# Frustum of a Cone

A solid is obtained by cutting a right circular cone by a plane parallel to the base of the cone. Such a solid is called a Frustum. We shall discuss problems on finding the volume and surface area of a frustum of a right circular cone.

**Bucket Flower Pot Drinking glass Table lamp**

**Definition **

Right circular cone VAB is cut by a plane Aâ€™Oâ€™Bâ€™ parallel to its circular base AOB. This gives the solid AAâ€™Oâ€™Bâ€™ BOA which is the frustum of right circular cone VAB. The circular faces AOB and Aâ€™Oâ€™Bâ€™ are called the circular ends of the Frustum.

Let us now define height, lateral (slant) height of the frustum.

**Height**

The height or thickness of a frustum is the perpendicular distance between its two ends.

OOâ€™ is the height of the frustum of a cone.

**Slant Height**

The slant height of a frustum of a right circular cone is the length of the line segment joining the extremities of two parallel radii, drawn in the same direction, of the circular ends.

AAâ€™ = BBâ€™ is the slant height of the frustum.

**Volume and Surface Area of a Frustum**

Let R and r be the radii of two circular ends and h be the height of a frustum of a right circular cone Let l be the slant height of the frustum. Clearly, triangles VOA and VOâ€™Aâ€™ are similar.

= =

= =

= =

= =

= =

= =

H = and L =

From right angled triangle AMAâ€™, we have

AAâ€™^{2 }= AM^{2 }+ Aâ€™M^{2 }

â‡’ l^{2} = (R - r)^{2 }+ h^{2 }

â‡’ l =

Let V be the volume of the frustum. Then,

V = Volume of cone VAB â€“ Volume of cone VAâ€™Bâ€™

â‡’ V = Ï€ R^{2}H - Ï€ r^{2}(H - h)

â‡’ V = Ï€ {R^{2}H - r^{2}(H - h)}

â‡’ V = Ï€ {(R^{2} Ã—) - (r^{2} Ã—)} [Q H = âˆ´ H - h= ]

â‡’ V = Ï€ ()h

â‡’ V = Ï€ (R^{2} +Rr+r^{2})h

Let S be the lateral surface area of the frustum. Then,

S = Lateral surface area of cone VAB â€“ Lateral surface area of cone VAâ€™Bâ€™

= Ï€ RL - Ï€ r(L - l)

= Ï€ R Ã— - Ï€ r [ L = ; L - l = ]

= Ï€ ()l

= Ï€ (R + r)l

Total surface area of the frustum = Lateral surface area + Surface area of its circular ends

= Ï€(R + r)l + Ï€R^{2} + Ï€r^{2}

= Ï€ [(R + r)l + R^{2 }+ r^{2}]

Thus, if R and r are the radii of the circular ends of a frustum of a cone, h is the height and l is the slant height, then

Volume of the frustum = Ï€ (R^{2} + Rr + r^{2})h

Lateral surface area = Ï€ (R + r)l

Total surface area = Ï€ [(R + r)l + R^{2 }+ r^{2}]

Slant height =

Let us now discuss some examples:

If the radii of the circular ends of a conical bucket are 28 cm and 7 cm, whose height is 45 cm. Find the capacity of the bucket (Use Ï€ = ).

Clearly, bucket forms a frustum of a cone such that the radii of its circular ends are R = 28 cm, r = 7 cm and height is = 45 cm.

Therefore,Capacity of the bucket = Volume of the frustum

= Ï€h (R^{2} + Rr + r^{2})

= Ã— Ï€ Ã— 45(28^{2 }+ 7^{2 }+ 28 Ã— 7)

= 22 Ã— 15 Ã— (28 Ã— 4 + 7 + 28)

= 330 Ã— 147 cm^{3}

= 48510 cm^{3}.

The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the lateral surface area and total surface area of the frustum.

R = 14 cm, r = 6 cm and h = 6 cm.

Let l be the slant height of the frustum. Then,

l =

â‡’ l =

= 10 cm.

â‡’ Lateral surface area = Ï€ (R + r)l

= Ï€ (14 + 6)10 cm^{2}

= 628.57 cm^{2}

Total surface area = Ï€ [(R + r)l + R^{2 }+ r^{2}]

= Ï€ (200 + 196 + 36) cm^{2}

= Ï€ Ã— 432 cm^{2}

= 1357.71 cm^{2}.

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

Let the height of the bucket be h cm.

We have, R = 28 cm, r = 21 cm and,

V = Volume of the bucket = 28.490 litres = 28.490 Ã— 1000 cm^{3} = 28490 cm^{3}.

V = 28490 cm^{3}.

â‡’ Ï€ (R^{2} + Rr + r^{2})h = 28490

= 28490

â‡’ h =

â‡’ h = 15 cm.

Thus, height of the bucket = 15 cm.

**Example**

A friction clutch is in the form of a frustum of a cone the diameters of the ends being 32 cm and 20 cm and length 8 cm. Find its bearing surface and volume.

**Solution **

Let ABBâ€™Aâ€™ be the friction clutch of slant height l cm.

We have,R = 16 cm, r = 10 cm and h = 8 cm.

âˆ´ l^{2} = h^{2 }+ (R - r)^{2 }

â‡’ l^{2} = 64 + (16 - 10)^{2 }

^{ }^{ }= 64 + (6)^{2}

^{ }= 64 + 36 = 100 cm

â‡’ l = 10 cm

Bearing surface of the clutch = Lateral surface of the frustum

= Ï€ (R + r)l

= Ï€ (16 + 10)10 cm^{2 }

^{ } = (26)10

= 817.14 cm^{2}.

Volume = Ï€ (R^{2} + Rr + r^{2})h

= Ã— Ï€ Ã— 8(16^{2 }+ 16 Ã— 10 + 10^{2}) cm^{3}

= Ã— Ï€ Ã— 8(256 + 160 + 100) cm^{3}

= 4324.57 cm^{3}.