Cone
- A cone is the solid bounded by a conical surface (lateral surface) whose directrix is a closed
curve, and a plane (base) which cuts all the elements.
vertex
h Volume, V = ⅓ B h
B
directrix
Right Circular Cone
- A right circular cone is a circular cone whose axis is perpendicular to its base.
vertex
Volume, V = ⅓ Bh
Lateral area, AL = ½ circumference of base x
L h slant height
B AL = ½ CL
r
Total surface area, As = B + AL
Properties:
a. The slant height of a right circular cone is the length of an element.
b. The altitude of a right circular cone is the distance between the vertex and the center of the
circle which forms its base.
c. All elements of a right circular cone are equal.
d. A section of a right circular cone parallel to the base is a circle whose center is on the axis
of the cone.
e. A section of a right circular cone which contains the vertex and two points of the base is an
isosceles triangle.
EXAMPLES
1. Find the volume of a cone to be constructed from a sector having a diameter of 72 cm and a
central angle of 150°.
150° S L = 36 h
R = 36
r
36 (150°) π
V = ⅓Bh ; S = Rθ = = 30 π
180°
C = S
2 π r = 30 π
r = 15 cm
h2 = 362 – 152
h = 32.726 cm
V = ⅓ [π(15)2](32.726) = 7710.88 cm3
2. Compute the surface area of the cone having a slant height of 5 cm and a diameter of the base
equal to 6 cm.
L=5
D=6
AL = ½ CL = ½ [2π (3)](5)
AL = 47.124 cm2
3. A pile of sand is in the form of a right circular cone of altitude 2.15 m and a slant height of 7.60
m. What is the weight of the sand, if the sand weighs 16.95 kN/m3.
L = 7.60 h = 2.15 m
W = γV ; V = ⅓ Bh ; r2 = 7.602 – 2.152
r = 7.29 m
V = ⅓ [ π (7.29) ] (2.15)
2
= 119.65 m3
W = 16.95 (119.65) = 2028.07 kN
CE Board Nov. 2000
The axis of the cone makes an angle of 60° with the horizontal. If the length of the axis is 30 cm
and its base radius is 20 cm, compute for the volume of the cone.
h
60°
r = 20
V = ⅓ Bh ; h = 30 sin 60° = 25.98 m
= ⅓ [ π (20)2] (25.98)
V = 10882.48 m3
CE Board Nov. 2002
A cylindrical cone having a diameter of its base equal to 1 m. has a height of 60 cm. If it is filled
with salt to a height of 36 cm, compute the volume of the salt.
0.50 m
r 0.60 m
0.36 m
r 0.36
V = ⅓ Bh ; B = π r2 ; =
0.50 0.60
r = 0.30
= ⅓ [ π (0.30)2] (0.36)
V = 0.0339 m3 (volume of salt)
OR
Vs (0.36)3
= ; Vc = ⅓ [ π (0.50)2] (0.60) = 0.157
Vc (0.60)3
Vs (0.36)3
= ; Vs = 0.0339 m3 (volume of salt)
3
0.157 (0.60)
CE Board May 2004
A closed conical vessel has its horizontal base at the bottom and its axis vertical. The water in it
stands 0.50 m deep. The base radius is 1 m and the altitude is 2.50 m. The vessel’s position was
changed such that the horizontal base was uppermost. Find the depth of the water from the
bottom.
1m
2m
V1 2.5 m
V2 h
V2 0.5 m
1m
V1 23
= ; Vc = ⅓ [ π (1)2] (2.50) = 2.618
Vc (2.50)3
V1 23
= ; V1 = 1.34
3
2.618 (2.50)
V2 = Vc – V1 = 2.618 – 1.34 = 1.278
1.278 h3
=
2.618 (2.50)3
1.278 (2.50)3
3
h = ; h = 1.97 m
2.618
CE Board May 2009
A plane parallel to the base of a right circular cone cuts the cone ‘y’ meter from the vertex. Find the
value of ‘y’ in meters if the volume of the smaller cone is ⅓ times the volume of the big and the
height of the big cone is 9 m.
y
V1 9m
r
R
V1 = ⅓ Vc
⅓ (π r2 y) = ⅓ [⅓ π R2 (9)]
⅓ r2 y = R2
3 R2 R r
y = ; =
r2 9 y
9r
R =
y
3 (9 r / y)2 243 r2 / y2 243
y = = =
r2 r2 y2
y3 = 243
y = 6.24 m
OR
V1 y3
= ; Vc = ⅓ [ π R2 (9)] = 3 π R2
Vc 93 V1 = ⅓ Vc = ⅓ (3 π R2) = π R2
π R2 y3
=
3πR 2
93
3
9
y3 =
3
y = 6.24 m
CE Board May 2011
A cone is inscribed in a 1 m x 1 m x 1 m cube. Find the volume of said cone.
1m
1m
1m
V = ⅓Bh
= ⅓ [ π (0.50)2] (1)
V = 0.26 m3
CE Board May 2012
A cone has a diameter of 80 mm and a perpendicular height of 120 mm. Calculate its volume.
V = ⅓Bh h = 120 mm
= ⅓ [ π⁄4 (80)2] (120)
= 201061.93 mm3
V = 201.06 cu. cm
80 mm
CE Board May 2012
If a cone has a height perpendicular to its base of 120 cm and its base has a diameter of 80 cm,
what is the curved surface area?
AL = ½ C L ; L2 = 1202 + 402
L = 126.49 cm L h = 120 cm
C = πD
= 80 π
AL = ½ (80 π)(126.49)
AL = 15895.20 cm2
80 cm
CE Board May 2014
You are given a particular cone whose diameter is 12 cm and whose height is 8 cm. Find its lateral
area.
AL = ½ C L ; L2 = 82 + 62
L = 10 cm L
C = πD h = 8 cm
= 12 π
AL = ½ (12 π)(10)
AL = 60 π cm2
12 cm
CE Board May 2014
A conical monument has a base with a diameter of 8.0 meters. Its height perpendicular to the base
is 12.0 meters. What is the area of its curved surface, in square meters?
AL = ½ C L ; L2 = 122 + 42
L = 12.65 m L
C = πD h = 12 m
= 8π
AL = ½ (8 π)(12.65)
AL = 158.96 m2
8m
CE Board May 2014
You are given two solids of revolutions, a right circular cone and a right circular cylinder, both of
which have the same radius and the same volume. The cone has a height of 18 cm. What is the
height of the cylinder in cm?
⅓ π r2 (18) = π r2 h
h = 6 cm 18 cm
h
r r
CE Board May 2015
A sector is bent to form a cone. If the angle of the sector is 60º with a radius of 6 cm, what would
be the radius of the cone if it has a slant height of 6 cm?
L=6
h
60° S
R=6
6 (60°) π
S = Rθ = = 2π
180°
C = S
2πr = 2 π
r = 1 cm
