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Geometry

Geometry. Volume of Prisms and Cylinders. Goals. Find the volume of prisms. Find the volume of cylinders. Solve problems using volume. Volume. The number of cubic units contained in a solid. Measured in cubic units. Basic Formula: V = Bh B = area of the base, h = height. Cubic Unit.

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Geometry

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  1. Geometry Volume of Prisms and Cylinders

  2. Goals • Find the volume of prisms. • Find the volume of cylinders. • Solve problems using volume.

  3. Volume • The number of cubic units contained in a solid. • Measured in cubic units. • Basic Formula: V = Bh • B = area of the base, h = height

  4. Cubic Unit V = s3 V = 1 cu. unit s 1 1 s 1 s

  5. Cavalieri’s Principle

  6. Prism: V = Bh B B h h h B

  7. Cylinder: V = r2h r B h h V = Bh

  8. 8 3 10 Example 1 Find the volume. Triangular Prism V = Bh Base = 40 V = 40(3) = 120 Abase = ½ (10)(8) = 40

  9. Example 2 Find the volume. V = Bh The base is a ? Hexagon 12 10

  10. 12 Example 2 Solution 12 12 ? ? ? 6 10

  11. Example 2 Solution V = Bh V = (374.1)(10) V  3741 12 374.1 10

  12. Example 3 A soda can measures 4.5 inches high and the diameter is 2.5 inches. Find the approximate volume. V = r2h V = (1.252)(4.5) V  22 in3 (The diameter is 2.5 in. The radius is 2.5 ÷ 2 inches.)

  13. Example 4 A wedding cake has three layers. The top cake has a diameter of 8 inches, and is 3 inches deep. The middle cake is 12 inches in diameter, and is 4 inches deep. The bottom cake is 14 inches in diameter and is 6 inches deep. Find the volume of the entire cake, ignoring the icing.

  14. Example 4 Solution VTop = (42)(3) = 48  150.8 in3 VMid = (62)(4) = 144  452.4 in3 VBot = (72)(6) = 294  923.6 in3 r = 4 8 3 r = 6 12 4 486  1526.8 in3 14 6 r = 7

  15. Concrete Pipe

  16. Example 5 A manufacturer of concrete sewer pipe makes a pipe segment that has an outside diameter (o.d.) of 48 inches, an inside diameter (i.d.) of 44 inches, and a length of 52 inches. Determine the volume of concrete needed to make one pipe segment. 48 44 52

  17. View of the Base Example 5 Solution Strategy: Find the area of the ring at the top, which is the area of the base, B, and multiply by the height.

  18. Example 5 Solution Strategy: Find the area of the ring at the top, which is the area of the base, B, and multiply by the height. Area of Outer Circle: Aout = (242) = 576 Area of Inner Circle: Ain = (222) = 484 Area of Base (Ring): ABase = 576 - 484 = 92 48 44 52

  19. Example 5 Solution V = Bh ABase = B = 92 V = (92)(52) V = 4784 V  15,029.4 in3 48 44 52

  20. Example 5 Alternate Solution 48 Vouter = (242)(52) Vouter = 94,096.98 Vinner = (222)(52) Vinner = 79,067.60 V = Vouter – Vinner V  15,029.4 in3 44 52

  21. Example 6 4 5 L A metal bar has a volume of 2400 cm3. The sides of the base measure 4 cm by 5 cm. Determine the length of the bar.

  22. Method 1 V = Bh B = 4  5 = 20 2400 = 20h h = 120 cm Method 2 V = L  W  H 2400 = L  4  5 2400 = 20L L = 120 cm Example 6 Solution 4 5 L

  23. Example 7 Diameter = 7 3 in V = 115 in3 A 3-inch tall can has a volume of 115 cubic inches. Find the diameter of the can.

  24. Summary • The volumes of prisms and cylinders are essentially the same: V = Bh & V = r2h • where B is the area of the base, h is the height of the prism or cylinder. • Use what you already know about area of polygons and circles for B.

  25. r B h h V = r2h V = Bh Add these to your formula sheet.

  26. 2.3 in 4 in 4.5 in 3.2 in 1.6 in Which Holds More? This one!

  27. What would the height of cylinder 2 have to be to have the same volume as cylinder 1? r = 3 r = 4 #2 #1 8 h

  28. Solution r = 4 #1 8

  29. Solution r = 3 #2 h

  30. Practice Problems

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