Lesson 23 Lesson 23: Student Outcomes Students use the known formula for the volume of a right rectangular prism (length width height). Students understand the volume of a right prism to be the area of the base times the height. Students compute volumes of right prisms involving fractional values for length. Lesson Notes Students extend their knowledge of obtaining volumes of right rectangular prisms via dimensional measurements to understand how to calculate the volumes of other right prisms. This concept will later be extended to finding the volumes of liquids in right prism-shaped containers and extended again (in Module 6) to finding the volumes of irregular solids using displacement of liquids in containers. The Problem Set scaffolds in the use of equations to calculate unknown dimensions. Classwork Opening Exercise (5 minutes) Opening Exercise The volume of a solid is a quantity given by the number of unit cubes needed to fill the solid. Most solids rocks, baseballs, people cannot be filled with unit cubes or assembled from cubes. Yet such solids still have volume. Fortunately, we do not need to assemble solids from unit cubes in order to calculate their volume. One of the first interesting examples of a solid that cannot be assembled from cubes but whose volume can still be calculated from a formula is a right triangular prism. What is the area of the square pictured on the right? Explain. The area of the square is 3333 uuuuuuuuuu because the region is filled with 3333 square regions that are 11 uuuuuuuu by 11 uuuuuuuu, or 11 uuuuuutt. Draw the diagonal joining the two given points; then, darken the grid lines within the lower triangular region. What is area of that triangular region? Explain. The area of the triangular region is 1111 uuuuuuuuuu. There are 1111 unit squares from the original square and 66 triangular regions that are 11 uuuuuutt. The 66 triangles can be paired together to form 33 uuuuuuuuuu. All together the area of the triangular region is (1111 + 33) uuuuuuuuuu = 1111 uuuuuuuuuu. How do the areas of the square and the triangular region compare? The area of the triangular region is half the area of the square region. Lesson 23: 315
Lesson 23 Exploratory Challenge (15 minutes): Exploratory Challenge is a continuation of the Opening Exercise. Exploratory Challenge: What is the volume of the right prism pictured on the right? Explain. The volume of the right prism is 3333 uuuuuuuuuu 33 because the prism is filled with 3333 cubes that are 11 uuuuuuuu long, 11 uuuuuuuu wide, and 11 uuuuuuuu high, or 11 uuuuuutt 33. Draw the same diagonal on the square base as done above; then, darken the grid lines on the lower right triangular prism. What is the volume of that right triangular prism? Explain. The volume of the right triangular prism is 1111 uuuuuuuuuu 33. There are 1111 cubes from the original right prism and 66 right triangular prisms that are each half of a cube. The 66 right triangular prisms can be paired together to form 33 cubes, or 33 uuuuuuuuss 33. All together the area of the right triangular prism is (1111 + 33) uuuuiiiiss 33 = 1111 uuuuuuuuss 33. In both cases, slicing the square (or square face) along its diagonal divided the area of the square into two equal-sized triangular regions. When we sliced the right prism, however, what remained constant? The height of the given right rectangular prism and the resulting triangular prism are unchanged at 1 unit. The argument used here is true in general for all right prisms. Since polygonal regions can be decomposed into triangles and rectangles, it is true that the polygonal base of a given right prism can be decomposed into triangular and rectangular regions that are bases of a set of right prisms that have heights equal to the height of the given right prism. How could we create a right triangular prism with five times the volume of the right triangular prism pictured to the right, without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has five times the volume of the triangular prism. If we stack five exact copies of the base (or bottom floor), the prism then has five times the number of unit cubes as the original, which means it has five times the volume, or 9999 uuuuuuuuss 33. Lesson 23: 316
