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An Affirmation. Please read the following and consider yourself in it. I am capable of learning. I can accomplish mathematical tasks. I am ultimately responsible for my learning. Content Standards MGSE9-12.G.GMD.1 Give informal arguments for geometric formulas.
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An Affirmation Please read the following and consider yourself in it. I am capable of learning. I can accomplish mathematical tasks. I am ultimately responsible for my learning.
Content Standards MGSE9-12.G.GMD.1 Give informal arguments for geometric formulas. a) Give informal arguments for the formulas of the circumference of a circle and area of a circle using dissection arguments and informal limit arguments. b) Give informal arguments for the formula of the volume of a cylinder, pyramid, and cone using Cavalieri’s principle. MGSE9-12.G.GMD.2 Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. MGSE9-12.G.GMD.3 Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. Visualize relationships between two-dimensional and three-dimensional objects MGSE9-12.G.GMD.4 Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. CCSS
Objectives: For each objective write down the important words/phrases each one contains. • G.GMD.1-2 – SWBAT apply Cavalieri’s principle IOT formulate an informal proof for the volume of a cylinder, pyramid, cone, sphere and other solids. • G.GMD.4 – SWBAT identify the shapes of two-dimensional cross-sections of three-dimensional objects IOT evaluate the volume of the solid. • G.GMD.4 – SWBAT identify three-dimensional objects generated by rotations of two-dimensional objects IOT derive formulas for volumes of solids. • G.GMD.3 – SWBAT apply and calculate volumes of cylinders, pyramids, cones, and spheres IOT solve problems. Then/Now
Important Formulas! This is our Anchor Slide All our work will come back to these formulas! Concept
lateral face • lateral edge • base edge • altitude • height • lateral area • axis • composite solid Vocabulary
Lateral Area of a Prism Find the lateral area of the regular hexagonal prism. The bases are regular hexagons. So the perimeter of one base is 6(5) or 30 centimeters. Lateral area of a prism P=30, h=12 Multiply. Answer: Example 1
Lateral Area of a Prism Find the lateral area of the regular hexagonal prism. The bases are regular hexagons. So the perimeter of one base is 6(5) or 30 centimeters. Lateral area of a prism P=30, h=12 Multiply. Answer: The lateral area is 360 square centimeters. Example 1
Surface Area of a Prism Find the surface area of the rectangular prism. Example 2
Surface Area of a Prism Surface area of a prism L=Ph Substitution Simplify. Answer: Example 2
Surface Area of a Prism Surface area of a prism L=Ph Substitution Simplify. Answer: The surface area is 360 square centimeters. Example 2
Lateral Area and Surface Area of a Cylinder Find the lateral area and the surface area of the cylinder. Round to the nearest tenth. L = 2rh Lateral area of a cylinder = 2(14)(18) Replace r with 14 and h with 18. ≈ 1583.4 Use a calculator. Example 3
Lateral Area and Surface Area of a Cylinder S = 2rh + 2r2Surface area of a cylinder ≈ 1583.4 + 2(14)2 Replace 2rh with 1583.4 and r with 14. ≈ 2814.9 Use a calculator. Answer: Example 3
Lateral Area and Surface Area of a Cylinder S = 2rh + 2r2Surface area of a cylinder ≈ 1583.4 + 2(14)2 Replace 2rh with 1583.4 and r with 14. ≈ 2814.9 Use a calculator. Answer: The lateral area is about 1583.4 square feet and the surface area is about 2814.9 square feet. Example 3
Find Missing Dimensions MANUFACTURINGA soup can is covered with the label shown. What is the radius of the soup can? L = 2rh Lateral area of a cylinder 125.6 = 2r(8) Replace L with 15.7 ● 8 and h with 8. 125.6 = 16r Simplify. 2.5 ≈ r Divide each side by 16. Example 4
Find Missing Dimensions Answer: The radius of the soup can is about 2.5 inches. Example 4
Find the lateral area of the cylinder. A. 65.3 ft2 B. 80.2 ft2 C. 122.5 ft2 D. 130.7 ft2 5-Minute Check 1
Find the lateral area of the cylinder. A. 65.3 ft2 B. 80.2 ft2 C. 122.5 ft2 D. 130.7 ft2 5-Minute Check 1
Find the lateral area of the prism. A. 150 in2 B. 126 in2 C. 108 in2 D. 96 in2 5-Minute Check 2
Find the lateral area of the prism. A. 150 in2 B. 126 in2 C. 108 in2 D. 96 in2 5-Minute Check 2
Find the surface area of the prism. A. 320 mm2 B. 400 mm2 C. 494.2 mm2 D. 502 mm2 5-Minute Check 3
Find the surface area of the prism. A. 320 mm2 B. 400 mm2 C. 494.2 mm2 D. 502 mm2 5-Minute Check 3
