Question
A right angled triangle whose sides are 3cm, 4cm and 5cm 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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