question_answer

Two circles touch internally. The sum of there are is \[=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{1}{30}\] and distance between their circlet 6 cm. Then, the radii of the circles are

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 = 4