Find the surface area of the cylinder. A. 962.5 cm2 B. 1005.3 cm2 C. 1243.6 cm2 D. 1407.4 cm2 5-Minute Check 4
Find the surface area of the cylinder. A. 962.5 cm2 B. 1005.3 cm2 C. 1243.6 cm2 D. 1407.4 cm2 5-Minute Check 4
The lateral area of a prism is 476 square meters. The area of one of its bases is 25 square meters. What is the surface area of the prism? A. 426 m2 B. 451 m2 C. 501 m2 D. 526 m2 5-Minute Check 5
The lateral area of a prism is 476 square meters. The area of one of its bases is 25 square meters. What is the surface area of the prism? A. 426 m2 B. 451 m2 C. 501 m2 D. 526 m2 5-Minute Check 5
regular pyramid • slant height • right cone • oblique cone • apothem Vocabulary
Lateral Area of a Regular Pyramid Find the lateral area of the square pyramid. Lateral area of a regular pyramid P = 2.5 ● 4 or 10, ℓ = 5 Answer: Example 1
Lateral Area of a Regular Pyramid Find the lateral area of the square pyramid. Lateral area of a regular pyramid P = 2.5 ● 4 or 10, ℓ = 5 Answer: The lateral area is 25 square centimeters. Example 1
ℓ = Simplify. 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 + 42a = 6, b = 4, and c = ℓ Example 2
S = Pℓ + B Surface area of a regular pyramid 1 1 __ __ = (32) + 64 P = 32, ℓ = , B = 64 2 2 Surface Area of a Square Pyramid Step 2 Find the perimeter and area of the base. P = 4 ● 8 or 32 m A = 82 or 64 m2 Step 3 Find the surface area of the pyramid. ≈ 179.4 Use a calculator. Answer: Example 2
S = Pℓ + B Surface area of a regular pyramid 1 1 __ __ = (32) + 64 P = 32, ℓ = , B = 64 2 2 Surface Area of a Square Pyramid Step 2 Find the perimeter and area of the base. P = 4 ● 8 or 32 m A = 82 or 64 m2 Step 3 Find the surface area of the pyramid. ≈ 179.4 Use a calculator. Answer: The surface area of the pyramid is about 179.4 square meters. Example 2
Find the surface area of the square pyramid to the nearest tenth. A. 96 in2 B. 124.3 in2 C. 138.5 in2 D. 156 in2 Example 2
Find the surface area of the square pyramid to the nearest tenth. A. 96 in2 B. 124.3 in2 C. 138.5 in2 D. 156 in2 Example 2
Surface Area of a Regular Pyramid Find the surface area of the regular pyramid. Round to the nearest tenth. Step 1 Find the perimeter of the base. P = 6 ● 10.4 or 62.4 cm Example 3
A central angle of the hexagon is or 60°, so the angle formed in the triangle is 30°. 360° ______ 6 Surface Area of a Regular Pyramid Step 2 Find the length of the apothem and the area of the base. Example 3
tan 30° = Write a trigonometric ratio to find the apothem a. a = Solve for a. A= Pa Area of a regular polygon ≈ (62.4)(9.0) Replace P with 62.4 and a width 9.0. Surface Area of a Regular Pyramid a≈ 9.0 Use a calculator. ≈ 280.8 Multiply. So, the area of the base is approximately 280.8 cm2. Example 3
S = Pℓ + B Surface area of regular pyramid ≈ (62.4)(15) + 280.8 P = 62.4, ℓ = 15, and B ≈ 280.8 Surface Area of a Regular Pyramid Step 3 Find the surface area of the pyramid. ≈ 748.8 Simplify. Answer: Example 3
S = Pℓ + B Surface area of regular pyramid ≈ (62.4)(15) + 280.8 P = 62.4, ℓ = 15, and B ≈ 280.8 Surface Area of a Regular Pyramid Step 3 Find the surface area of the pyramid. ≈ 748.8 Simplify. Answer: The surface area of the pyramid is about 748.8 cm2. Example 3
Find the surface area of the regular pyramid. Round to the nearest tenth. A. 198 in2 B. 228.5 in2 C. 255.5 in2 D. 316.3 in2 Example 3
Find the surface area of the regular pyramid. Round to the nearest tenth. A. 198 in2 B. 228.5 in2 C. 255.5 in2 D. 316.3 in2 Example 3
ICE CREAM A sugar cone has an altitude of 8 inches and a diameter of inches. Find the lateral area of the sugar cone. If the cone has a diameter of then the radius Use the altitude and the radius to find the slant height with the Pythagorean Theorem. Lateral Area of a Cone Example 4
Lateral Area of a Cone Step 1 Find the slant height ℓ. ℓ2 = 82 + 1.252 Pythagorean Theorem ℓ2 ≈ 65.56Simplify. ℓ≈ 8.1Take the square root of each side. Step 2 Find the lateral area L. L= rℓLateral area of a cone ≈(1.25)(8.1) r = 1.25 and ℓ ≈ 8.1 ≈ 31.8 Answer: Example 4
Lateral Area of a Cone Step 1 Find the slant height ℓ. ℓ2 = 82 + 1.252 Pythagorean Theorem ℓ2 ≈ 65.56Simplify. ℓ≈ 8.1Take the square root of each side. Step 2 Find the lateral area L. L= rℓLateral area of a cone ≈(1.25)(8.1) r = 1.25 and ℓ ≈ 8.1 ≈ 31.8 Answer: The lateral area of the sugar cone is about 31.8 square inches. Example 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. 39.7 in.2 B. 43.6 in.2 C. 48.2 in.2 D. 53.1 in.2 Example 4