Lesson 23 What could we do to cut the volume of the right triangular prism pictured on the right in half without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has half the volume of the given triangular prism. If we slice the height of the prism in half, each of the unit cubes that make up the triangular prism will have half the volume as in the original right triangular prism. The volume of the new right triangular prism is 99 uuuuuuuuss 33. What can we conclude about how to find the volume of any right prism? The volume of any right prism can be found by multiplying the area of its base times the height of the prism. If we let VV represent the volume of a given right prism, let BB represent the area of the base of that given right prism, and let h represent the height of that given right prism, then: VV = BBh. Have students complete the sentence below in their student materials. Scaffolding: Students often form the misconception that changing the dimensions of a given right prism will affect the prism s volume by the same factor. Use this exercise to show that the volume of the cube is cut in half because the height is cut in half. If all dimensions of a unit cube were cut in half, the resulting volume would be 1 2 1 1 = 1, which is not equal to 2 2 8 1 2 unit3. To find the volume (VV) of any right prism Multiply the area of the right prism s base (BB) times the height of the right prism (hh),. Example (5 minutes): The Volume of a Right Triangular Prism Students calculate the volume of a triangular prism that has not been decomposed from a rectangle. Example: The Volume of a Right Triangular Prism Find the volume of the right triangular prism shown in the diagram using. VV = 11 llll hh VV = 11 44 mm 11 mm 66 11 mm VV = mm 11 mm 66 11 mm VV = 11 mm 66 11 mm VV = 66 11 mm33 The volume of the triangular prism is 66 11 mm33. Lesson 23: 317
Lesson 23 Exercise (10 minutes): Multiple Volume Representations Students find the volume of the right pentagonal prism using two different strategies. Exercise: Multiple Volume Representations The right pentagonal prism is composed of a right rectangular prism joined with a right triangular prism. Find the volume of the right pentagonal prism shown in the diagram using two different strategies. Strategy #1 The volume of the pentagonal prism is equal to the sum of the volumes of the rectangular and triangular prisms. VV = VV rrrrrrrrrrrrrrrrrrrrrr pppppppppp + VV tttttttttttttttttttt pppppppppp VV = (llll)hh VV = 44 m 66 11 m 66 11 mm VV = ( m 2 + m 2 ) 66 11 mm VV = m 2 66 11 m VV = 11 llll hh VV = 11 44 m 11 m 66 11 m VV = m 11 m 66 11 m VV = (11 m2 ) 66 11 m VV = 111111 m 3 + 1111 m 3 VV = 111111 m 3 VV = 66 11 m3 So the total volume of the pentagonal prism is 111111 m 3 + 66 11 m3 = 111111 11 m3. Strategy #2 The volume of a right prism is equal to the area of its base times its height. The base is a rectangle and a triangle. BB = AA rrrrrrrrrrrrrrrrrr + AA tttttttttttttttt AA rrrrrrrrrrrrrrrrrr = 44 m 66 11 m AA tttttttttttttttt = 11 44 m 11 m VV = m 66 11 m AA rrrrrrrrrrrrrrrrrr = m + m AA tttttttttttttttt = m 11 m VV = 111111 m33 + 1111 11 m33 AA rrrrrrrrrrrrrrrrrr = m AA tttttttttttttttt = 11 m VV = 111111 11 m33 BB = m + 11 m = m The volume of the right pentagonal prism is 111111 11 m33. Scaffolding: An alternative method that will help students visualize the connection between the area of the base, the height, and the volume of the right prism is to create pentagonal floors or layers with a depth of 1 unit. Students can physically pile the floors to form the right pentagonal prism. This example involves a fractional height so representation or visualization of a floor with a height of 1 unit is necessary. 2 See below. Lesson 23: 318
