Geometry Volume of Prisms and Cylinders

9/11/2015 Goals Find the volume of prisms. Find the volume of cylinders. Solve problems using volume.

9/11/2015 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

9/11/2015 Cubic Unit V = 1 cu. unit s s s V = s 3

9/11/2015 Cavalieri’s Principle

9/11/2015 B B B h h h Prism: V = Bh

9/11/2015 Cylinder: V =  r 2 h B h r h V = Bh

9/11/2015 Example 1Find the volume Triangular Prism V = Bh Base = 40 V = 40(3) = 120 A base = ½ (10)(8) = 40

9/11/2015 Example 2Find the volume V = Bh The base is a ? Hexagon

9/11/2015 Example 2 Solution ? ? ? 6

9/11/2015 Example 2 Solution V = Bh V = (374.1)(10) V  3741

9/11/2015 Example 3 A soda can measures 4.5 inches high and the diameter is 2.5 inches. Find the approximate volume. V =  r 2 h V =  ( )(4.5) V  22 in 3 (The diameter is 2.5 in. The radius is 2.5 ÷ 2 inches.)

9/11/2015 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.

9/11/2015 Example 4 Solution r = 4 r = 6 r = 7 V Top =  (4 2 )(3) = 48   in 3 V Mid =  (6 2 )(4) = 144   in 3 V Bot =  (7 2 )(6) = 294   in   in 3

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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

9/11/2015 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. View of the Base

9/11/2015 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: A out =  (24 2 ) = 576  Area of Inner Circle: A in =  (22 2 ) = 484  Area of Base (Ring): A Base = 576   = 92 

9/11/2015 Example 5 Solution V = Bh A Base = B = 92  V = (92  )(52) V = 4784  V  15,029.4 in

9/11/2015 Example 5 Alternate Solution V outer =  (24 2 )(52) V outer = 94, V inner =  (22 2 )(52) V inner = 79, V = V outer – V inner V  15,029.4 in

9/11/2015 Example 6 A metal bar has a volume of 2400 cm 3. The sides of the base measure 4 cm by 5 cm. Determine the length of the bar. 4 5 L

9/11/2015 Example 6 Solution Method 1 V = Bh B = 4  5 = = 20h h = 120 cm Method 2 V = L  W  H 2400 = L  4  = 20L L = 120 cm 4 5 L

9/11/2015 Example 7 3 in V = 115 in 3 A 3-inch tall can has a volume of 115 cubic inches. Find the diameter of the can. Diameter = 7

9/11/2015 Summary The volumes of prisms and cylinders are essentially the same: V = Bh &V =  r 2 h 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.

9/11/2015 V = Bh V =  r 2 h B h h r Add these to your formula sheet.

9/11/2015 Which Holds More? 3.2 in1.6 in 4 in4.5 in 2.3 in This one!

9/11/2015 What would the height of cylinder 2 have to be to have the same volume as cylinder 1? r = 4 h r = 3 8 #1 #2

9/11/2015 Solution r = 4 8 #1

9/11/2015 Solution h r = 3 #2

9/11/2015 Practice Problems