Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) =...

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Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2) × (4.5) = 20

Transcript of Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) =...

Page 1: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4

V = Bh = [(12)(5)]/2 × (4) = 120

Find the volume of each prism.

V = Bh = π(52) × (10) = 250π

V = Bh = π(62) × (9) = 324π

V = Bh = (7.3)(6.2) × (4.5) = 203.7

Page 2: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4Surface Areas of Pyramids & Cones

Standard 9.0Students compute the surface areas of

pyramids and cones and commit to memory the formulas for pyramids.

Learning Target:I will be able to solve problems involving the surface area of pyramids and cones.

Ch 10.5Ch 12.4

Page 3: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

regular pyramidThe altitude is perpendicular to the base at its center AND the base is a regular polygon.The lateral faces form congruent isosceles triangles.

slant heightThe height of each lateral face represented by l.This is different than the altitude.

Ch 12.4

Page 4: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4

Theorem 12-7

Page 5: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Lateral Area of a Regular Pyramid

Find the lateral area of the square pyramid.

Lateral area of a regularpyramid

P = 2.5 × 4 , ℓ = 5

Answer: The lateral area is 25 cm2.

Ch 12.4

Multiply.= 25

= (10)(5)12

Page 6: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

A. 54 in2

B. 64 in2

C. 108 in2

D. 132 in2

Find the lateral area of the square pyramid.

Ch 12.4

Lateral area of a pyramid

P = (4 × 4) , ℓ = 8

Multiply.

= (16)(8)

= 64

L = Pℓ1212

Page 7: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4

Theorem 12-8

Page 8: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

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 = 62 + 42 a = 6, b = 4, and c = ℓ

ℓ = Simplify.

Ch 12.4

Page 9: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Surface Area of a Square Pyramid

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

__12

__12

= (32) + 64 P = 32, ℓ = , B = 64 ≈ 179.4 Use a calculator.

Ch 12.4

Find the surface area of the square pyramid to the nearest tenth.

Page 10: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

A. 96 in2

B. 124.3 in2

C. 138.5 in2

D. 156 in2

Find the surface area of the square pyramid to the nearest tenth.

Ch 12.4

Use a calculator.

= (12) (√55) + 36

≈ 138.5

P = (6 × 4) , ℓ = √(82 - 32) , B = s2 , s = 6

= (24) (√55) + (62)

S = Pℓ + B Surface area of a pyramid

__1212

Simplify.

Page 11: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Example 3

Surface Area of a Regular Pyramid

Step 1 Find the perimeter of the base.

P = 6 × 10.4 = 62.4 cm

Find the surface area of the regular pyramid. Round to the nearest tenth.

Ch 12.4

9

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.

= 280.8 Multiply.

Page 12: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Surface Area of a Regular Pyramid

Ch 12.4

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) + 280.8 P = 62.4, ℓ = 15, and B = 280.8

= 748.8 Simplify.

Page 13: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

A. 198 in2

B. 228.5 in2

C. 255.5 in2

D. 316.3 in2

Find the surface area of the regular pyramid. Round to the nearest tenth.

Ch 12.4

= 255.5

P = (6 × 6) , ℓ = 9, B = aP , a = 2.6 2

= (36) (9) + (93.5)

S = Pℓ + B Surface area of a pyramid

__1212

Simplify.

2.6

Page 14: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4

Theorem 12-9 & 12-10

Page 15: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

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.

Ch 12.4

Step 1 Find the slant height ℓ.

ℓ2 = 82 + 1.252 Pythagorean Theorem

ℓ2 ≈ 65.56 Simplify.

ℓ ≈ 8.1 Take the square root ofeach side.

Page 16: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Lateral Area of a Cone

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.

Ch 12.4

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

Page 17: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

A. 12.6π in.2

B. 13.9π in.2

C. 15.3π in.2

D. 16.9π in.2

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.

Ch 12.4

Lateral area of a pyramid

P = 2π r, r = 2, ℓ = √(62 + 22)

Multiply.

= (2π)(2) (√52)

= 2π√40

L = Pℓ1212

Use a calculator.= 12.6π

Page 18: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

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

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 20.2 square centimeters. This is close to the estimate, so the answer is reasonable.

Ch 12.4

S = Pℓ + B Surface area of a cone__12

12

= (2π r) ℓ + π r2

Page 19: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

A. 18.5π cm2

B. 19.5π cm2

C. 20.2π cm2

D. 22.5π cm2

Find the surface area of the cone in terms of π

Ch 12.4

S = Pℓ + B Surface area of a cone__12

P = 2π r , B = π r2

= (3)(4.5) + (3)2 r = 3 , ℓ = 4.5

= 22.5π Simplify

12

= (2π r) ℓ + π r2

Page 20: Ch 12.4 V = Bh = [(12)(5)]/2 × (4) = 120 Find the volume of each prism. V = Bh = π(5 2 ) × (10) = 250π V = Bh = π(6 2 ) × (9) = 324π V = Bh = (7.3)(6.2)

Ch 12.4