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Frustum of cone is the part of a crown that is obtained by cutting it into two plains.. The upper part of the cone remains of the same shape while the lower part makes a frustum.
Hence frustum of cone may be defined as the portion of the cone between the cutting plane and the base of the cone.
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Keywords : Frustum, cone, area, pyramid, perimeter, circumference, volume, radius.
Also read: Quadrilateral Formula
Frustum of Cone
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When a solid, generally a cone, is cut in such a way that the base of the solid and the plane that cuts the solid are parallel to each other, part of the solid remaining between the parallel cutting plane and the base is known as a frustum of that solid.
Depending on the shape from which it is obtained, there are three different types of frustum -
- Frustum of cone
- Frustum of a triangular pyramid
- Frustum of a square pyramid
The frustum of a cone is the part of the cone when it is split into 2 elements with a plane that's parallel to the bottom of the cone.
Example - An ice-cream cone completely filled with ice cream. When the cone is sliced, the section left between the base and parallel plane is the frustum.
The video below explains this:
Surface Area and Volume Detailed Video Explanation:
Surface Area of a Frustum of Cone
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There are two types of areas in frustum of cone -
- Curved Surface Area (CSA) or Lateral Surface Area (LSA): Area of the curved surface of the frustum.
- Total Surface Area (TSA): Sum of areas of all its faces (i.e., CSA + Sum of areas of the circular bases).
CSA of frustum of cone = πl [ (R2 - r2) / r ] (OR) πL (R + r)
Total surface area (TSA) = Area of bases + Lateral (or curved) surface area
CSA of frustum of cone = πl [ (R2 - r2) / r ] (OR) πL (R + r)
We know that the base areas of the frustum of the cone are πR2 and πr2.
Thus, the sum of base areas is π (R2 + r2).
TSA of frustum of cone = π (R2 + r2) + πl [ (R2 - r2) / r ]
Or
TSA of frustum of cone = πL (R + r) + π (R2 + r2)
There are two methods to derive the curved surface area of a frustum of cone formula.
Method 1
The circumference of frustum of cone are
C1= 2πR
C2 = 2πr
Substituting these values in the curved surface area of frustum formula,
CSA = (1/2) ×( C1+C2)(C1+C2) × L
CSA = (1/2) × (2πR + 2πr) × L
CSA = πL (R + r)
Method 2
The curved surface area of the full cone is πR(L + l).
The curved surface area of the cone (with apex) that is cut is, πrl.
The curved surface area (CSA) of frustum of the cone = Curved surface area of the full cone - Curved surface area of the cone that is cut
CSA = πR (L + l) - πrl... (1)
(L + l) / l = R / r ... (2)
L + l = Rl / r ... (3)
Substituting this in (1),
CSA = πR (Rl / r) - πrl
V = πl [ (R2 - r2) / r ]
From (2),
(L / l) + 1 = R / r
L / l = (R / r) - 1
L / l= (R - r) / r
Reciprocating on both sides,
l / L = r / (R - r)
l = (L r) / (R - r)
Substituting this in the above formula,
CSA = (π) [ (L r) / (R - r) ] [ (R2 - r2) / r ]
Using the algebraic formulas, a2 - b2 = (a - b) (a + b) and by applying this formula to R2 - r2,
CSA = (π ) [ (L r) / (R - r) ] [ (R - r) (R + r) / r ]
CSA = πL (R + r)
Also read: Surface area of right sided cone
Volume of Frustum of Cone
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There are two methods to get the volume of the cone.
Method 1
The base areas of the frustum of the cone are:
S1 = πR2
S2 = πr2
Substituting these values in the volume of frustum formula,
V = H3(S1+S2+√S1S2)
H3(S1+S2+S1S2)
V = H/3(πR2+πr2+√πR2⋅πr2)
V=πH/3(R2 + Rr + r2)
Method 2
The volume of the full cone is πR2 (H + h) / 3.
The volume of the cone (with apex) = πr2h / 3.
We have,
The volume of frustum of the cone( V) = The volume of the full cone - The volume of the cone that is cut
V = πR2 (H + h) / 3 - πr2h / 3 ... (1)
The triangles OBC and PQC are similar,and thus,
(H + h) / h = R / r ... (2)
H + h = Rh / r ... (3)
Substituting this in (1),
V = πR2 · (Rh / r) - πr2h / 3
V = πh/3 [ (R3 - r3) / r ]
We derived one formula of volume of the frustum of the cone.
Now we will derive another formula from this.
From (2),
(H / h) + 1 = R / r
H / h = (R / r) - 1
H / h = (R - r) / r
Reciprocating on both sides,
h / H = r / (R - r)
h = (H r) / (R - r)
Substituting this in the above formula,
V = (π / 3) [ (H r) / (R - r) ] [ (R3 - r3) / r ]
Using one of the algebraic formulas, a3 - b3 = (a - b) (a2 + ab + b2).
