2. Do Now!

3. Lesson 9-1: Area of 2-D Shapes2 Lesson 9-1 Area of 2-D Shapes

4. Lesson 9-1: Area of 2-D Shapes3 Squares and Rectangles s s A = s² 6 6 A = 6² = 36 sq. units L W A = LW 12 5 A = 12 x 5 = 60 sq. units Example: Area of Rectangle: A = LW Area of Square: A = s²

5. Lesson 9-1: Area of 2-D Shapes 4 Circles r 9 cm A =  (9)² = 81  sq. cm Area of Circle: A =  r² Example:

6. Lesson 9-1: Area of 2-D Shapes 5 Triangles and Trapezoids h h h bb b1b1 b2b2 h is the distance from a vertex of the triangle perpendicular to the opposite side. h is the distance from b1 to b2, perpendicular to each base

7. Lesson 9-1: Area of 2-D Shapes 6 Example: Triangles and Trapezoids 7 6 8 12 6

8. Lesson 9-1: Area of 2-D Shapes 7 Parallelograms, Rhombi, &Kites Area of Parallelogram: A = b h 6 9 A = 9 x 6 = 54 sq. units 8 1010 A = ½ (8)(10) = 40 sq units h b Example:

9. VOLUME OF SOLIDS By: SAMUEL M. GIER

10. Imagine transferring a 12-ounce soft drink from the bottle to an ice plastic container. Will the volume change? If the ice plastic is twisted a little, will the volume change?

11. VOLUME OF SOLIDS DEFINITION: -is the amount of space enclosed in a solid figure. The volume of a solid is the number of cubic units contained in the solid.

12. VOLUME OF SOLIDS In finding volume of solids, you have to consider the area of a base and height of the solid. If the base is triangular, you have to make use of the area of a triangle, if rectangular, make use of the area of a rectangle and so on.

13. VOLUME OF A CUBE The volume V of a cube with edge s is the cube of s. That is, V = s 3 s=h s

14. Example 1. Find the volume of a cube whose sides 8 cm. 8 cm  Solution:  V = s³  = (8cm) ³  V = 512 cm³

15. VOLUME OF A RECTANGULAR PRISM

16. Rectangular prism The volume V of a rectangular prism is the product of its altitude h, the length l and the width w of the base. That is, V = lwh l w h

17. Example 2. Find the volume of a rectangular prism. 8 cm  Solution:  V = lwh  = (8cm)(4 cm) (5 cm)  V = 160 cm³ 4 cm 5cm

18. VOLUME OF A SQUARE PRISM

19. Square prism It is a prism whose bases are squares and the other faces are rectangles. Square base Height (H)

20. Square prism s s h The volume V of a square prism is the product of its altitude H and the area of the base, s². That is, V = s²H

21. Example 3. Find the volume of a square prism. 4 cm  Solution:  V = s²H  = (4 cm) ² (5 cm)  = (16 cm²) 5 cm  V = 80 cm³ 4 cm 5cm

22. VOLUME OF A TRIANGULAR PRISM

23. Triangular prism H The volume V of a triangular prism is the product of its altitude H and the area of the base(B), ½bh. That is, V = (½bh) H. h b Base

24. Example 4. Find the volume of a Triangular prism Solution: V = (½bh) H = ½ (3.9 cm)(4.5 cm)(2.8 cm) = ½ (49.14 cm³) V = 24.57 cm³ 3.9 cm 4.5cm 2.8 cm

25. ANOTHER KIND OF POLYHEDRON PYRAMIDS

26. Altitude Or Height -Height of the pyramid Slant Height (height of the Triangular face)

27. TYPES OF PYRAMIDS Pyramids are classified according to their base. 1.SquarePyramid- the base is square. 2.Rectangular Pyramid- the base is rectangle. 3.Triangular Pyramid- the base is triangle.

28. VOLUME of PYRAMIDS Consider a pyramid and a prism having equal altitudes and bases with equal areas. If the pyramid is filled with water or sand and its contents poured into a prism, only one- third of the prism will be filled. Thus the volume of a pyramid is ⅓ the volume of the prism. w = 4 cm l = 9 cm h = 6 cm base

29. Volume of pyramids The volume V of a pyramid is one third the product of its altitude h and the area B of its base. That is, V = ⅓Bh. SQUARE PYRAMID V = ⅓(s²)H RECTANGULAR PYRAMID V = ⅓(lw)H TRIANGULAR PYRAMID V = ⅓(½bh)H

30. EXAMPLE 5: FIND THE VOLUME OF A RECTANGULAR PYRAMID Solution: V = ⅓(lwH) = ⅓(6 cm)(4 cm)(10 cm) = ⅓ (240 cm³) V = 80 cm³ W= 4 cm Height ( 10 cm ) l= 6 cm

