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, $\Big(\frac{1}{3}\text{ pl }\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }7\text{r}^2\text{h}\Big):\Big(\frac{1}{3}\text{ pl }19\text{r}^2\text{h}\Big)$
$= 1 : 7 : 19$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Prove the following identities:
$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\big(\cos^2\theta-\sin^2\theta\big)$
→$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\big(\cos^2\theta-\sin^2\theta\big)$
Find the area of $\triangle ABC$ with vertices $A (0,-1), B (2,1)$ and $C (0,3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4: 1$
→Find the term of the arithmetic progression $9, 12, 15, 18, .....$ which is $39$ more than its $36^{th}$ term.
→Prove that $\big(5-2\sqrt3\big)$ is an irrational number.
→If at some time of the day the ratio of the height of a vertically standing pole to the length of its shadow on the ground is $\sqrt{3}:1$ then find the angle of elevation of the sun at that times.
→Find all zeros of the polynomial $2 x^4+7 x^3-19 x^2-14 x+30$, if two of its zeros are $\sqrt{2}$ and $\sqrt{2}.$
→In the given figure, $O$ is the centre of a circle $PT$ and $PQ$ are tangents to the circle from an external point P. If $\text{TPQ} = 70^\circ$ then$ \angle\text{TRQ} .$
→Find the perimeter of the $s$ ded region in the figure, if $A B C D$ is a square of side $14 \ cm$ and $A P B$ and $C P D$ are
→From a solid cylinder of height $2.8\ cm$ and diameter $4.2\ cm$, conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
→In tha given fingure, $ABCD$ is a trapezium of area $24.5cm^2$. $24.5cm^2$ If $\text{AD}||\text{BC},\ \angle\text{DAB}=90^\circ,\ \text{AD}=10\text{cm},\ \text{BC}=4\text{cm}$ and ABE is quadrant of a circle then find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→