AM)}^{2}}=16\]\[\Rightarrow \]\[AM=4\] \[\Rightarrow \] \[AB=2\times AM\] \[\Rightarrow \] \[AB=2\times 4\,cm\]\[\Rightarrow \]\[AB=8\,cm\] Hence, the required length of chord is 8 cm. |
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