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Chapter 12: Areas and Volumes of Solids

Chapter 12: Areas and Volumes of Solids. Margot Romano and Gaby Henriquez. 12-1 Prisms. The two shaded faces of the prisms shown are its bases An altitude of a prism is a segment joining the two base planes and perpendicular to both

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Chapter 12: Areas and Volumes of Solids

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  1. Chapter 12: Areas and Volumes of Solids Margot Romano and Gaby Henriquez

  2. 12-1 Prisms • The two shaded faces of the prisms shown are its bases • An altitude of a prism is a segment joining the two base planes and perpendicular to both • The faces of a prism that are not its bases are called lateral faces • Adjacent lateral faces intersect in parallel segments called lateral edges • If the lateral faces of a prism are rectangles the prism is a right prism; otherwise the prism is an oblique prism

  3. The surface are of a solid is measured in square units • T.A = L.A + 2B • The lateral area (L.A.) of a prism is the sum of the areas of its lateral faces • The total area (T.A.) is the sum of the areas of all its faces • Theorem 12-1 • The lateral area of a right prism equals the perimeter of a base times the height of the prism (L.A. = ph) • Examples: • L.A = ah + bh + ch + dh + eh • = (a + b + c + d + e)h = Perimeter(h) = ph

  4. Theorem 12-2 • The volume of a right prism equals the area of a base times the height of the prism (V=Bh) • Example 1 • Volume = Base Area X height = (4 x 2)(3) = 24 cubic units • Example 2 • A right trapezoid prism is show. Find the (a) lateral area, (b) total area, and (c) volume • (A)First find the perimeter of a base. p = 5 + 6 + 5 + 12 = 28 (cm) L.A. = ph = 28 x 10 = 280 (cm) • (B) First find the area of a base B = ½ x 4 x (12 + 6) = 36 (cm) T.A = L.A + 2B = 280 + 2 x 36 = 352 (cm) • (C) V = Bh= 36 x 10 = 360 (cm)

  5. Practice Problems • A brick with dimensions 20 cm, 10 cm, and 5 cm weighs 1.2 kg. A second brick of the same material has dimensions 25 cm, 15 cm, and 4 cm. What is its weight? • Find its lateral area, total area, and volume • All nine edges of a right triangular prism are congruent. Find the length of these edges if the volume is 54√3cm.

  6. 12-2: Pyramids • In the diagram shown, point v is the vertex of the pyramid and pentagon ABCDE is the base. • The segment from the vertex perpendicular to the base is the altitude and its length is the height (h) of the pyramid • The 5 triangular faces with V in common, such as VAB, are lateral faces. These faces intersect in segments called lateral edges.

  7. Properties of regular pyramids • The base is a regular polygon • All lateral edges are congruent • All lateral faces are congruent isosceles triangles. The height of a lateral face is called the slant height of the pyramid (denoted by l) • The altitude meets the base at its center, O. • Examples: • Theorem 12-3 • the lateral area of a regular pyramid equals half the perimeter of the bases times the slant height (L.A=1/2pl) • Examples L.A. = (1/2bl)n = ½(nb)l Since nb = p L.A = ½pl

  8. Theorem 12-4 • The volume of a pyramid equals one third the area of the base times the height of the pyramid (V= 1/3Bh) • Examples: • Find the area of the hexagonal base • Divide the base into six equilateral triangle. • Find the area of one triangle and multiply by 6 • Base area = B = 6(1/2 x 6 x 3√30 = 54√3 • Then V = 1/3Bh = 1/3x 54√3 x 12 = 216√3

  9. Practice Problems • A pyramid has a base area of 16 cm and a volume of 32 cm. Find its height • If h = 4 and l = 5, find OM, OA and BC. Also find the lateral area and the volume.

