The document defines and provides information about cones, conical frustums, and how to calculate their volume and total surface area. It explains that a cone tapers from a circular base to a point, and can be a right cone or oblique cone. Formulas are given for calculating the volume and total surface area of cones and conical frustums. Examples are worked through applying the formulas to calculate volume and surface area of different cone and frustum shapes.

2. What is a Cone?
 A solid or hollow object that tapers from a
circular or roughly circular base to a point.

3. Types of Cones
 Right Cone - A cone that has its apex aligned
directly above the center of its base.
 Oblique Cone – A cone that has its apex not
aligned above the center of its base
Right Cone Oblique Cone
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4. Volume of Cone
Formula:
Volume = 𝜋𝑟2
ℎ
r = radius of circular base
h = height of the cone
Can be used for both right and oblique
cone.

5. Total Surface Area
Formula:
𝑇𝑆𝐴 = 𝜋rs + 𝜋𝑟2
= 𝜋𝑟(𝑠 + 𝑟)
r = radius of circular base
s = slant height of the cone
This formula CANNOT be used for oblique
cone. (There are NO formula to find TSA of
oblique cone)

6. Example 1
Find out the Volume and the
Total Surface Area of the
cone.
< Given: 𝜋 =
22
7
>
Volume =
1
3
𝜋𝑟2
ℎ
=
1
3
×
22
7
× 72
×10
= 513.333𝑐𝑚3

7. s = 102 + 72
= 149
Total Surface Area
= 𝜋𝑟 𝑠 + 𝑟
=
22
7
× 7( 149 + 7)
= 22 149 + 7
= 422.544𝑐𝑚2

8. What is a Conical Frustum?
A conical frustum is a frustum created by
slicing the top off a cone (with the cut
made parallel to the base).

9. Volume of Conical Frustum
METHOD 1
 Formula:
V =
𝜋ℎ
3
(𝑅2
+ 𝑅𝑟 + 𝑟2
)
h = height of the frustum
r = radius of the circular top of the frustum
R = radius of the circular base of the frustum

10. Total Surface Area of Conical
Frustum
𝑠 = (𝑅 − 𝑟)2 + ℎ2
 Formula:
Lateral Surface Area:
TSA = 𝜋𝑠 (𝑅 + 𝑟)
Circular Surface Area:
(Top) CSA = 𝜋𝑟2
(Bottom) CSA = 𝜋𝑅2
Total Surface Area:
TSA = 𝜋𝑠 𝑅 + 𝑟 + 𝜋𝑟2 + 𝜋𝑅2
= 𝜋 (𝑅 − 𝑟)2 + ℎ2 𝑅 + 𝑟 + 𝜋𝑟2 + 𝜋𝑅2

11. Example 2
Find out the Volume and the
Total Surface Area of the
frustum.
< Given: 𝜋 =
22
7
>
Volume =
𝜋ℎ
3
(𝑅2 + 𝑅𝑟 + 𝑟2)
=
22
7
×14
3
(72
+ 35 +
52)
=
44
3
× 109
= 1598.667𝑐𝑚3

12. S = (7 − 5)2+142
= 10 2
Total Surface Area
= 𝜋𝑠 𝑅 + 𝑟 + 𝜋𝑅2
+ 𝜋𝑟2
=
22
7
× 10 2 7 + 5 +
22
7
×
72
+
22
7
× 52
= 533.361 + 154 + 78.571
= 765.932𝑐𝑚2

13. Example 3
Diagram shows cone A and
frustum B. Ratio of slant height
cone A to slant height frustum
B is 1:2 . Given that radius of
cone A is 4cm and its volume
is 16π𝑐𝑚3
. Find out the
volume and the total surface
area of frustum B in terms of
π.

14. Height of cone A:
Volume =
1
3
𝜋𝑟2ℎ
16𝜋𝑐𝑚3 =
1
3
× 𝜋 × 42 × ℎ
16𝑐𝑚3 =
16
3
ℎ
ℎ = 16 ×
3
16
ℎ = 3𝑐𝑚
Slant height of cone:
ℓ = 𝑟2 + ℎ2
= 42 + 32
= 25
= 5𝑐𝑚

15. Ratio of slant height cone A to slant height frustum B

16. Solution 1:
Volume of frustum B
=
1
3
𝜋𝑅2 𝐻 −
1
3
𝜋𝑟2ℎ
=
1
3
𝜋 × 122 × 9 −
1
3
𝜋 × 42 × 3
= 432𝜋 − 16𝜋
= 416𝜋𝑐𝑚3 Solution 2:
Volume of frustum B
=
𝜋 𝐻 − ℎ
3
(𝑅2
+ 𝑅𝑟 + 𝑟2
)
=
𝜋 9 − 3
3
122 + 48 + 42
= 2𝜋 × 208
= 416𝜋𝑐𝑚3

17. Total Surface Area of frustum B
= 𝜋 (𝑅 − 𝑟)2 + (𝐻 − ℎ)2 𝑅 + 𝑟 + 𝜋𝑟2 +
𝜋𝑅2
= 𝜋 12 − 4 2 + 9 − 3 2 12 + 4 +
𝜋(4)2 + 𝜋(12)2
= 𝜋10 16 + 16𝜋 + 144𝜋
= 160𝜋 + 160𝜋
= 320𝜋𝑐𝑚2