A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with the apex of the cone and it forms a right angle. A cone is a threedimensional shape having a circular base and narrowing smoothly to a point above the base. This point is known as apex.
We come across many geometrical shapes while dealing with geometry. We usually study about two dimensional and threedimensional figures in school. Twodimensional shapes have length and breadth. They can be drawn on paper, for example – circle, rectangle, square, triangle, polygon, parallelogram etc. The threedimensional figures cannot be drawn on paper since they have an additional third dimension as height or width. The examples of 3D shapes are a sphere, hemisphere, cylinder, cone, pyramid, prism etc.
Right Circular Cone Definition
A right circular cone is one whose axis is perpendicular to the plane of the base. We can generate a right cone by revolving a right triangle about one of its legs.
In the figure, you can see a right circular cone, which has a circular base of radius r and whose axis is perpendicular to the base. The line which connects the vertex of the cone to the centre of the base is the height of the cone. The length at the outer edge of the cone, which connects a vertex to the end of the circular base is the slant height.
Right Circular Cone Formula
For a right circular cone of radius ‘r’, height ‘h’ and slant height ‘l’, we have;

Surface Area of a Right Circular Cone
The surface area of any right circular cone is the sum of the area of the 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
= π r^{2} + π r l
= πr(r + l)
Here, l = √(r^{2}+h^{2})
Where ‘r’ is the radius, ‘l’ is the slant height and ‘h’ is the height of the cone.
Volume of a Right Circular Cone
The volume of a cone is onethird of the product of the area of the base and the height of the cone. The volume 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 = ⅓ (Base area × Height)
Where Base Area = π r^{2}
Hence, Volume = ⅓ π r^{2}h
Properties of Right Circular Cone
 It has a circular base whose centre joins its vertex, showing the axis of the respective cone.
 The slant height of this cone is the length of the sides of the cone taken from vertex to the outer line of the circular base. It is denoted by ‘l’.
 The altitude of a right cone is the perpendicular line from the vertex to the centre of the base. It coincides with the axis of the cone and is represented by ‘h’.
 If a right triangle is rotated about its perpendicular, considering the perpendicular as the axis of rotation, the solid constructed here is the required cone. The surface area generated by the hypotenuse of the triangle is the lateral surface area.
 Any section of right circular cone parallel to the base forms a circle lies on the axis of the cone.
 A section which contains the vertex and two points of the base of a right circular cone is an isosceles triangle.
Frustum of a Right Circular Cone
A frustum is a portion of the cone between the base and the parallel plane when a right circular cone is cut off by a plane parallel to its base.
Equation of Right Circular Cone
The equation of the right circular cone with vertex origin is:
(x^{2}+y^{2}+z^{2})cos^{2}θ=(lx+my+nz)^{2} 
Where θ 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, the axis is the line x = y/3 =z/2 and which makes a semivertical angle of 60 degrees.
The direction cosines of the axis are:
[(1/1^{2}+3^{2}+2^{2}) , (3/1^{2}+3^{2}+2^{2}), (2/1^{2}+3^{2}+2^{2})] = (1/√14, 3/√14, 2/√14)The semivertical angle is, θ = 60°
Therefore, the equation of the right circular cone with vertex (0, 0, 0) is:
(x^{2}+y^{2}+z^{2}) ¼ = 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}12xy24yz8xz = 0; which is the required equation.
Examples
Q. 1: Solve the surface area of right cone for the given radius is 6 cm and slant height is 10 cm.
Solution: Given,
Radius (r) = 6 cm
Slant height (l) = 10 cm
Surface area of a cone = π r(r + l)
Solving surface area,
SA = 3.14 × 6(6 + 10)
SA = 3.14 × 6 × 16
SA = 301.44
Therefore, Surface area of right cone is 301.4 square cm.
Question 2: Solve the Volume of the right cone for the given radius 6 cm and height 10 cm.
Solution: Given,
Radius = 6 cm
Height = 10 cm
Volume of right cone = 1/3π r^{2} h
Volume, V = ⅓ × π × (6)^{2} × 10
V = 0.333 × 3.14 × 36 × 10
V = 376.42
Therefore, Volume of right cone is 376.42 cubic cm.