By applying this formula to R3 - r3,
V = (π / 3) [ (H r) / (R - r) ] [ (R - r) (R2 + Rr + r2) / r ]
V = πH/3 (R2 + Rr + r2)
Also read: Unit Conversion
Important formula
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- Total surface area (TSA) = Area of bases + Lateral (curved) surface area
- The volume of right circular cone = 1⁄3 r2h
- Volume of frustum of cone = V=πh/3(R2 + Rr + r2)
Things to remember
- A cone has three parts- The curved portion, apex which is pointed at the top and the base which is the flat portion at the bottom.
- The curved surface area of frustum of a cone: CSA = πl(R – r)
- The lateral surface of the cone is the same as its curved surface area.
- Examples of a frustum of a cone include- Glass tumbler, space capsule, lampshade, and buckets.
Also read: Isosceles Triangle Theorems
Sample questions
Ques. A cone is cut by a plane horizontally. The radius of the circular top and base of the frustum is 10m and 3m, respectively. The height of the frustum is 24m. If the height of the cone is 28m, then find the lateral surface area of the frustum. (3 Marks)
Ans. Let, Radii are r1 = 10m and r2 = 3m
Height, h = 24m
Slant height l = √[(r1 – r2)2 + h2]
= √{(10 – 3)2 + 242}
= √(49 + 576)
= √625 m
= 25 m
Lateral surface of frustum = π(r1 + r2)l
LSA = π(10 + 3)25
= 325π sq.m.
Ques. Find the volume of the frustum of a cone of height 20 in, large base radius to be 25 in, and small base radius to be 4 in. Find the answer in terms of π. (3 Marks)
Ans. Height of the frustum of the cone H = 20 in.
Large base radius of the frustum R = 25 in.
Its slant height, L = 29 in.
Its small base radius, r = 4 in.
The volume of the given frustum of the cone is,
V = πH/3 (R2 + Rr + r2)
Substituting the values of R, r, and H here,
V = π(20)/3 [ (25)2 + (25)(4) + 42]
V = 4940π in3
The volume of the frustum of the cone = 4940 π in3.
Ques. If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm, find the capacity of the bucket. (3 Marks)
Ans. The bucket forms a frustum of a cone such that the radii of its circular ends are r1 = 28 cm, r2 = 7 cm, and height h = 45 cm.
Let V be the capacity of the bucket.
Then, by frustum of cone formula,
V = Volume of the frustum
V=1/3πH(r2+rr′+r′2)
V=1/3×22/7×45(282+28×7+72)
= 330 x 147 cm3
= 48510 cm3.
Ques. The slant height of the frustum of a cone is 4 cm, and the perimeter of its circular bases is 18 cm and 6 cm, respectively. Find the curved surface area of the frustum. (3 Marks)
Ans. Let r1 and r2 be the radii of the circular base of the frustum
l be the slant height
h be the height.
We have l = 4cm,
2 π r1 = 18
r1 = 9/π
2 π r2 = 6
r2 = 3/π
By frustum of a cone formula,
Curved surface area = π(r1 + r2)l = π((9/π + 3/π) × 4
= 48 cm2
Ques. The bases of a frustum of a square pyramid are of lengths 10 units and 7 units. Its slant height is 12 units. Find the total surface area of the frustum of the pyramid. (3 Marks)
Ans. The height of the frustum of a square pyramid, L = 12 units.
The perimeters of bases of the frustum are
P1= 4(10) = 40 units.
P2= 2(7) = 28 units.
The total surface area of the frustum of a square pyramid is,
TSA = (1/2) × P1+P2 × L + Area of bases
TSA = (1/2) × (40 + 28) × 12 + 102 + 72
TSA = 557 square units.
Hence the TSA of the frustum of a square pyramid = 557 sq.
Ques. What is the total surface area of the frustum of a right circular cone of height 20 in, large base radius to be 25 in, and slant height to be 29 in. Express the answer in π. (4 Marks)
Ans. Large base radius of the frustum is, R = 25 in.
Let its small base radius be 'r'.
The height of the frustum of the cone is H = 20 in.
Slant height is L = 29 in.
We know that,
L2 = H2 + (R - r)2
292 = 202 + (25 - r)2
841 = 400 + (25 - r)2
441 = (25 - r)2
Taking square root on both sides,
21 = 25 - r
r = 4
Thus, the total surface area of the given frustum of right circular cone is,
TSA = πL (R + r) + π (R2 + r2)
TSA = π (29) (25 + 4) + π (252 + 42) = 1482π
Hence The TSA of frustum of the cone = 1482π in2.
Ques. What is the total surface area of a base of cone which has a radius of 22 in, a height of 24 in, and a slant height of 26 in? (4 Marks)
Ans. The large base radius of the frustum is R = 22 in.
Assume that its small base radius is 'r'.
The height of the frustum of the cone is H = 24 in.
Its slant height is L = 26 in.
We know that,
L2 = H2 + (R - r)2
262 = 242 + (22 - r)2
676 = 576 + (22 - r)2
100 = (22 - r)2
Taking square root on both sides,
10 = 22 - r
r = 12
Thus, the total surface area of the given frustum of a right circular cone is,
TSA = πL (R + r) + π (R2 + r2)
TSA = π (26) (22 + 12) + π (222 + 122) = 1512π
The TSA of the frustum of the cone = 1512π in2.
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