A) 1 : 2
B) 3 : 1
C) 1 : 1
D) 4 : 3
Correct Answer: D
Solution :
\[2r=h+\sqrt{{{r}^{2}}+{{h}^{2}}}\] \[\Rightarrow \] \[2r-h=\sqrt{{{r}^{2}}+{{h}^{2}}}\] On squaring, \[4{{r}^{2}}+{{h}^{2}}-4rh={{r}^{2}}+{{h}^{2}}\] \[\Rightarrow \]\[4{{r}^{2}}-{{r}^{2}}=4rh\] \[\Rightarrow \]\[3{{r}^{2}}=4rh\Rightarrow 3r=4h\] \[\Rightarrow \]\[\frac{r}{h}=\frac{4}{3}\]
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