A) \[b=\sqrt{2}a\]
B) \[a=\sqrt{2}b\]
C) \[a=2b\]
D) \[b=2a\]
Correct Answer: A
Solution :
[a] In \[\Delta AOB,\] AO = BO = r [radius of circle] \[{{b}^{2}}={{r}^{2}}+{{r}^{2}}\] \[b=\sqrt{2{{r}^{2}}}\] \[b=\sqrt{2}\,r\] (i) In \[\Delta COD,\] \[\angle COD=60{}^\circ \] Then, \[\angle OCD+\angle ODC=180-\angle COD\] \[=180-60=120{}^\circ \] Also \[\angle OCD=\angle OCD=\]angle opposite to equal sides \[\therefore \] \[\angle OCD=\angle ODC=60{}^\circ \] So, \[\Delta COD\] is equilateral and \[r=a\] ...(ii) From Eqs. (i) and (ii), \[b=r\sqrt{2}\,a\] |
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