A right angled triangle whose sides are \[{\text{3}}cm,4cm\,\,{\text{and }}5cm\] revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed .
Answer
Verified361.8k+ views
Hint:
Here in the figure we have shown the triangle which has revolved about the two of its side and the two cones formed. Cone (a) is formed when the triangle is revolved around the 4 cm side and (b) is the cone when the triangle is revolved around 3 cm of the side.
Hypotenuse is the longest side of a right-angled triangle so it is obvious when the question says the triangle revolved around its sides, it means the sides with length 3 cm and 4 cm. Therefore, for the two cones that form:
1) First cone: Height = 4 cm, Radius of base = 3 cm
2) Second cone: Height = 3 cm, Radius of base = 4 cm
Drawing a diagram is helpful in case you are unable to visualize the cones.
Here in the figure we have shown the triangle which has revolved about the two of its side and the two cones formed. Cone (a) is formed when the triangle is revolved around the 4 cm side and (b) is the cone when the triangle is revolved around 3 cm of the side.
Now, try to solve the question by yourself before looking at the complete solution.
Complete step-by-step answer:
As we know the right angled triangle can be revolved from $3cm\,\,\& \,\,4cm$.
Since the sides making the right angle are $3cm{\text{ and }}4cm$.
When it is revolved about the side of length ${\text{3}}cm$ then the radius of the cone is ${\text{3}}cm$ and the height of the cone is $4cm$.
Then the volume of the cone is,
$\dfrac{1}{3}\pi ({3^2})(4) = 12\pi \,\,\,\,\,\,...({\text{i}})$
When it is revolved about the side of length $4cm$ then the radius of the cone is $4cm$ and the height of the cone is $3cm$.
Then the volume of the cone is,
$\dfrac{1}{3}\pi ({4^2})(3) = 16\pi \,\,\,...{\text{(ii)}}$
Subtracting (i) from (ii) we get,
${\text{16}}\pi {\text{ - 12}}\pi {\text{ = 4}}\pi $
Hence the answer is $4\pi $.
Note:
As we know the right angled triangle can be revolved from $3cm\,\,\& \,\,4cm$.
Since the sides making the right angle are $3cm{\text{ and }}4cm$.
When it is revolved about the side of length ${\text{3}}cm$ then the radius of the cone is ${\text{3}}cm$ and the height of the cone is $4cm$.
Then the volume of the cone is,
$\dfrac{1}{3}\pi ({3^2})(4) = 12\pi \,\,\,\,\,\,...({\text{i}})$
When it is revolved about the side of length $4cm$ then the radius of the cone is $4cm$ and the height of the cone is $3cm$.
Then the volume of the cone is,
$\dfrac{1}{3}\pi ({4^2})(3) = 16\pi \,\,\,...{\text{(ii)}}$
Subtracting (i) from (ii) we get,
${\text{16}}\pi {\text{ - 12}}\pi {\text{ = 4}}\pi $
Hence the answer is $4\pi $.
Note:
In this problem we have revolved the right angled triangle about the sides other than hypotenuse. Since those are the sides, if we revolve the triangle around them we will get two cones of different volume. Then the side about which the triangle is revolved will become the radius of the cone and the side other than hypotenuse will be the height of the cone. Then after finding their volume we subtract the both volumes to get the solution to this problem. The slant height of the cone will be the same in both the cases since it is hypotenuse.
Last updated date: 18th Sep 2023
•
Total views: 361.8k
•
Views today: 4.61k
