 # 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.

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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

 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

 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)

1. 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 2. All nine edges of a right triangular prism are congruent. Find the length of these edges if the volume is 54√3cm.

 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.

 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

 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 = 1/3 x 54√3 x 12 = 216√3

1. A pyramid has a base area of 16 cm and a volume of 32 cm. Find its height 2. If h = 4 and l = 5, find OM, OA and BC. Also find the lateral area and the volume.

 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

 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 a. L.A = 2πrh = 2π x 5 x 4 = 40π b. T.A = L.A + 2B = 40π + 2(π x 5^2) = 90π c. V = π(r^2)h = π x 5^2 x 4 = 100π

 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 a)L.A. = πrl = π x 3 x 3 √5 = 9π√5 b)T.A = L.A + B = 9π√5 + π x 3^2 = 9π√5 + 9π c)V = 1/3π (r^2)h = 1/3π x 3^2 x 6 = 18π

 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.

 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)

 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?

 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

 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.

 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

 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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