Circular Cones
Circular Cones
The figure shows a circular cone.
The base (yellow region) of a cone
is a circle.
The distance between the vertex and
the base is called the height of the
cone.
The distance between the vertex and any point on the
circumference of the base is called the slant height of the
cone.
A circular cone with equal slant
heights is called a right circular cone.
The centre of the base of a right circular cone is the foot of
the perpendicular from the vertex.
Volumes of Right Circular Cones
Consider a regular pyramid.
If the number of sides of its base increases gradually,
what kind of shape will the base become? Circle
The regular pyramid eventually becomes a right circular cone.
From the formula for the volume of pyramids,
we can deduce the formula for the volume of a right
circular cone:
1

 base area  height
of a pyramid
Volume ofVolume
a right circular
cone
3
In fact, for any circular cones,
1 2
volume of a circular cone   r h
3
base area   r 2
Refer to the figure on the right.
Volume of the right circular cone
1
    5 2  9 cm3
3
 75  cm3
r = 5 cm,
h = 9 cm
9 cm
5 cm
Follow-up question
Find the volume of the right circular
cone in the figure. (Give your answer in
terms of  . )
16 cm
Solution
Volume of the circular cone
2
1
 12 
       16 cm3
3
 2 
 192  cm 3
12 cm
Example 7
In the figure, the base diameter and the height of the
right circular cone are 14 cm and 6 cm respectively.
Find the volume of the cone in terms of .
Solution
1
Base radius   14 cm  7 cm
2
1
Volume of the cone     7 2  6 cm 3
3
 98 cm 3
Example 8
The base radius and the slant height of the right
circular cone in the figure are 2 cm and 10.1 cm
respectively. Find the volume of the cone correct
to the nearest cm3.
Solution
Height of the cone  10.12  2 2 cm (Pyth. theorem)
 98.01 cm
 9.9 cm
∴
1
Volume of the cone     2 2  9.9 cm 3
3
 41 cm 3 (cor. to the nearest cm 3 )
Example 9
The figure shows two inverted right
conical vessels, where the larger
vessel is fully filled with water. If the
water in the larger vessel can just fill
up 4 smaller identical vessels, find the
height of the larger vessel.
Solution
1

   3 2  4 cm 3
Volume of smaller vessel
3
 12 cm 3
Volume of larger vessel  4  volume of smaller vessel
 4  12 cm 3
 48 cm 3
Let h cm be the height of the larger vessel.
1
48     4 2  h
3
h9
∴
The height of the larger vessel is 9 cm.
Frustum of a Right Circular Cone
If the top of a right circular cone is cut away by a plane
which is parallel to the base of the cone, the remaining part
is called a frustum of the right circular cone.
The removed part is a right circular cone of smaller volume.
volume of the volume of the
Volume of the
frustum of a right  larger right  smaller right
circular cone circular cone
circular cone
Refer to the figure on the right.
1

   3 2  12 cm 3
Volume of cone VCB
3
 36 cm 3
V
12 cm
C
If the volume of right circular cone VDA
is 210 cm3, then
volume of frustum ABCD
 volume of cone VDA  volume of cone VCB
3
 (210  36 ) cm
 174  cm3
D
3 cm
P
Q
B
A
Follow-up question
The figure shows a frustum ABCD of a
right circular cone. If the volume of
cone VCB is 190 cm3, find the volume 12 cm
of frustum ABCD, correct to 3
significant figures.
D
Solution
1
Volume of cone VDA     5 2  12 cm 3
3
 100 cm3
Volume of frustum ABCD
 volume of cone VDA  volume of cone VCB
 (100  190) cm3
 124 cm3 (cor. to 3 sig. fig.)
V
C
P
B
A
Q
5 cm
Example 10
The figure shows a frustum ABCD
of a right circular cone.
A cylindrical hole of the same radius
as that of the upper base is drilled
out from the frustum to form a new
solid.
(a)
Find the length of VP.
(b)
Find the volume of the frustum
ABCD.
(c)
Find the volume of the new solid.
(Give your answers in terms of  if necessary.)
Solution
(a)
Let VP  x cm, then VQ  (x  6) cm.
(AAA)
∵ △VPB ~ △VQA
VP PB

