Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The diameter of an ice cream cone is 7 cm and its height is 12 cm. Find the volume of ice-cream that the cone can contain.

Last updated date: 12th Sep 2024
Total views: 439.2k
Views today: 8.39k
Verified
439.2k+ views
Hint: We will proceed by finding the volume of cone by using the formula ${\text{V = }}\dfrac{{1}}{{3}}{\pi }{{\text{r}}^{\text{2}}}{\text{h}}$. As diameter is given, we will find the radius by dividing diameter by 2. We will substitute the values of radius and height in the formula above to find the volume of the cone.

Complete step by step solution: The ice cream cone is in the shape of a cone. The volume of ice-cream that the cone can contain is equal to the volume of the cone. The volume of the cone can be determined after knowing its physical dimensions i.e. radius and height.
Let ${\text{h}}$ be the height of the cone and ${\text{r}}$ be its radius.
As the height of the ice cream cone is given as 12 cm.
So, ${\text{h = 12cm}}$
The diameter of the cone is given as 7 cm.
We know that the diameter of a cone is twice its radius, implying, the radius will be half of the diameter.
So, ${\text{r = }}\dfrac{7}{2}{\text{cm}}$
Now, V be the volume of a cone having height ${\text{h}}$ and radius ${\text{r}}$
We know the formula for the volume of a cone, i.e.
So, $V = \dfrac{1}{3}\pi {r^2}h$
Substituting, ${\text{r = }}\dfrac{{\text{7}}}{{\text{2}}}{\text{cm}}$ and ${\text{h = 12cm}}$ in the above formula,
${\text{V = }}\dfrac{{\text{1}}}{{\text{3}}}{\pi }{\left( {\dfrac{{\text{7}}}{{\text{2}}}} \right)^{\text{2}}}{\text{12c}}{{\text{m}}^{\text{3}}}$
$V = \dfrac{1}{3}\pi \dfrac{{49}}{4}12$
${\text{V = 49}\pi \text{c}}{{\text{m}}^{\text{3}}}$
So, the volume of cone is $49\pi c{m^3}$
Therefore, the volume of ice-cream that the cone can contain is $49\pi c{m^3}$
Note: Students must remember the formula of volume of different 3D objects carefully and do not get confused with radius and diameter of an object. The volumes of various 3D objects are:
Sphere with radius ${\text{r = }}\dfrac{{\text{4}}}{{\text{3}}}{\pi}{{\text{r}}^{\text{3}}}$
Cube with edge a: ${{\text{a}}^{\text{3}}}$
Cuboid with dimensions l, b, h: ${\text{l} \times \text{b} \times \text{h}}$
Cylinder with height h and radius r: ${\pi}{{\text{r}}^{\text{2}}}{\text{h}}$
These formulae must be remembered to solve the problems.