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Last modified on September 29th, 2022

A right circular cone is one whose axis is perpendicular to its base. So the vertex (tip or apex) is aligned straight over the base center.

It is somewhat like a pyramid, having a circular base, similar to a square pyramid with a square base.

**Vertex****(or Apex)**– The sharp tip aligned above the base.**Radius (r)**– It is the distance between the center of its circular base to any point on the circumference of the base.**Height (h)**– The axis coincides with the height. It is the perpendicular distance between the vertex to the center of its circular base.**Slant Height (s)**– It is the distance from its vertex to the point on the outer edge of its circular base.

A typical example of a right circular cone is an ice-cream cone. It is an inverted right circular cone.

We can achieve a right circular cone by rotating a right triangle about one of its leg except the hypotenuse. Thus, the perpendicular leg is also the axis of rotation. The triangle’s hypotenuse is the slant height of the cone. It generates the lateral surface area of a right circular cone.

A section parallel to the base, forms a circle on the axis. (A horizontal section parallel to the base of the right circular cone gives the cross-section of the cone.) When cut through the cross-section of the cone, the remaining basal portion is the frustum of a cone (truncated cone).

Any two equal lines drawn from the vertex to the base at two points generate a section which is an isosceles triangle.

In a right circular cone, the axis is perpendicular to the base. It joins the center and the vertex. So it coincides with the height.

In an oblique cone, the axis is not perpendicular to the base. So its height is measured separately considering a perpendicular line drawn from the vertex to the base level.

The formula is:

**Find the surface area of a right circular cone with a slant height of 30 mm and a radius of 14 mm.**

Solution:

As we know,**Surface Area (SA) = πr ^{2} + πrs**, here r = 14 mm, s = 30 mm, π = 3.141

∴ SA = 3.141 × 14

≈ 1935.22 mm

**Calculate the lateral surface area of a right circular cone with a slant height of 24 cm and a radius of 9.5 cm.**

Solution:

As we know,**Lateral Surface Area (LSA) = πrs**, here r = 9.5 cm, s = 24 cm, π = 3.141

∴ SA = 3.141 × 9.5 × 24

= 716.14 cm^{2}

Finding the **SLANT HEIGHT** of a right circular cone when the **Height** and **Radius** are known

**Find the slant height of a right circular cone with a radius of 4 cm and height of 3 cm.**

Solution:

Here we will apply the Pythagorean’s Theorem as the slant height is the hypotenuse.

${s=\sqrt{h^{2}+r^{2}}}$, here h = 3 cm, r = 4 cm

${\ therefore s=\sqrt{3^{2}+4^{2}}}$

= 5 cm

The formula is:

**Find the volume of a right circular cone with a radius of 5 in and height of 21 in.**

Solution:

As we know,**Volume** ${\left( V\right) =\dfrac{1}{3}\pi r^{2}h}$, here r = 5 in, h = 21 in, π = 3.141

∴ ${V=\dfrac{1}{3}\times \pi \times 5^{2}\times 21}$

= 549.78 mm^{3}

The equation of a right circular cone is :

(x^{2}+y^{2}+z^{2})cos^{2}θ = (lx + my + nz)^{2}, here θ = semi-vertical angle, l, m, and n = direction cosines of the axis.

Let us solve an example involving the concept.

**Find the equation of the right circular cone with a vertex at the origin. Its axis is the line x = y/2 = z/3 which makes a semi-vertical angle of 60°.**

Solution:

The direction cosines of the axis are:

[${\dfrac{1}{\sqrt{1^{2}+2^{2}+3^{2}}}}$ , ${\dfrac{2}{\sqrt{1^{2}+2^{2}+3^{2}}}}$, ${\dfrac{3}{\sqrt{1^{2}+2^{2}+3^{2}}}}$] = ${\left( \dfrac{1}{\sqrt{1+}},\dfrac{3}{\sqrt{14}},\dfrac{2}{\sqrt{14}}\right)}$

The semi-vertical angle is, θ = 60°

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

${\dfrac{x^{2}+y^{2}+z^{2}}{14}}$ = ${\dfrac{(x+2y+3z)^{2}}{14}}$

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

5x^{2}– y^{2 }– 11z^{2} – 8xy – 24yx – 6xz = 0