A) 4 cm and 9 cm
B) 5 cm and 10 cm
C) 4 cm and 8 cm
D) 4 cm and 10 cm
Correct Answer: D
Solution :
Let the radius of outer circle = R and radius of inner circle = r \[\therefore \] \[\pi {{R}^{2}}+\pi {{r}^{2}}=116\pi \] \[{{R}^{2}}+{{r}^{2}}=116\] ?(i) If O and O' be the centre of these circles, then\[OO'=(R-r)\] Also \[(R-r)=6\] (given) So, \[{{(R+r)}^{2}}+{{(R-r)}^{2}}=2\,\,({{R}^{2}}+{{r}^{2}})\] \[\Rightarrow \]\[{{(R+r)}^{2}}=2\,\,(116)-36=196\]\[\Rightarrow \]\[R+r=\sqrt{196}\] So,\[R+r=14\] ?(i) \[R-r=6\] ?(ii) From Eqs. (i) and (ii), R = 10, r = 4You need to login to perform this action.
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