1. Ch 12.4 Find the volume of each prism. V = Bh V = Bh
= [(12)(5)]/2 × (4) = 120
V = Bh
= π(52) × (10) = 250π
V = Bh
= π(62) × (9) = 324π
V = Bh
= (7.3)(6.2) × (4.5) = 203.7

2. Ch 12.4 Surface Areas of Pyramids & Cones
Learning Target:
I will be able to solve problems involving the surface area of pyramids and cones.
Standard 9.0
Students compute the surface areas of pyramids and cones and commit to memory the formulas for pyramids.

3. Ch 12.4
regular pyramid
The altitude is perpendicular to the base at its center AND the base is a regular polygon.
The lateral faces form congruent isosceles triangles.
slant height
The height of each lateral face represented by l.
This is different than the altitude.
Vocabulary

4. Ch 12.4
Theorem 12-7
Concept

5. Ch 12.4 Find the lateral area of the square pyramid.
Lateral Area of a Regular Pyramid
Find the lateral area of the square pyramid.
Lateral area of a regular pyramid
= (10)(5) 1 2
P = 2.5 × 4 , ℓ = 5
= 25
Multiply.
Answer: The lateral area is 25 cm2.
Example 1

6. Ch 12.4 Find the lateral area of the square pyramid. A. 54 in2
B. 64 in2
C. 108 in2
D. 132 in2 1 2
L = Pℓ
Lateral area of a pyramid 1 2
= (16)(8)
P = (4 × 4) , ℓ = 8
= 64
Multiply.
Example 1

7. Ch 12.4
Theorem 12-8
Concept

8. Ch 12.4
Surface Area of a Square Pyramid
Find the surface area of the square pyramid to the nearest tenth.
Step 1 Find the slant height
c2 = a2 + b2 Pythagorean Theorem
ℓ2 = a = 6, b = 4, and c = ℓ
ℓ = Simplify.

9. Ch 12.4
Surface Area of a Square Pyramid
Find the surface area of the square pyramid to the nearest tenth.
Step 2 Find the perimeter and area of the base.
P = 4 × 8 = 32 m B = 82 = 64 m2
Step 3 Find the surface area of the pyramid.
S = Pℓ + B Surface area of a regular pyramid __ 1 2 __ 1 2
= (32) P = 32, ℓ = , B = 64
≈ Use a calculator.
Example 2

10. Ch 12.4
Find the surface area of the square pyramid to the nearest tenth.
A. 96 in2
B in2
C in2
D. 156 in2
S = Pℓ + B Surface area of a pyramid __ 1 2 1 2
= (24) (√55) + (62)
P = (6 × 4) ,
ℓ = √( ) ,
B = s2 , s = 6
= (12) (√55) + 36
Simplify.
≈ 138.5
Use a calculator.
Example 2

11. Ch 12.4
Surface Area of a Regular Pyramid
Find the surface area of the regular pyramid. Round to the nearest tenth. 9
Step 1 Find the perimeter of the base.
P = 6 × 10.4 = 62.4 cm
Step 2 Find the area of the base.
B = Pa Area of a regular polygon
= (62.4)(9.0) P = 10.4 × 6, a = 9.
= Multiply.
Example 3

12. Ch 12.4
Surface Area of a Regular Pyramid
Find the surface area of the regular pyramid. Round to the nearest tenth. 9
Step 3 Find the surface area of the pyramid.
S = Pℓ + B Surface area of regular pyramid
= (62.4)(15) P = 62.4, ℓ = 15, and B = 280.8
= Simplify.
Example 3

13. Ch 12.4
Find the surface area of the regular pyramid. Round to the nearest tenth.
2.6
A. 198 in2
B in2
C in2
D in2
S = Pℓ + B Surface area of a pyramid __ 1 2 1 2
= (36) (9) + (93.5)
P = (6 × 6) , ℓ = 9,
B = aP , a = 2.6 2
= 255.5
Simplify.
Example 3

14. Ch 12.4
Theorem 12-9 & 12-10
Concept

15. Ch 12.4
Lateral Area of a Cone
ICE CREAM A sugar cone has an altitude of 8 inches and a diameter of 2.5 inches. Find the lateral area of the sugar cone.
If the cone has a diameter of 2.5 inches then the radius is 2.5 ÷ 2. Use the altitude and the radius to find the slant height with the Pythagorean Theorem.
Step 1 Find the slant height ℓ.
ℓ2 = Pythagorean Theorem
ℓ2 ≈ Simplify.
ℓ ≈ 8.1 Take the square root of each side.
Example 4

16. Ch 12.4
Lateral Area of a Cone
ICE CREAM A sugar cone has an altitude of 8 inches and a diameter of 2.5 inches. Find the lateral area of the sugar cone.
Step 1 Find the slant height ℓ: ℓ ≈ 8.1
Step 2 Find the lateral area L.
L = rℓ Lateral area of a cone
= (1.25)(8.1) r = 1.25 and ℓ ≈ 8.1
= 10.1π Multiply
Answer: The lateral area of the sugar cone is about 31.8 in2.
Example 4

17. Ch 12.4
HATS A conical birthday hat has an altitude of 6 inches and a diameter of 4 inches. Find the lateral area of the birthday hat.
A. 12.6π in.2
B. 13.9π in.2
C. 15.3π in.2
D. 16.9π in.2 1 2
L = Pℓ
Lateral area of a pyramid 1 2
= (2π)(2) (√52)
P = 2π r, r = 2,
ℓ = √( )
= 2π√40
Multiply.
= 12.6π
Use a calculator.
Example 4

18. Ch 12.4 Find the surface area of the cone in terms of π.
Surface Area of a Cone
Find the surface area of the cone in terms of π.
Estimate: S ≈ (3 × 1.5 × 3) + (3 × 2) = 19.5 cm2
S = Pℓ + B Surface area of a cone __ 1 2 1 2
= (2π r) ℓ + π r2
P = 2π r , B = π r2
= (1.4)(3.2) + (1.4)2 r = 1.4 and ℓ = 3.2
= 6.56π Simplify.
Answer: The surface area of the cone is about square centimeters. This is close to the estimate, so the answer is reasonable.
Example 5

19. Ch 12.4 Find the surface area of the cone in terms of π A. 18.5π cm2
B. 19.5π cm2
C. 20.2π cm2
D. 22.5π cm2
S = Pℓ + B Surface area of a cone __ 1 2 1 2
= (2π r) ℓ + π r2
P = 2π r , B = π r2
= (3)(4.5) + (3)2 r = 3 , ℓ = 4.5
= 22.5π Simplify
Example 5

20. Ch 12.4
Concept