Lesson 23 Closing (2 minutes) What are some strategies that we can use to find the volume of three-dimensional objects? Find the area of the base, then multiply times the prism s height; decompose the prism into two or more smaller prisms of the same height, and add the volumes of those smaller prisms. The volume of a solid is always greater than or equal to zero. If two solids are identical, they have equal volumes. If a solid SS is the union of two non-overlapping solids AA and BB, then the volume of solid SS is equal to the sum of the volumes of solids AA and BB. Exit Ticket (8 minutes) Lesson 23: 319
Lesson 23 Name Date Lesson 23: Exit Ticket The base of the right prism is a hexagon composed of a rectangle and two triangles. Find the volume of the right hexagonal prism using the formula VV = BBh. Lesson 23: 320
Lesson 23 Exit Ticket Sample Solutions The base of the right prism is a hexagon composed of a rectangle and two triangles. Find the volume of the right hexagonal prism using the formula. The area of the base is the sum of the areas of the rectangle and the two triangles. BB = AA rrrrrrrraaaaaaaaaa + AA tttttttttttttttt AA rrrrrruuaauuaaaarr = llll AA rrrrrruuaauuaaaarr = 11 44 iiii. 11 11 iiii. AA rrrrrruuaauuaaaarr = 99 44 33 iinn AA AA rrrrrruuaauuaaaarr = 88 iinn AA tttttttttttttttt = 11 llll AA tttttttttttttttt = 11 11 11 iiii. 33 iiii. 44 tttttttttttttttt = 11 33 33 iinn 44 AA tttttttttttttttt = 99 1111 iinn BB = 88 iinn + 99 1111 iinn BB = 88 iinn + 99 88 iinn BB = 3333 88 iinn BB = 99 iinn VV = 99 iinn 33 iiii. VV = iinn33 VV = 1111 11 iinn33 The volume of the hexagonal prism is 1111 11 iinn33. Problem Set Sample Solutions 1. Calculate the volume of each solid using the formula (all angles are 9999 degrees). a. 88 cccc VV = (88 rrmm 77 rrmm) 1111 11 77 cccc cccc VV = 5555 1111 11 ccmm33 VV = 666666 ccmm 33 + ccmm 33 1111 11 cccc VV = 777777 ccmm 33 The volume of the solid is 777777 ccmm 33. b. VV = 33 44 iiii. 33 44 iiii. 33 44 iiii. VV = 99 1111 33 44 iinn33 VV = 6666 iinn33 The volume of the cube is 6666 iinn33. 33 44 iiii. 33 44 iiii. 33 44 iiii. Lesson 23: 321
Lesson 23 c. BB = AA rrrrrruuaauuaaaarr + AA ssssuuaarrrr BB = llll + ss BB = 11 iiii. 44 11 iiii. + 11 11 iiii. BB = 1111 iinn + 11 11 44 iinn + 11 11 iiii. 11 11 iiii. BB = 1111 11 44 iinn + 11 11 iinn + 33 44 iinn BB = 1111 11 44 iinn + 33 44 iinn + 11 11 iinn BB = 1111 iinn + 11 11 iinn BB = 1111 11 iinn VV = 1111 11 iinn 11 iiii. VV = 1111 iinn33 + 11 44 iinn33 VV = 66 iinn 33 + 11 iinn33 + 11 44 iinn33 VV = 66 33 44 iinn33 The volume of the solid is 66 33 44 iinn33. d. BB = (AA llll rrrrrruuaauuaaaarr ) (AA ssss rrrrrruuaauuaaaarr ) BB = (llww) 11 (llll) BB = (66 yyyy. 44 yyyy. ) 11 11 yyyy. yyyy. 33 BB = yydd yydd + 33 yydd BB = yydd yydd 33 yydd BB = yydd 33 yydd BB = 11 33 yydd VV = 11 33 yydd 33 yyyy. VV = 1111 yydd 33 + 11 33 yydd yyyy. 33 VV = 1111 yydd 33 + 99 yydd33 VV = 1111 99 yydd33 The volume of the solid is 1111 99 yydd33. e. VV = BBhh pppppppppp BB = 11 bbhh tttttttttttttttt BB = 11 44 rrmm 44 rrmm VV BB = 44 ccmm BB = 88 ccmm = 88 ccmm 66 77 1111 rrmm VV = 4444 ccmm 33 + 5555 1111 ccmm33 VV = 4444 ccmm 33 + 55 ccmm 33 + 66 1111 ccmm33 VV = 5555 ccmm 33 + 33 55 ccmm33 VV = 5555 33 55 ccmm33 The volume of the solid is 5555 33 55 ccmm33. Lesson 23: 3
