Presentation on theme: "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."— Presentation transcript:
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.