∴
VQ QA
6
x

x6 8
8 x  6 x  36
2 x  36
∴
x  18
VP  18 cm
(b)
1

   8 2  (18  6) cm 3
Volume of cone VAD
3
 512 cm 3
1
2
3
Volume of cone VBC     6  18 cm
3
 216 cm3
Volume of the frustum
 volume of cone VAD  volume of cone VBC
 (512  216 ) cm 3
 296 cm 3
(c)
Volume of the cylindrical hole
   6 2  6 cm 3
 216 cm3
Volume of the new solid
 volume of frustum ABCD  volume of cylindrical hole
 (296  216 ) cm3
 80 cm3
Total Surface Areas of Right Circular Cones
The surface of a right circular cone is composed of a base
and a curved surface.
Total surface area of  curved surface area  base area
a right circular cone
How can we calculate the curved
surface area of a right circular cone?
Therefore, the curved surface
Finally,
wethe
getcone
a sector.
Let’s
the
following
areaconsider
of
is equal tocone.
the
area of the corresponding sector.
1. Cut along the dotted line.
2. Unfold the curved surface.
Curved surface area of the cone  area of sector VAB
    2
360
By considering the arc length of AB, we
can express  in terms of r and  .

AB  circumference of the base
(a) Consider the base of the cone.
 2 r
(b) Consider sector VAB.

AB 

360
From the results obtained in (a) and (b), we have:

360
 2  2r
  360r

 2

Curved surface area of the cone 
  2
360°
360°r


  2
360°
 r
For a right circular cone with base radius r and slant height ,
curved surface area   r 
Total surface area of a right circular cone
= curved surface area + base area
Total surface area of a right circular cone   r   r 2
Refer to the figure.
4 cm
9 cm
Total surface area of the right circular cone
 curved surface area  base area
 (  4  9    4 2 ) cm2
 52 cm2
r = 4 cm,  = 9 cm
Follow-up question
12 cm
The figure show a right circular cone.
(a) Find the slant height of the cone.
(b) Find the total surface area of the cone.
B

.
(Give your answer in terms of )
A
5 cm
C
Solution
(a) Consider △ABC.
Slant height of the cone 
AB 2  AC 2
 12 2  5 2 cm
 13 cm
(Pyth. theorem)
Follow-up question (cont’d)
12 cm
The figure show a right circular cone.
(a) Find the slant height of the cone.
(b) Find the total surface area of the cone.
B

.
(Give your answer in terms of )
Solution
(b) Total surface area of the right circular cone
 curved surface area  base area
 (  5  13    5 2 ) cm 2
 90 cm 2
A
5 cm
C
Example 11
In the figure, the base diameter and the slant
height of the right circular cone are 16 cm and
10 cm respectively. Find the total surface area
of the cone in terms of .
Solution
Base radius 
1
 16 cm  8 cm
2
Curved surface area of the cone    8  10 cm 2
 80 cm 2
Base area of the cone    8 2 cm 2
 64 cm 2
∴
Total surface area of the cone  curved surface area  base area
 (80  64 ) cm 2
 144 cm 2
Example 12
The figure shows a right circular cone of base diameter
18 cm and height 40 cm.
(a)
(b)
Find the curved surface area of the cone.
Find the volume of the cone.
(Give your answers in terms of .)
Solution
(a)
Base radius 
1
 18 cm  9 cm
2
Slant height  40 2  9 2 cm (Pyth. theorem)
 1681 cm
 41 cm
∴
Curved surface area of the cone    9  41 cm 2
 369 cm 2
(b)
1
Volume of the cone     9 2  40 cm 3
3
 1080 cm 3
Example 13
The figure shows a right conical
vessel of base radius 4 cm and
capacity 64 cm3. The vessel is
cut along OA to form a sector of
radius OA.
(a)
(b)
Find the length of OA.
Find the curved surface area of the vessel.
(c) Find .
(Give your answers correct to 3 significant figures.)
Solution
(a)
Let h cm be the height of the vessel.
1
64     4 2  h
3
h  12
∴ The height of the vessel is 12 cm.
Consider the right-angled triangle as shown.
OA  4 2  12 2 cm (Pyth. theorem)
 160 cm
 12.6 cm (cor. to 3 sig. fig.)
(b) Curved surface area of the vessel    4  160 cm 2
 158.95 cm 2
 159 cm 2
(c)
∵
∴
(cor. to 3 sig. fig.)
Area of the sector  curved surface area of the vessel

360
   ( 160 ) 2  158.95
  114 (cor. to 3 sig. fig.)