Question
A right angled triangle with sides 3cm and 4cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.
✓
Answer
In right angled $\triangle\text{ABC}$ ,$ \angle\text{B}=90°$
AB = 3cm and BC = 4cm
$\therefore$ Diagonal CA $=\sqrt{\text{AB}^2+\text{BC}^2}$ (pythagoras theorem)
$=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\text{cm}$
We get double cones as shown in the figure BB' is joined which bisects by AC ar O at right angles
In $\triangle\text{ABC},\angle\text{B}=90^\circ$
$\therefore$ area $(\triangle\text{ABC})=\frac{3\times4}{2}=6\text{cm}^2$
and area $(\triangle\text{ABC})=\frac{1}{2}\text{AC}\times\text{BO}$
$\Rightarrow6=\frac{1}{2}\times5\times\text{BO}$
$\Rightarrow\text{BO}=\frac{6\times2}{5}=\frac{12}{5}=2.4\text{cm}$
$\therefore$ Radius of cone along AO is BO which iscm and Also along CO is BO which is 2.4cm Now volumes of two cones so formes
$=\frac{1}{3}\pi\text{r}^2\times\text{AO}+\frac{1}{3}\pi\text{r}^2\times\text{CO}$
$=\frac{1}{3}\pi\text{r}^2(\text{AO}+\text{CO})=\frac{1}{3}\pi\text{r}^2\times\text{AC}$
$=\frac{1}{3}\times\frac{22}{7}\times(2.4)^2\times5\text{cm}^3$
$=\frac{22}{21}\times5.76\times5\times\text{cm}^3=\frac{110\times1.92}{7}\text{cm}^2$
$=\frac{211.20}{7}\text{cm}^3$
$=\frac{211.20}{7}\text{cm}^3$
$=\frac{21120}{7\times100}=\frac{1056}{35}\text{cm}^3$
$=30\frac{6}{35}\text{cm}^3$
AB = 3cm and BC = 4cm
$\therefore$ Diagonal CA $=\sqrt{\text{AB}^2+\text{BC}^2}$ (pythagoras theorem)
$=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\text{cm}$
We get double cones as shown in the figure BB' is joined which bisects by AC ar O at right angles
In $\triangle\text{ABC},\angle\text{B}=90^\circ$
$\therefore$ area $(\triangle\text{ABC})=\frac{3\times4}{2}=6\text{cm}^2$
and area $(\triangle\text{ABC})=\frac{1}{2}\text{AC}\times\text{BO}$
$\Rightarrow6=\frac{1}{2}\times5\times\text{BO}$
$\Rightarrow\text{BO}=\frac{6\times2}{5}=\frac{12}{5}=2.4\text{cm}$
$\therefore$ Radius of cone along AO is BO which iscm and Also along CO is BO which is 2.4cm Now volumes of two cones so formes
$=\frac{1}{3}\pi\text{r}^2\times\text{AO}+\frac{1}{3}\pi\text{r}^2\times\text{CO}$
$=\frac{1}{3}\pi\text{r}^2(\text{AO}+\text{CO})=\frac{1}{3}\pi\text{r}^2\times\text{AC}$
$=\frac{1}{3}\times\frac{22}{7}\times(2.4)^2\times5\text{cm}^3$
$=\frac{22}{21}\times5.76\times5\times\text{cm}^3=\frac{110\times1.92}{7}\text{cm}^2$
$=\frac{211.20}{7}\text{cm}^3$
$=\frac{211.20}{7}\text{cm}^3$
$=\frac{21120}{7\times100}=\frac{1056}{35}\text{cm}^3$
$=30\frac{6}{35}\text{cm}^3$
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