Two circles touch externally. The sum of their areas is \[130\pi \,c{{m}^{2}}\] and the distance between their centres is 14 cm. The radius of the smaller circle is [SSC (CGL) Pre 2015] |
A) 2 cm
B) 3 cm
C) 4 cm
D) 5 cm
Correct Answer: B
Solution :
Here, |
Now, \[\pi \,{{(14-R)}^{2}}+\pi \,{{R}^{2}}=130\pi \] |
\[196+{{R}^{2}}-28R+{{R}^{2}}=130\] |
\[\Rightarrow \] \[2{{R}^{2}}-28R+66=0\] |
\[\Rightarrow \] \[{{R}^{2}}-14R+33=0\] |
\[\Rightarrow \] \[{{R}^{2}}-11R-3R+33=0\] |
\[\Rightarrow \] \[R\,(R-11)-3\,(R-11)=0\]\[\Rightarrow \]\[R=11\] |
\[\therefore \] Radius of smaller circle \[=14-11=3\,cm\] |
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