Lesson 23 f. VV = BBhh pppppppppp BB = 11 bbhh tttttttttttttttt BB = 11 33 99 iiii. 11 iiii. BB = 11 11 33 iiii. 99 iiii. BB = 11 11 33 99 iinn 44 VV = 5555 55 iinn 55 iiii. VV = 5555 iinn33 BB = 55 44 iinn BB = 5555 55 iinn The volume of the solid is 5555 iinn 33. g. BB = AA rrrrrruuaauuaaaarr + AA tttttttttttttttt BB = llll + 11 bbbb BB = 55 11 44 rrmm 44 rrmm + 11 44 rrmm 11 11 44 rrmm BB = ( ccmm + 11 ccmm ) + rrmm 11 11 44 rrmm BB = ccmm + ccmm + 11 ccmm VV = 11 ccmm 99 rrmm VV = ccmm33 + 99 ccmm33 VV = ccmm33 + 44 ccmm 33 + 11 ccmm33 VV = 11 ccmm33 BB = ccmm + 11 ccmm BB = 11 ccmm The volume of the solid is 11 ccmm33. h. BB = AA rrrrrruuaauuaaaarr + AA tttttttttttttttt BB = llll + 11 bbbb BB = 11 iiii. 11 55 iiii. + 11 11 88 iiii. 11 iiii. 55 VV VV = 11 88 iinn iiii. = 11 44 iinn33 BB = 11 1111 iinn + 11 4444 iinn BB = 44 4444 iinn + 11 4444 iinn The volume of the solid is 11 44 iinn33. BB = 55 4444 iinn BB = 11 88 iinn Lesson 23: 323
Lesson 23 2. Let ll represent length, ww the width, and hh the height of a right rectangular prism. Find the volume of the prism when: a. ll = 33 rrmm, ww = 11 rrmm, and hh = 77 rrmm. VV = llllll VV = 33 rrmm 11 rrmm 77 rrmm VV = 11 ccmm33 VV = 5555 11 ccmm33 The volume of the prism is 5555 11 ccmm33. b. ll = 11 44 rrmm, ww = 44 rrmm, and hh = 11 11 rrmm. VV = llllll VV = 11 44 rrmm 44 rrmm 11 11 rrmm VV = 11 11 ccmm33 The volume of the prism is 11 11 ccmm33. 3. Find the length of the edge indicated in each diagram. a. Let hh represent the number of inches in the height of the prism. AAAAAAAA = iinn 9999 11 iinn33 = iinn hh 9999 11 iinn33 = hh iinn hh = 9999. 55 iiii. hh = 44. iiii. The height of the right rectangular prism is 44 11 44 iiii.? What are possible dimensions of the base? 1111 iiii. by iiii., or iiii. by 11 iiii. VVVVVVVVVVVV = 9999 11 iinn33 b. Let hh represent the number of meters in the height of the triangular base of the prism. VV = 11 bbhh tttttttttttttttt hh pppppppppp 44 11 mm33 = 11 33 mm hh 66 mm 44 11 mm33 = 11 1111 mm hh? 44 11 mm33 = 99hh mm 99hh = 44. 55 mm hh = 00. 55 mm The height of the triangle is 11 mm. Lesson 23: 324
Lesson 23 4. The volume of a cube is 33 33 88 iinn33. Find the length of each edge of the cube. VV = ss 33, and since the volume is a fraction, the edge length must also be fractional. 33 33 88 iinn33 = 88 iinn33 33 33 88 iinn33 = 33 iiii. 33 iiii. 33 iiii. 33 33 88 iinn33 = 33 iiii. 33 The lengths of the edges of the cube are 33 iiii. = 11 11 iiii. 5. Given a right rectangular prism with a volume of 77 11 fftt33, a length of 55 ffff., and a width of ffff., find the height of the prism. VV = (llll)hh Let hh represent the number of feet in the height of the prism. 77 11 fftt33 = (11ffuu.. ) hh 77 11 fftt33 = 1111 fftt hh 77. 55 fftt 33 = 1111hh fftt hh = 00. 7777 ffff. The height of the right rectangular prism is 33 ffff. (or 99 iiii.). 44 Lesson 23: 325
Lesson 24 Lesson 24: Student Outcomes Students use the formula for the volume of a right rectangular prism to answer questions about the capacity of tanks. Students compute volumes of right prisms involving fractional values for length. Lesson Notes Students extend their knowledge about the volume of solid figures to the notion of liquid volume. The Opening Exercise for Lesson 24 requires a small amount of water. Have an absorbent towel available to soak up the water at the completion of the exercise. Classwork Opening Exercise (3 minutes) Pour enough water onto a large flat surface to form a puddle. Have students discuss how to determine the volume of the water. Provide 2 minutes for student discussion, and then start the class discussion. Discussion (3 minutes) Why can t we easily determine the volume of the water in the puddle? The puddle does not have any definite shape or depth that we can easily measure. How can we measure the volume of the water in three dimensions? The volume can be measured in three dimensions if put into a container. In a container, such as a prism, water takes on the shape of the container. We can measure the dimensions of the container to determine an approximate volume of the water in cubic units. Exploratory Challenge (8 minutes): Measuring a Container s Capacity Students progress from measuring the volume of a liquid inside a right rectangular prism filled to capacity to solving a variety of problems involving liquids and prism-shaped containers. Ask questions to guide students in discovering the need to account for the thickness of the container material in determining the inside volume of the container. For instance, ask, Is the length of the inside of the container 12 inches? Why not? What is the width of the inside container? The depth? Why did you have to subtract twice the thickness to get the length and width, but only one times the thickness to get the depth? Lesson 24: 326
