A right angled sector of radius \[r\,\,cm\]is rolled up into a cone in such a way that two binding radii are joined together. Then, the curved surface area of the cone is [SSC (10+2) 2011] |
A) \[\pi {{r}^{2}}c{{m}^{2}}\]
B) \[4\pi {{r}^{2}}c{{m}^{2}}\]
C) \[\frac{\pi {{r}^{2}}}{4}c{{m}^{2}}\]
D) \[2\pi {{r}^{2}}c{{m}^{2}}\]
Correct Answer: C
Solution :
Length of arc, \[x=\frac{\theta }{360{}^\circ }\times 2\pi r\] |
\[x=\frac{\theta }{360{}^\circ }\times 2\pi r\] |
\[\Rightarrow \] \[x=\frac{90{}^\circ }{360{}^\circ }\times 2\pi r=\frac{\pi r}{2}\] |
Now, length of arc will be circumference of base and let radius of base be \[{{r}_{1}}.\] |
\[\therefore \] \[2\pi {{r}_{1}}=x,\]\[2\pi {{r}_{1}}=\frac{\pi r}{2}\] |
\[{{r}_{1}}=\frac{r}{4}\]and slant height \[=r\] |
\[\therefore \]Curved surface area\[=\pi {{r}_{1}}l=\pi \times \frac{r}{4}\times r=\frac{\pi {{r}^{2}}}{4}c{{m}^{2}}\] |
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