31. EXAMPLE 6: FIND THE VOLUME OF A SQUARE PYRAMID Solution: V = ⅓(s²H) = ⅓(6 cm)²(8 cm) = ⅓ (36 cm²)(8 cm) = ⅓ (288 cm³) V = 96 cm³ 6 cm Height ( 8 cm ) 6 cm

32. EXAMPLE 7: Find the volume of a regular triangular pyramid. 8 cm 6 cm Solution: V = ⅓( ½bh)H = ⅓[ ½( 6 cm )(3 cm) ( 8 cm)] = ⅓( 9 cm²) (8 cm) = ⅓ (72 ) cm³ V = 24 cm³ h=3 cm

33. COMMON SOLIDS CYLINDERS

34. CYLINDER -is a space figure with two circular bases that are parallel and congruent. Circular base.. Height Radius

35. CYLINDERS Guide questions: What is the geometric figure represented by the bases of the cylinder? How do you compute its area?

36. Volume of a cylinder ANSWERS Circles (circular bases) A =  r² Height radius

37. Volume of a cylinder How can the volume of a cylinder be computed? V = Bh, where B is the area of the base and h is the height of the cylinder. by substitution, V= πr² h

38. EXAMPLE 8: Find the volume of a cylinder. Use π = 3.14 Solution: V = πr²h =(3.14)(5 cm)² 10 cm = 3.14( 25cm²) (10 cm) = 3.14( 250 cm³) V = 785 cm³ 5 cm 10 cm

39. VOLUME OF A CONE REFLECTIVE TRAFFIC CONE

40. CONE -is a space figure with one circular base and a vertex Vertex Circular base. Height Of the cone Radius Slant Height Of the cone

41. VOLUME of a CONE Consider a CONE and a CYLINDER having equal altitudes and bases with equal areas. If the CONE is filled with water or sand and its contents poured into a CYLIDER, only one- third of the CYLINDER will be filled. Thus the volume of a CONE is ⅓ the volume of the CYLINDER. h r

42. Volume of a cone How can the volume of a cone be computed? V = ⅓ Bh, where B is the area of the base and h is the height of the cone. by substitution, V= ⅓ πr² h

43. Find the volume of a cone. Use π = 3.14 Solution: V = ⅓ πr²h = ⅓ (3.14)(5 cm) ²(10 cm) = ⅓ (3.14)(25 cm ² ) (10 cm) = ⅓ (785 cm³) V= 261.67 cm ³ 5 cm 10 cm

44. Find the volume of a cone. Use π = 3.14 Solution: Step 1. find h. Using Pythagorean theorem, h²= 5² - 3² =25-9 h² = 16 h = 4 cm 3 cm 5 cm Solution: V = ⅓ πr²h V = ⅓ πr²h = ⅓ (3.14)(3 cm) ²(4 cm) = ⅓ (3.14)(3 cm) ²(4 cm) = ⅓ (3.14)(9 cm ² ) (4 cm) = ⅓ (3.14)(9 cm ² ) (4 cm) = ⅓ (113.04 cm³) = ⅓ (113.04 cm³) V= 37.68 cm ³ V= 37.68 cm ³ 4 cm

45. VOLUME OF A SPHERE THE EARTH BALLS

46. SPHERE A sphere is a solid where every point is equally distant from its center. This distance is the length of the radius of a sphere. radius

47. VOLUME OF A SPHERE The formula to find the VOLUMEof a sphere is V = πr³, where r is the length of its radius. BALL How can the volume of a sphere be computed?

48. Archimedes of SyracuseArchimedes of Syracuse (287-212 BC) is regarded as the greatest of Greek mathematicians, and was also an inventor of many mechanical devices (including the screw, the pulley, and the lever). He perfected integration using Eudoxus' method of exhaustion, and found the areas and volumes of many objects.

49. Archimedes of SyracuseArchimedes of Syracuse (287-212 BC) A famous result of his is that the volume of a sphere is two-thirds the volume of its circumscribed cylinder, a picture of which was inscribed on his tomb.

50. The height (H)of the cylinder is equal to the diameter (d) of the sphere. radius r H = d r

51. Volume (Sphere)= ⅔ the volume of a circumscribed cylinder radius r H = d r

52. Volume (Sphere)= ⅔  r²h = ⅔  r² (2r) =  r³ radius r H = d= 2r r

53. 1. Find the volume of a sphere. Use π = 3.14 Solution: V = 4/3 πr³ = 4/3(3.14)(10 cm)³ = 12.56 (1000 cm³) 3 = 12,560 cm³) 3 V= 4,186.67 cm³ 10 cm

54. 2. Find the volume of a sphere. Use π = 3.14 Solution: V = 4/3 πr³ = 4/3(3.14)(7.8 cm)³ = 12.56 (474.552 cm³) 3 = 5960.37312 cm³ 3 V= 1,986.79 cm³ 7.8 cm