Answer:
Let \[BC=r\text{ }cm\] & \[DE=R\text{ }cm\] Since B is mid-point of AD & \[BC\parallel DE\] \[\therefore \] C is mid-point of AE or \[AC=CE\] Also \[\Delta \,ABC\sim \Delta \,ADE\] \[\therefore \] \[\frac{AB}{AD}=\frac{BC}{DE}=\frac{AC}{AE}=\frac{1}{2}\] \[BC=\frac{1}{2}DE\] \[r=\frac{1}{2}R\] or \[R=2r\] Now, \[\frac{Volume\,\,of\,\,cone}{Volume\,\,of\,\,frustum}=\frac{\frac{1}{3}\pi {{r}^{2}}\left( \frac{h}{2} \right)}{\frac{1}{3}\pi \left( \frac{h}{2} \right)\left( {{R}^{2}}+{{r}^{2}}+Rr \right)}\] \[=\frac{{{r}^{2}}}{\left( {{R}^{2}}+{{r}^{2}}+Rr \right)}=\frac{{{r}^{2}}}{4{{r}^{2}}+{{r}^{2}}+2r\cdot r}\] \[=\frac{{{r}^{2}}}{7{{r}^{2}}}=\frac{1}{7}\] or \[1:7\]
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