A) \[\text{4 cm},\text{1}0\text{ cm}\]
B) \[\text{2 cm},6\text{ cm}\]
C) \[\text{3 cm},5\text{ cm}\]
D) \[\text{5 cm},8\text{ cm}\]
Correct Answer: A
Solution :
| [a] Let \[{{r}_{1}}\] be the radius of the bigger circle and \[{{r}_{2}}\] be the radius of the smaller circle. |
| It is given that the two circles touch each other internally. |
| \[\therefore \] Difference between their radii |
| = Distance between the centres of the two circles. |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{r}_{1}}-{{r}_{2}}=6\,cm\] ...(1) \[[{{r}_{1}}>{{r}_{2}}]\] |
| Also, sum of their areas \[=116\pi \,c{{m}^{2}}\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\pi r_{1}^{2}+\pi r_{2}^{2}=116\pi \] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,r_{1}^{2}+r_{2}^{2}=116\] .(2) |
| From eq. (1) and (2), we get |
| \[{{({{r}_{2}}+6)}^{2}}+r_{2}^{2}=116\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r_{2}^{2}+36+12{{r}_{2}}+r_{2}^{2}=116\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2r_{2}^{2}+12{{r}_{2}}-80=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,r_{2}^{2}+6{{r}_{2}}-40=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{r}_{2}}+10{{r}_{2}}-4{{r}_{2}}-40=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{r}_{2}}({{r}_{2}}+10)-4({{r}_{2}}+10)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,({{r}_{2}}+10)\,\,({{r}_{2}}-4)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{r}_{2}}=-10,4\] |
| Since, the radius of a circle cannot be negative. |
| So, \[{{r}_{2}}=4cm\] \[[{{r}_{2}}\ne \,\,-10]\] |
| \[\therefore \,\,\,\,\,\,{{r}_{1}}=6+{{r}_{2}}=6+4=10\,cm\] |
| Thus, the radii of the circles are \[4\,cm\] and \[10\,cm\]. |
You need to login to perform this action.
You will be redirected in
3 sec
