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Volume and Surface area of Solids — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsVolume and Surface area of Solids5 Marks
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

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