1. Gaby Pavia and Gaby Pages

2. Section 12-1 Bases: congruent polygons lying in parallel planes Altitude: segment joining the two base planes and perpendicular to both the length of the altitude is the height of the prism Lateral Edges: the intersection of adjacent lateral faces. Are parallel segments. Lateral Faces: the faces of a prism that are not its bases - The Lateral Faces of a prism are parallelograms Right Prism: a prism whose lateral faces are rectangles. Oblique Prism: a prism whose lateral face are parallelograms but not rectangles.

3. Examples of Prisms Right Prism Oblique Prism

4. Different Areas of a Prism Lateral Area of a Prism: the sum of the areas of its lateral faces Total Area of a Prism: the sum of the areas of all its faces

5. Finding Lateral Area of a Right Prism: Theorem 12-1: The lateral area of a right prism equals the perimeter of a base times the height of the prism. L.A = P x H ( P = perimeter of the base, H = height) Formula applies to any right prism

6. EXAMPLE : Find the Lateral Area of the Prism SOLUTION: L.A = P x H Perimeter : 10 x 2 + 6 x 2 = 32 Height : 5 L.A = 32 x 5 = 160 units 2

7. Finding the Total Area of a Right Prism: Formula : T.A = L.A + 2B B = the area of the base

8. EXAMPLE: finding total area SOLUTION: T.A = L.A + 2B L.A = P x H = Perimeter: 5 + 5 + 6 = 16 Height = 12 L.A = 192 Area of Base = (6 x 4) / 2 =12 T.A = 192 + 12 = 204 units 2

9. Finding the Volume of a Right Prism Theorem 12-2: The volume of a prism equals the area of a base times the height of the prism. V = B x H

10. Example: find the volume of this right prism SOLUTION: V = BH Area of Base: (8 x 15)/ 2 = 120 / 2 = 60cm 2 Height of Prism : 9cm V = 60 x 9 = 540cm 3

11. Section 12.2 : Pyramids Point S is the vertex of the pyramid Segment SO is the altitude of the pyramid The five triangular faces with point S in common are the lateral faces - ∆SBC, ∆SBA, ∆SAE, ∆SED, and ∆SDC The lateral faces intersect in segments called lateral edges : - Segments : SA, SE, SD, SC, and SB Segment SF is the slant height of the pyramid. It is the height of a lateral face, and detonated by l A pentagonal pyramid:

12. Regular Pyramids All regular pyramids have the same following 4 properties: 1 ) The base is a regular polygon 2 ) All lateral edges are congruent 3) All lateral faces are congruent isosceles triangles 4) The altitude meets the base at its center

13. A regular square pyramid has base edges 10cm and lateral edges 13cm. Find the slant height SOLUTION: Slant Height: 1) Drop an altitude to point F 2) Use Pythagorean Theorem: l = 13 2 – 5 2 = 144 = 12

14. Lateral Area of a pyramid Theorem 12-3 : The lateral area of a regular pyramid equals half the perimeter of the base times the slant height L.A : (1/2) P x l There are two methods to find the lateral area of a regular pyramid with n lateral faces Method 1: Find the area of one lateral face and multiply by n Method 2: Use the formula ( L.A = (1/2) P x L)

15. Example: Find the lateral area of the square pyramid Method 1 Solution: Area of Lateral Face : (6 x 8) / 2 = 48/2 = 24 24 x 4 = 96 units 2 Method 2 : L.A = (1/2) P x L Perimeter: 8 x 4 = 32 Slant Height : 6 L.A = (1/2)32 x 6 = 16 x 6 = 96 units 2

16. Finding the Volume of a Pyramid Since the volume of a prism is base x height, the volume of the pyramid must be less than the base x height. Theorem 12-4: The volume of a pyramid equals one third the area of the base times the height of the pyramid V= (1/3) x Area of Base x Height

17. Section 12-3: Cylinders Cylinder- a prism with bases as circles. In a right cylinder, the segments joining the centers of both circles is an altitude. The length of an altitude is the height of the cylinder. The radius of the base is the radius of the the cylinder.

18. Finding L.A, T.A and volume

19. Example 1: Find the Lateral Area SOLUTION: (Use 3.14 as π) Circumference of the Base: 2πr = 2π (5) = 31.4 Height : 8 L.A = 2 πr x height = 3.14 x 8 = 251.2 units 2

20. Example 2: Find the volume SOLUTION: Area of base: πr 2 = π (5) 2 = π x 25 = 78.5 Height : 15 V = πr 2 x height = V= 78.5 x 15 = 1177.5 units 3

21. Example 3 5 cm 4 cm A cylinder has radius 5 cm and height 4 cm. Find (a) the lateral area, (b) total area, and (c) volume of the cylinder.

22. 12-3 Cylinders and cones (cont.) Cones- similar to a pyramid except that its base is a circle instead of a polygon Slant height, l

23. Finding L.A, T.A and volume

24. Example Find the (a) lateral area, (b) total area, and (c) volume of the cone showed 3 6 l

25. Section 12.4 : Spheres Recall from Chapter 9 that a sphere is the set of all points that are a given distance from a given point

26. Finding the Area of a Sphere: Theorem 12-9: The area of a sphere equals 4π times the square of the radius A = 4πr 2

27. EXAMPLE: Find the area of the sphere ( use 3.14 as π ) SOLUTION: A = 4πr 2 A = 4 x π x 2 2 = 4 x π x 4 = 16π = 50.24 cm 2

28. Finding the Volume of a Sphere: Theorem 12-10: The volume of a sphere equals 4/3π times the cube of the radius V = 4/3π x r 3

29. EXAMPLE: Find the volume of the sphere. (Use 3.14 as π) SOLUTION: V = 4/3π x r 3 V = 4/3 π x 5 3 = 4/3π x 125 = 4.19 x 125 = 523.75 units 3

30. Section 12.5: Similar Solids Similar solids are solids that have the same shape but not the same size. To decide whether two solids are similar: - determine whether the bases are similar * all circles and all squares are similar* - determine whether corresponding lengths are proportional

31. Determining ratios

32. Example 6 10