  10. 12-3: Cylinders and Cones • A cylinder is like a prism except its bases are circles instead of polygons • In a right cylinder, the segment joining the centers of the circular bases is an altitude. • The length of the altitude is called the height (h) of the cylinder • A radius of a base is also called the radius (r) of a cylinder

  11. Theorem 12-5 • The lateral area of a cylinder equals the circumference of a base times the height of the cylinder (L.A= 2πrh) • Theorem 12-6 • The volume of a cylinder equals the area of a base times the height of the cylinder V=π(r^2)h Examples: A cylinder has a radius 5 cm and height 4 cm. Find the (a) lateral, (b) total area, and (c) volume of the cylinder • L.A = 2πrh = 2π x 5 x 4 = 40π • T.A = L.A + 2B = 40π + 2(π x 5^2) = 90π • V = π(r^2)h = π x 5^2 x 4 = 100π

  12. A cone is like a pyramid except that its base is a circle instead of a polygon. • Theorem 12-7 • The lateral area of a cone equals half the circumference of a base times the slant height • Theorem 12-8 • The volume of a cone equals one third the area of the base times the height of the cone • Examples: • Find the (a) lateral area, (b) total area, and (c) volume of the cone shown. • First use the Pythagorean Theorem to find l • L = (√6^2 + 3^2) = (√45) = 3√5 • L.A. = πrl = π x 3 x 3 √5 = 9π√5 • T.A = L.A + B = 9π√5 + π x 3^2 = 9π√5 + 9π • V = 1/3π (r^2)h = 1/3π x 3^2 x 6 = 18π

  13. Practice Problems • 1. The volume of a cylinder is 64π. If r = h, find r. • 2. A cone and a cylinder both have height 48 and radius 15. Give the ratio of their volumes without calculating the two volumes. • 3. A cone is inscribed in a regular square pyramid with slant height 9 cm and base edge 6 cm. Make a sketch. Then find the volume of the cone.

  14. 12-4: Spheres • A sphere is a set of all points that are a given distances from a given point • Theorem 12-9 • The area of a sphere equals 4pi times the square of the radius • Theorem 12-10 The volume of a sphere equals 4/3pi times the cube of the radius • Example: • Find the area and the volume of a sphere with radius 2 cm • A = 4π(r^2) = 4π x 2^2 = 16π (cm^2) • V = 4/3π(r^3) = 4/3π x 2^3 = 32π/3 (cm^3)

  15. Practice Problems • 1. Find the area of the circle formed when a plane passes 2 cm from the center of a sphere with radius 5 cm. • 2. A circle with a diameter of 9 in. is rotated about diameter. Find the area and volume of the solid formed. • 3. A solid metal ball with radius 8 cm is melted down and recast as a solid cone with the same radius. What is the height of the cone?

  16. 12-5: Areas and Volumes of Similar Solids • Similar solids are solids that have the same shape but not necessarily the same size. • Theorem 12-11 • If the scale factor of two similar solids is a:b then • The ratio of corresponding perimeters is a:b • The ratio of the base areas, of the lateral areas, and of the total areas is a squared :b squared • The ratio of the volumes is a cubed: b cubed • Examples: • For the similar solids shown, find the ratios of the (a) base perimeters, (b) lateral areas, and (c) volumes • The scale factor is 6 : 10, or 3 : 5 • Ratio of base perimeters = 3 : 5 • Ratio of lateral areas = (3^2) : (5^2)= 9 : 25 • Ratio of volume = (3^3) : (5^3) = 27 : 125

  17. Practice Problems • 1. Assume that the Earth and the moon are smooth spheres with diameters 12,800 km and 3,200 km, respectively. Find the ratios of the following • Lengths of their equators • Areas • volumes • 2. Two similar cones have radii of 4 cm and 6 cm. The total area of the smaller cone is 36π cm^2. Find the total area of the larger cone. • 3. A pyramid with height 15 cm is separated into two pieces by a plane parallel to the base and 6 cm above it. What are the volumes of these two pieces if the volume of the original pyramid is 250 cm^3.

  18. Answer Key • 12-1 • 2) 1.8 kg • 3) 6cm • 12-2 • 1) 6cm • 2) 3; 6; 6√3 • 45√3; 36√3 • 12-3 • 1) 4 • 2) 1 : 3 • 3) 18π√2

  19. Answer Key • 12-4 • 1) 21π cm^3 • 2) 81π in^2 ; 121.5π in^3 • 3) 32 cm • 12-5 • 1) a, 4 : 1 b, 16 : 1 c, 64 : 1 • 2) 81π cm^2 • 3) 54 cm^3; 196 cm^3

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