20S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Volumes of Prisms Learning Outcome B-4 2D3D-L2 Objectives: To solve problems involving the evaluation of volumes of prisms.

20S Applied Math Mr. Knight – Killarney School Slide 2 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Prisms In Lesson I, you reviewed how to calculate the area of a geometric figure. For a rectangle such as the one below you counted the squares or used the formula A = lw to get 8 sq. units. Volume is the amount of 3D space an object occupies. You can measure it by counting the number of cubes that the figure contains. Example: The top layer of the object has 12 cubes. The object has two identical layers. Therefore, the object's volume can be calculated as follows: No. of cubes/layer x No. of layers, e.g., 12 x 2 = 24 cubes. Counting cubes works for finding the volume of rectangular objects. However, for non-rectangular objects, you will need another method. You can find the volume of any prism by multiplying the area of the base by the height. Theory – Volume of Prisms

20S Applied Math Mr. Knight – Killarney School Slide 3 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Theory – Formula for Calculating Prism Volume Example: Examine the base of the diagram. The area is 12 square units. A = lw = 4(3) =12 sq. units If this area is maintained throughout the height of the figure as it is in this case, the volume can be obtained by simply multiplying the area of the base by the height. V = Bh where B is area of the base and h is height of the figure. For the above example: V = 12(2) = 24 u 3 (cubic units) Similarly, you can find the volume for any "prism" (a 3D figure where the shape of the base is maintained throughout the height) by multiplying the area of the base by the height.

20S Applied Math Mr. Knight – Killarney School Slide 4 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 1 - Calculating Prism Volume Find the volume of the triangular prism. Solution The triangular base of the prism, B, is equal to the area of the triangle. Triangular area of base or B = ½(8)(5) = 20 cm 2 Volume of prism = Bh V = (20)(10) = 200 cm 3

20S Applied Math Mr. Knight – Killarney School Slide 5 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 2 - Calculating Prism Volume A one-half inch diameter pipe, which feeds a hot water tap, is 14' from the hot water tank. How much water will be wasted each time the hot water tap is run until hot water arrives at the tap? (Recall 12" = 1'). Solution To help visualize the problem and begin a solution, think about standing the pipe on end. The 14 feet of pipe equal 168 inches in length (12" x 14'). The pipe cross-section is a circle, so we want to find the area of the base. The formula is  r 2. Area =  (¼) 2 Base (B) =  (¼) 2 V = Bh = πr 2 h = π(¼) 2 (14 x 12) = π (1/16)(168) = 10.5 π V = 33.0 in 3

20S Applied Math Mr. Knight – Killarney School Slide 6 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 3 - Calculating Prism Volume A paint shop has a room that measures 30' by 30' and has a height of 10'. The room needs its air replaced every two hours in order for it to be ventilated safely. If it were to be ventilated by a fan rated in ft 3 /min, what minimum-sized fan would be required? Solution The volume of the room is, V= Bh = (30 x 30) 10 = 9000 ft ft 3 of air needs to be moved every two hours 4500 ft 3 every hour (since there are 60 min in an hour) 4500 ÷ 60 = 75 ∴ 75 ft 3 /min

20S Applied Math Mr. Knight – Killarney School Slide 7 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Theory - Volume of Pyramids A pyramid is a 3D figure in which the shape of the base reduces to a single point throughout the height of the figure. The volume of these figures is one-third of the volume of its corresponding prism. V = 1 / 3 Bh Triangular Pyramid (base is a triangle) Cone (base is a circle) Rectangular Pyramid (base is a rectangle)

20S Applied Math Mr. Knight – Killarney School Slide 8 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 1 - Calculating Volume of Pyramids Find the volume of the following: Solution (base = rectangle (square) V = 1 / 3 Bh (area of base = B = L x W) = 1 / 3 x 3 x 3 x 7 = 21 in 3

20S Applied Math Mr. Knight – Killarney School Slide 9 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 2 - Calculating Volume of Pyramids Find the volume of the following: Solution (base = parallelogram) V = 1 / 3 Bh (area of base = B = bh) = 1 / 3 x 8 x 8 x 27 (this can be (8)(18)(27) ÷ 3) = 1296 cm 3

20S Applied Math Mr. Knight – Killarney School Slide 10 Unit: Real Number System Lesson: 2D3D-L2 Volumes of Prism Example 3 - Calculating Volume of Pyramids If a cone inserted within a cylinder is filled with water, what is the volume of air left in the cylinder? Solution Volume of cylinder V = Bh (B = circular base of cylinder = πr 2 ) = πr²h = π(12) 2 (42) = ft 3 Volume of cone V = (B = circular base of cone if it were turned over) = 1 / 3 π(12) 2 (42) = ft 3 Volume of air left in cylinder V = = ft 3