| There are two circles of radii \[{{r}_{1}}\]and \[{{r}_{2}}({{r}_{1}}<{{r}_{2}}).\] The area of the bigger circle is \[\frac{693}{2}c{{m}^{2}}.\]The difference of their circumferences is \[22\,\,cm.\]What is the sum of the diameters of the two circles? |
A) 17.5 cm
B) 22 cm
C) 28.5 cm
D) 35 cm
Correct Answer: C
Solution :
| According to the question, |
| \[\pi r_{2}^{2}=\frac{693}{2}\]\[\Rightarrow \]\[r_{2}^{2}=\frac{693}{2}\times \frac{7}{22}\] |
| \[\Rightarrow \]\[{{r}_{2}}=\sqrt{\frac{693\times 7}{2\times 22}}\] |
| \[\therefore \]\[{{d}_{2}}=2\sqrt{\frac{693\times 7}{2\times 22}}=21\,\,cm\] |
| Again, according to the question, |
| \[2\pi {{r}_{2}}-2\pi {{r}_{1}}=22\] |
| \[\Rightarrow \] \[\pi \times 21-2\pi {{r}_{1}}=22\] |
| \[\Rightarrow \] \[\frac{22}{7}\times 21-2\pi {{r}_{1}}=22\] |
| \[\Rightarrow \] \[66-2\pi {{r}_{1}}=22\] |
| \[\Rightarrow \]\[2\pi {{r}_{1}}=44\]\[\Rightarrow \]\[2\times \frac{22}{7}\times {{r}_{1}}=44\] |
| \[{{r}_{1}}=7\,\,cm\] |
| \[\therefore \] \[{{d}_{1}}=14\,\,cm\] |
| \[\therefore \] \[{{d}_{1}}+{{d}_{2}}=14+21=35\,\,cm\] |
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