## Frustum or right Circular cone-Geometry Formulas

### Frustum or right Circular cone

Frustum of a right circular cone is that portion of right circular cone included between the base and a section parallel to the base not passing through the vertex.

### Properties of Frustum of Right Circular Cone:

• The altitude of a frustum of a right circular cone is the perpendicular distance between the two bases. It is denoted by h.
• All elements of a frustum of a right circular cone are equal. It is denoted by L.

### Curved surface area of a frustum :

C.S.A. of a frustum = π(R + r)l sq. units

Where,$$l=\sqrt {h^2 + (R-r)^2}$$

### Example:

The slant height of a frustum of a cone is 5 cm and the radii of its ends are 4 cm and 1 cm. Find its curved surface area.

### Solution:

Given that, l= 5 cm, R = 4 cm, r = 1 cm

Now, C.S.A. of the frustum = π (R + r) l sq. units

$=\frac{22}{7}×\left(4+1\right)×5$
$=\frac{550}{7}$

Therefore, C.S.A. = 78.57 cm2

### Total surface area of a frustum:

T.S.A. of a frustum = π(R + r)l + πR2 + πr 2 sq. units

Where, $$l=\sqrt {h^2 + (R-r)^2}$$

### Example:

An industrial metallic bucket is in the shape of the frustum of a right circular cone whose top and bottom diameters are 10 m and 4 m and whose height is 4 m. Find the total surface area of the bucket.

### Solution:

Given that, diameter of the top =10 m; radius of the top R = 5 m.

diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h= 4 m

$l=\sqrt{{h}^{2}+\left(R–r{\right)}^{2}}$
$l=\sqrt{{4}^{2}+\left(5–2{\right)}^{2}}$
$l=\sqrt{16+9}=\sqrt{25}$
$\text{T.S.A}=\pi \left(R+r\right)l+\pi {R}^{2}+\pi {r}^{2}$
$=\frac{22}{7}\left[\left(5+2\right)5+25+4\right]$
$=\frac{1408}{7}$
$=201.14$

T.S.A. = 201.14 m2

### Volume of a frustum:

V = (πh/3)(R2+Rr+r2)

### Example:

Find the volume of a frustum of a right circular cone with height 20, lower base radius 34 and top radius 19.

### Solution:

Given.

h = 20, R = 34, r = 19

V =( (π×20)/3)(342+34×19+192)

V = 14420 π cubic units