Question
A right angled triangle whose sides are $3\ cm, 4\ cm$ and $5\ cm$ is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed. Also, find their curved surfaces.
✓
Answer
We consider the following figure as follows
Let the angle B is right angle and the sides of the triangle are $AB = 4cm, BC = 3cm,$
AC = 5cm.
When the triangle is revolved about the side $AB$, then the base-radius, height and slant height of the produced cone becomes $BC, AB$ and $AC$ respectively. Therefore, the volume of the produced cone is
$\text{V}_1=\frac{1}{3}\pi\times\text{BC}^2\times\text{AB}$
$=\frac{1}{3}\pi\times(3)^2\times4$
$=12\pi$ cubic cm
In this case, the curved surface area of the cone is
$\text{S}_1=\pi\times\text{BC}\times\text{AC}$
$=\pi\times3\times5$
$=15\pi$ square cm
When the triangle is revolved about the side $BC$, then the base-radius, height and slant height of the produced cone becomes $AB, BC$ and $AC$ respectively. Therefore, the volume of the produced cone is
$\text{V}_2=\frac{1}{3}\pi\times\text{AB}^2\times\text{BC}$
$=\frac{1}{3}\pi\times(4)^2\times3$
$=16\pi$ cubic cm
Let the angle B is right angle and the sides of the triangle are $AB = 4cm, BC = 3cm,$
AC = 5cm.
When the triangle is revolved about the side $AB$, then the base-radius, height and slant height of the produced cone becomes $BC, AB$ and $AC$ respectively. Therefore, the volume of the produced cone is
$\text{V}_1=\frac{1}{3}\pi\times\text{BC}^2\times\text{AB}$
$=\frac{1}{3}\pi\times(3)^2\times4$
$=12\pi$ cubic cm
In this case, the curved surface area of the cone is
$\text{S}_1=\pi\times\text{BC}\times\text{AC}$
$=\pi\times3\times5$
$=15\pi$ square cm
When the triangle is revolved about the side $BC$, then the base-radius, height and slant height of the produced cone becomes $AB, BC$ and $AC$ respectively. Therefore, the volume of the produced cone is
$\text{V}_2=\frac{1}{3}\pi\times\text{AB}^2\times\text{BC}$
$=\frac{1}{3}\pi\times(4)^2\times3$
$=16\pi$ cubic cm
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