1. 3-D Shapes Topic 14: Lesson 9
3-D Shapes Topic 14: Lesson 9 Surface Area of Cones Holt Geometry Texas ©2007

2. Objectives and Student Expectations
TEKS: G2B, G3B, G6B, G8D, G11D
The student will make conjectures about 3-D figures and determine the validity using a variety of approaches.
The student will construct and justify statements about geometric figures and their properties.
The student will use nets to represent and construct 3-D figures.
The student will find surface area and volume of prisms, cylinders, cones, pyramids, spheres, and composite figures.
The student will describe the effect on perimeter, area, and volume when one or more dimensions of a figure are changed.

3. The vertex is the point opposite the base
The vertex is the point opposite the base. The height is the segment with endpoints at the vertex and the center of the base. The axis of a cylinder is perpendicular to the base. The axis of an oblique cone is not perpendicular to the base.

4. The slant height of a right cone is the distance from the vertex of a right cone to a point on the edge of the base. The altitude of a cone is a perpendicular
segment from the vertex of the cone to the plane of the base.

5. Example: 1
Find the lateral area and surface area of a right cone with radius 9 cm and slant height 5 cm.
L = rℓ
Lateral area of a cone
L = (9)(5)
L = 45 cm2
Substitute 9 for r and 5 for ℓ.
S = rℓ + r2
Surface area of a cone
S = 45 + (9)2
S = 126 cm2
Substitute 5 for ℓ and 9 for r.

6. Example: 2 Find the lateral area and surface area of the cone.
Use the Pythagorean Theorem to find ℓ.
L = rℓ
Lateral area of a right cone
L = (8)(17)
L = 136 in2
Substitute 8 for r and 17 for ℓ.
S = rℓ + r2
Surface area of a cone
S = 136 + (8)2
S = 200 in2
Substitute 8 for r and 17 for ℓ.

7. Example: 3 Find the surface area of the composite figure.
Left-hand cone:
The lateral area of the cone is
L = rl = (6)(12) = 72 in2.
Right-hand cone:
Using the Pythagorean Theorem, l = 10 in.
The lateral area of the cone is
L = rl = (6)(10) = 60 in2.
Composite figure: 
S = (left cone lateral area) + (right cone lateral area)
S = 60 in2 + 72 in2 = 132 in2

8. Example: 4 Find the lateral area and surface area of the right cone.
Use the Pythagorean Theorem to find ℓ. ℓ
L = rℓ
Lateral area of a right cone
L = (8)(10)
L = 80 cm2
Substitute 8 for r and 10 for ℓ.
S = rℓ + r2
Surface area of a cone
S = 80 + (8)2
S = 144 cm2
Substitute 8 for r and 10 for ℓ.

9. Example: 5
If the pattern shown is used to make a paper cup, what is the diameter of the cup?
The radius of the large circle used to create the pattern is the slant height of the cone.
The area of the pattern is the lateral area of the cone. The area of the pattern is also of the area of the large circle, so

10. Example: 5 cont.
If the pattern shown is used to make a paper cup, what is the diameter of the cup?
Substitute 4 for ℓ, the slant height of the cone and the radius of the large circle.
r = 2 in.
Solve for r.
The diameter of the cone is 2(2) = 4 in.

11. Example: 6
What if…? If the radius of a large circle were 12 in., what would be the radius of the cone?
The radius of the large circle used to create the pattern is the slant height of the cone.
The area of the pattern is the lateral area of the
cone. The area of the pattern is also of the area of
the large circle, so
Substitute 12 for ℓ, the slant height of the cone and the radius of the large circle.
r = 9 in.
Solve for r.
The radius of the cone is 9 in.

12. Example: 7
A frustrum is a part of a cone that has been cut off by a plane that is parallel to the base of the cone. Find the surface area of the frustrum in the figure shown if the radius of the smaller cone is 3 cm with slant height 5 cm, and the radius of the larger cone is 9 cm with slant height 15 cm. Round to the nearest tenth.
frustrum

13. Example: 7 continued frustrum
The radius of the smaller cone is 3 cm with slant height 5 cm; radius of larger cone is 9 cm with slant height 15 cm.