Lesson 24 Exploratory Challenge: Measuring a Container s Capacity A box in the shape of a right rectangular prism has a length of 1111 iiii., a width of 66 iiii., and a height of 88 iiii. The base and the walls of the container are 11 iiii. thick, and its top is open. What is the capacity of the right rectangular prism? 44 (Hint: The capacity is equal to the volume of water needed to fill the prism to the top.) If the prism is filled with water, the water will take the shape of a right rectangular prism slightly smaller than the container. The dimensions of the smaller prism are a length of 1111 11 iiii., a width of 55 11 iiii., and a height of 77 33 44 iiii. VV = (llll)hh VV = 1111 11 iiii. 55 11 iiii. 77 33 44 iiii. VV = iiii. 1111 VV = 44 VV = 77777777 1111 iinn 3333 44 iiii. iinn33 VV = 444444 33 1111 iinn33 iiii. 3333 44 iiii. The capacity of the right rectangular prism is 444444 33 1111 iinn33. Example 1 (5 minutes): Measuring Liquid in a Container in Three Dimensions Students use the inside of right prism-shaped containers to calculate the volumes of contained liquids. Example 1: Measuring Liquid in a Container in Three Dimensions A glass container is in the form of a right rectangular prism. The container is 1111 cccc long, 88 cccc wide, and 3333 cccc high. The top of the container is open, and the base and walls of the container are 33 mmmm (or 00. 33 cccc) thick. The water in the container is 66 cccc from the top of the container. What is the volume of the water in the container? Because of the walls and base of the container, the water in the container forms a right rectangular prism that is 99. 44 cccc long, 77. 44 cccc wide, and. 77 cccc tall. VV = (llll)hh VV = (99. 44 cccc 77. 44 cccc). 77 cccc VV = 9999 7777 cccc cccc 1111 1111 1111 cccc VV = 66666666 111111 ccmm 1111 cccc VV = 11111111111111 ccmm 33 11111111 VV = 11111111. 555555 ccmm 33 The volume of the water in the container is 11111111. 66 ccmm 33. Lesson 24: 327
Lesson 24 Example 2 (8 minutes) Students determine the depth of a given volume of water in a container of a given size. Example 2 77. LL of water are poured into a container in the shape of a right rectangular prism. The inside of the container is 5555 cccc long, cccc wide, and cccc tall. How far from the top of the container is the surface of the water? (11 LL = 11111111 ccmm 33 ) 77. LL = 77777777 ccmm 33 VV = (llll)hh 77777777 ccmm 33 = (5555 cccc)( cccc)hh 77777777 ccmm 33 = 11111111 ccmm hh 11 77777777 ccmm 33 11111111 ccmm = 11111111 11 ccmm hh 11111111 ccmm 77777777 cccc = 11 hh 11111111 77. cccc = hh The depth of the water is 77. cccc. The height of the container is cccc. The surface of the water is cccc 77. cccc = 1111. 88 cccc from the top of the container. Example 3 (8 minutes) Students find unknown measurements of a right prism given its volume and two dimensions. Example 3 A fuel tank is the shape of a right rectangular prism and has LL of fuel in it. It is determined that the tank is 33 full. The 44 inside dimensions of the base of the tank are 9900 cccc by 5555 cccc. What is the height of the fuel in the tank? How deep is the tank? (11 LL = 11111111 ccmm 33 ) Let the height of the fuel in the tank be hh cccc. LL = ccmm 33 VV = (llll)hh ccmm 33 = (9999 cccc 5555 cccc) hh ccmm 33 = (44444444 ccmm ) hh 11 ccmm 33 44444444 ccmm = 44444444 11 ccmm hh 44444444 ccmm cccc = 11 hh 44444444 66 cccc = hh The height of the fuel in the tank is 66 cccc. The height of the fuel is 33 the depth of the tank. Let dd represent the depth of 44 the tank in in centimeters. 66 cccc = 33 44 dd 66 cccc 44 33 = 33 44 44 33 dd 88 cccc = dd The depth of the fuel tank is 88 cccc. Lesson 24: 328
