Question
A right triangle whose sides are 15cm and 20cm (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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