Question
A right circular cone is divived into three parts by trisecting its height by two planes drawn parallel to the base. show that the volumes of the three portions starting from top are in ratio $1 : 7 : 19.$
✓
Answer
The height of a Cone, 3h is trisected by $2$ planes to the base of the cone at equal distances.
So, the cone is divided into a smaller cone & $2$ frustums of the cone. The height of each piece is 'h' unit.
Since, right triangle $ABG$ ~ tri $ACF$ ~ tri $ADE$ ( by $AAA$ similarity criterion. So corresponding sides are to be proportional.
So, $\frac{\text{AB}}{\text{AC}}=\frac{\text{h}}{2\text{h}}=\frac{1}{2}=\frac{\text{BG}}{\text{CF}}=\frac{\text{r}}{2\text{r}}$
$\frac{\text{AB}}{\text{AD}}=\frac{\text{h}}{3\text{h}}=\frac{1}{3}=\frac{\text{BG}}{\text{DF}}=\frac{\text{r}}{3\text{r}}$
Now, we find the volume of each piece.. a smaller cone & 2 frustums
Volume of Cone $\text{ABG}=\frac{1}{3}\text{pl }\text{r}^2\text{ h}\ ....(1)$
Volume of middle frustum =
$\frac{1}{3}\text{ pl}\text{ (r}^2+4\text{r}^2+2\text{r}^2)\text{h}$
$=\frac{1}{3}\text{pl }7\text{r}^2\text{ h}\ ....(2)$
Volume of next frustum =
$=\frac{1}{3}\text{ pl}\ (4\text{r}^2+9\text{r}^2+6\text{r}^2)\text{h}$
$=\frac{1}{3}\text{ pl }19\text{r}^2\text{h}\ ....(3)$
Now, by finding the ratio of (1),(2)&(3)
we get, $\Big(\frac{1}{3}\text{ pl }\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }7\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }19\text{r}^2\text{h}\Big)$
$= 1 : 7 : 19$
So, the cone is divided into a smaller cone & $2$ frustums of the cone. The height of each piece is 'h' unit.
Since, right triangle $ABG$ ~ tri $ACF$ ~ tri $ADE$ ( by $AAA$ similarity criterion. So corresponding sides are to be proportional.
So, $\frac{\text{AB}}{\text{AC}}=\frac{\text{h}}{2\text{h}}=\frac{1}{2}=\frac{\text{BG}}{\text{CF}}=\frac{\text{r}}{2\text{r}}$
$\frac{\text{AB}}{\text{AD}}=\frac{\text{h}}{3\text{h}}=\frac{1}{3}=\frac{\text{BG}}{\text{DF}}=\frac{\text{r}}{3\text{r}}$
Now, we find the volume of each piece.. a smaller cone & 2 frustums
Volume of Cone $\text{ABG}=\frac{1}{3}\text{pl }\text{r}^2\text{ h}\ ....(1)$
Volume of middle frustum =
$\frac{1}{3}\text{ pl}\text{ (r}^2+4\text{r}^2+2\text{r}^2)\text{h}$
$=\frac{1}{3}\text{pl }7\text{r}^2\text{ h}\ ....(2)$
Volume of next frustum =
$=\frac{1}{3}\text{ pl}\ (4\text{r}^2+9\text{r}^2+6\text{r}^2)\text{h}$
$=\frac{1}{3}\text{ pl }19\text{r}^2\text{h}\ ....(3)$
Now, by finding the ratio of (1),(2)&(3)
we get, $\Big(\frac{1}{3}\text{ pl }\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }7\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }19\text{r}^2\text{h}\Big)$
$= 1 : 7 : 19$
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