10-5 Review for Mastery Surface Area of Pyramids and Cones
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10 -5 Area of Pyramids and Cones Objectives Larn and apply the formula for the surface area of a pyramid. Learn and employ the formula for the surface surface area of a cone. Holt Geometry
10 -five Surface Area of Pyramids and Cones The vertex of a pyramid is the point opposite the base of the pyramid. The base of a regular pyramid is a regular polygon, and the lateral faces are congruent isosceles triangles. The slant summit of a regular pyramid is the altitude from the vertex to the midpoint of an edge of the base. The distance of a pyramid is the perpendicular segment from the vertex to the airplane of the base of operations. Holt Geometry
10 -v Surface Area of Pyramids and Cones Holt Geometry
ten -v Surface Area of Pyramids and Cones Example i A: Finding Lateral Area and Surface Area of Pyramids Find the lateral area and surface surface area of a regular foursquare pyramid with base edge length 14 cm and slant pinnacle 25 cm. Round to the nearest tenth, if necessary. Lateral area of a regular pyramid P = 4(xiv) = 56 cm Surface area of a regular pyramid B = 142 = 196 cm 2 Holt Geometry
10 -5 Surface area of Pyramids and Cones Example 1 B: Finding Lateral Surface area and Surface Area of Pyramids Find the lateral area and surface surface area of the regular pyramid. Step 1 Discover the base perimeter and apothem. The base perimeter is half dozen(ten) = 60 in. The apothem is Holt Geometry , so the base of operations surface area is
10 -5 Surface Area of Pyramids and Cones Example 1 B Continued Find the lateral expanse and surface area of the regular pyramid. Stride 2 Find the lateral area. Lateral area of a regular pyramid Substitute 60 for P and 16 for ℓ. Holt Geometry
10 -v Surface Surface area of Pyramids and Cones Instance 1 B Continued Find the lateral area and surface area of the regular pyramid. Step 3 Detect the surface expanse. Surface expanse of a regular pyramid Substitute Holt Geometry for B.
10 -5 Surface Surface area of Pyramids and Cones The vertex of a cone is the betoken opposite the base. The axis of a cone is the segment with endpoints at the vertex and the eye of the base of operations. The axis of a correct cone is perpendicular to the base. The axis of an oblique cone is not perpendicular to the base. Holt Geometry
ten -5 Area of Pyramids and Cones The camber summit of a right cone is the distance from the vertex of a correct cone to a point on the edge of the base of operations. The altitude of a cone is a perpendicular segment from the vertex of the cone to the airplane of the base. Holt Geometry
x -5 Surface Area of Pyramids and Cones Example 2 A: Finding Lateral Area and Surface Area of Right Cones Find the lateral surface area and surface area of a right cone with radius 9 cm and slant peak 5 cm. L = rℓ Lateral expanse of a cone = (9)(5) = 45 cm two Substitute 9 for r and v for ℓ. S = rℓ + r two Area of a cone = 45 + (9)2 = 126 cm 2 Substitute v for ℓ and 9 for r. Holt Geometry
10 -5 Surface Expanse of Pyramids and Cones Example 2 B: Finding Lateral Surface area and Surface Surface area of Right Cones Detect the lateral expanse and surface expanse of the cone. Utilise the Pythagorean Theorem to detect ℓ. L = rℓ Lateral area of a right cone = (8)(17) Substitute 8 for r and 17 for ℓ. 2 = 136 in S = rℓ + r 2 = 136 + (viii)2 = 200 in 2 Holt Geometry Surface area of a cone Substitute 8 for r and 17 for ℓ.
x -5 Surface Surface area of Pyramids and Cones Instance three Continued iii in. 2 in. original dimensions: base edge length and slant height divided by 5: Find that . If the base border length and slant tiptop are divided by five, the expanse is divided by 52, or 25. Holt Geometry
x -v Surface Surface area of Pyramids and Cones Check It Out! Example 3 The base edge length and camber height of the regular square pyramid are both multiplied past. Describe the effect on the surface area. Holt Geometry
10 -five Surface Area of Pyramids and Cones Bank check Information technology Out! Example three Continued 8 ft 10 ft original dimensions: 1 Southward = 2 Pℓ + B = 585 cm 2 multiplied by 2-thirds: Southward= one Pℓ 2 +B = 260 cm 2 Past multiplying the dimensions by 2-thirds, the surface area was multiplied by. Holt Geometry
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