Question
A right triangle whose sides are $15\ cm$ and $20\ cm$ (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate)
✓
Answer
We have,
In $\Delta\text{ABC},\ \angle\text{B}=90^\circ,\ \text{AB}=\text{l}_1=15\text{cm}$ and $\text{BC}=\text{l}_2=20\text{cm}$
Let $\text{OD}=\text{OB}=\text{r},\ \text{AO}=\text{h}_1$ and $\text{CO}=\text{h}_2$
Using Pythagoras theorem,
$\text{AC}=\sqrt{\text{AB}^2+\text{BC}^2}$
$=\sqrt{15^2+20^2}$
$=\sqrt{225+400}$
$=\sqrt{625}$
$\Rightarrow\text{h}=25\text{cm}$
As, ar $(\Delta\text{ABC})=\frac{1}{2}\times\text{AC}\times\text{BO}=\frac{1}{2}\times\text{AB}\times\text{BC}$
$\Rightarrow\text{AC}\times\text{BO}=\text{AB}\times\text{BC}$
$\Rightarrow25\text{r}=15\times20$
$\Rightarrow\text{r}=\frac{15\times20}{25}$
$\Rightarrow\text{r}=12\text{cm}$
Now,
Volume of the double cone so formed = Volume of cone $1 +$ Volume of cone $2$
$=\frac{1}{3}\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\frac{1}{3}\pi\text{r}^2(\text{h}_1+\text{h}_2)$
$=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\times3.14\times12\times12\times25$
$=3768\text{cm}^3$
Also,
Surace area of the solid so formed $= CAS$ of cone $1 + CSA$ of cone $2$
$=\pi\text{rl}_1+\pi\text{rl}_2$
$=\pi\text{r}(\text{l}_1+\text{l}_2)$
$=\frac{22}{7}\times12\times(15+20)$
$=\frac{22}{7}\times12\times35$
$=1320\text{cm}^2$
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