We come across with many geometrical shapes while dealing with geometry. We usually study about two dimensional and three dimensional figures in school. Two dimensional shapes have length and breadth. They can be drawn on paper, for example - circle, rectangle, square, triangle, polygon, parallelogram etc. The three dimensional figures cannot be drawn on paper, since they have additional third dimension as height or width. The examples are 3D shapes are sphere, hemisphere, cylinder, cone, pyramid, prism etc.

Here, we are going to study about the cone, especially a right circular cone. A cone is a three-dimensional shape having a circular base and tapering smoothly to a point above the base. This point is known as apex. A right circular cone is a cone where the axis of cone is the line joining the vertex to the midpoint of the circular base. That is, the center point of the circular base is joined with the apex of the cone. It forms a right angle. Therefore, it is called as a right circular cone.

In the figure shown above, we can see that a cone that has a radius, 'r', height, 'h' and slant height, 'l'. Since the height and the radius forms a right angle, it is called as a Right Circular Cone.

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A right circular cone is one whose axis is perpendicular to the plane of the base. We can generated a right cone by revolving a right triangle about one of its legs.

For a right circular cone of radius r, height h and slant height s, we have

Lateral area of a right circular cone = $\pi$ r s

Total surface area of a right circular cone = $\pi$(r + s) r

Volume of a right circular cone = $\frac{1}{3}$ $\pi$ r$^2$ h

Area is measured in square units and volume is measured in cubic units.

The equation of the right circular cone with vertex origin is

$(x^2 + y^2 + z^2) cos^2 \theta$ = $(lx + my +nz)^2$

Where, $\theta$ is the semivertical angle and <l, m, n> are direction cosines of the axis.

Let us find the equation of the right circular cone whose vertex is the origin, axis is the line x = $\frac{y}{3}$ = $\frac{z}{2}$, and which makes semivertical angle of 60$^o$.

The direction cosines of the axis are <$\frac{1}{1^2 + 3^2 + 2^2}$, $\frac{3}{1^2 + 3^2 + 2^2}$, $\frac{2}{1^2 + 3^2 + 2^2}$> <=> <$\frac{1}{\sqrt 14}$, $\frac{3}{\sqrt 14}$, $\frac{2}{\sqrt 14}$>

Semivertical angle is, $\theta$ = 60$^o$

Therefore, the equation of the right circular cone with vertex (0, 0, 0) is

$(x^2 + y^2 + z^2)$$\frac{1}{4}$ = $\frac{1}{14}$ (x + 3y + 2z)$^2$

7$(x^2 + y^2 + z^2)$ = 2$(x^2 + 9y^2 + 4z^2 + 6xy + 12yz + 4xz)$

$5x^2 - 11y^2 - z^2 - 12xy - 24yz - 8xz$ = 0 (By solving)

Which is the required equation of the cone. Given below are some examples on right circular cone.### Solved Examples

**Question 1: **Solve the surface area of right cone for the given radius is 6 cm and slant height is 10 cm.

** Solution: **

**Question 2: **Solve the Volume of right cone for the given radius 6 cm and height 10 cm.

** Solution: **

**Question 3: **Find the volume of the largest right circular cone that can be fitted in a cube with edge 10 cm.

** Solution: **

For a right circular cone of radius r, height h and slant height s, we have

Lateral area of a right circular cone = $\pi$ r s

Total surface area of a right circular cone = $\pi$(r + s) r

Volume of a right circular cone = $\frac{1}{3}$ $\pi$ r$^2$ h

Area is measured in square units and volume is measured in cubic units.

The surface area of a right circular cone is the sum of area of base and lateral surface area of a cone. The surface area is measured in terms of square units.

Surface area of a cone = Base Area + Lateral Area of a cone

= $\pi$ r$^2$ + $\pi$ rs

= $\pi$r(r + s)

Surface Area of a Right Circular Cone can be calculated by the following formula,

Area of a right circular Cone = $\pi$r(r + s)

Here, s = $\sqrt{r^2 + h^2}$

Where, r - Radius

h - Height and

s - slant height of a cone.

The volume of a cone is one third of the product of the area of base and the height of the cone. The volume of a right circular cone is measured in terms of cubic units.

Volume of a right circular cone can be calculated by the following formula,

Volume of a right circular cone = $\frac{1}{3}$ (Base area $\times$ Height)

Where, Base Area = $\pi$ r$^2$

**Volume of a Circular Cone **= $\frac{1}{3}$ $\times$ $\pi$ $\times$ Radius$^2$ $\times$ Height

The equation of the right circular cone with vertex origin is

$(x^2 + y^2 + z^2) cos^2 \theta$ = $(lx + my +nz)^2$

Where, $\theta$ is the semivertical angle and <l, m, n> are direction cosines of the axis.

Let us find the equation of the right circular cone whose vertex is the origin, axis is the line x = $\frac{y}{3}$ = $\frac{z}{2}$, and which makes semivertical angle of 60$^o$.

The direction cosines of the axis are <$\frac{1}{1^2 + 3^2 + 2^2}$, $\frac{3}{1^2 + 3^2 + 2^2}$, $\frac{2}{1^2 + 3^2 + 2^2}$> <=> <$\frac{1}{\sqrt 14}$, $\frac{3}{\sqrt 14}$, $\frac{2}{\sqrt 14}$>

Semivertical angle is, $\theta$ = 60$^o$

Therefore, the equation of the right circular cone with vertex (0, 0, 0) is

$(x^2 + y^2 + z^2)$$\frac{1}{4}$ = $\frac{1}{14}$ (x + 3y + 2z)$^2$

7$(x^2 + y^2 + z^2)$ = 2$(x^2 + 9y^2 + 4z^2 + 6xy + 12yz + 4xz)$

$5x^2 - 11y^2 - z^2 - 12xy - 24yz - 8xz$ = 0 (By solving)

Which is the required equation of the cone. Given below are some examples on right circular cone.

**Given:**

Radius (r) = 6 cm

Slant height (s) = 10 cm

Surface area of a cone = $\pi$ r(r + s)

Solving surface area,

SA = 3.14 $\times$ 6(6 + 10)

SA = 3.14 $\times$ 6 $\times$ 16

SA = 301.44

Therefore, **Surface area of right cone is 301.4 square cm.**

**Given:**

Radius = 6 cm

Height = 10 cm

**Volume of right cone** = $\frac{1}{3}$ $\pi$ r$^2$ h

Solving Volume,

V = $\frac{1}{3}$ $\times$ $\pi$ $\times$ (6)$^2$ $\times$ 10

V = 0.333 $\times$ 3.14 $\times$ 36 $\times$ 10

V = 376.42

Therefore, **Volume of right cone is 376.42 cubic cm.**

For the largest right circular cone to be fitted in a cube, the base of the cone will touch all the vertical faces of the cube.

Therefore, the diameter of base of cone = Side of cube = 10 cm

=> Radius of the base of cone = $\frac{10}{2}$ = 5

Radius of the base of cone (r) = 5 cm

Also, height of the cone = Side of cube = 10 cm

Height of the cone (h) = 10 cm

Now, volume of the required cone = $\frac{1}{3}$ $\pi$ r$^2$ h

= $\frac{1}{3}$ $\times$ 3.14 $\times$ 5$^2$ $\times$ 10

= $\frac{1}{3}$ $\times$ 3.14 $\times$ 25 $\times$ 10

= 259.05

Hence, the volume of the largest right circular cone is 259.05 cm$^3$.

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