Lesson 24 Closing (2 minutes) How do containers, such as prisms, allow us to measure the volumes of liquids using three dimensions? When liquid is poured into a container, the liquid takes on the shape of the container s interior. We can measure the volume of prisms in three dimensions, allowing us to measure the volume of the liquid in three dimensions. What special considerations have to be made when measuring liquids in containers in three dimensions? The outside and inside dimensions of a container will not be the same because the container has wall thickness. In addition, whether or not the container is filled to capacity will affect the volume of the liquid in the container. Exit Ticket (8 minutes) Students may be allowed to use calculators when completing this Exit Ticket. Lesson 24: 329
Lesson 24 Name Date Lesson 24: Exit Ticket Lawrence poured 27.328 L of water into a right rectangular prism-shaped tank. The base of the tank is 40 cm by 28 cm. When he finished pouring the water, the tank was 2 3 full. (1 L = 1000 cm3 ) a. How deep is the water in the tank? b. How deep is the tank? c. How many liters of water can the tank hold in total? Lesson 24: 330
Lesson 24 Exit Ticket Sample Solutions Lawrence poured. 333333 LL of water into a right rectangular prism shaped tank. The base of the tank is 4444 cccc by cccc. When he finished pouring the water, the tank was 33 full. (11 LL = 11111111 ccmm33 ) a. How deep is the water in the tank?. 333333 LL = ccmm 33 VV = (llll)hh ccmm 33 = (4444 cccc cccc) hh ccmm 33 = 11111111 ccmm hh 11 ccmm 33 11111111 ccmm = 11111111 11 ccmm hh 11111111 ccmm cccc = 11 hh 11111111 cccc = hh 11111111 cccc = hh 55 The depth of the water is 55 cccc. b. How deep is the tank? The depth of the water is the depth of the tank. Let dd represent the depth of the tank in centimeters. 33 55 cccc = 33 dd 55 cccc 33 = 33 33 dd 3333 cccc + 33 cccc = 11dd 55 3333 33 cccc = dd 55 The depth of the tank is 3333 33 55 cccc. c. How many liters of water can the tank hold total? VV = (llll)hh VV = (4444 cccc cccc) 3333 33 55 cccc VV = 11111111 ccmm 3333 33 55 cccc VV = 4444444444 ccmm 33 + 666666 ccmm 33 VV = 4444444444 ccmm 33 4444444444 ccmm 33 = 4444. 999999 LL The tank can hold up to 4444. 00 LL of water. Lesson 24: 331
Lesson 24 Problem Set Sample Solutions 1. Mark wants to put some fish and decorative rocks in his new glass fish tank. He measured the outside dimensions of the right rectangular prism and recorded a length of 5555 cccc, width of 4444 cccc, and height of 3333 cccc. He calculates that the tank will hold 8888. 7777 LL of water. Why is Mark s calculation of volume incorrect? What is the correct volume? Mark also failed to take into account the fish and decorative rocks he plans to add. How will this affect the volume of water in the tank? Explain. = (llll)hh VV = 5555 cccc 4444 cccc 3333 cccc VV = ccmm 3333 cccc VV = 8888888888 ccmm 33 8888888888 ccmm 33 = 8888. 7777 LL Mark measured only the outside dimensions of the fish tank and did not account for the thickness of the sides of the tank. If he fills the tank with 8888. 7777 LL of water, the water will overflow the sides. Mark also plans to put fish and rocks in the tank which will force water out of the tank if it is filled to capacity. 2. Leondra bought an aquarium that is a right rectangular prism. The inside dimensions of the aquarium are 9999 cccc long, by 4444 cccc wide, by 6666 cccc deep. She plans to put water in the aquarium before purchasing any pet fish. How many liters of water does she need to put in the aquarium so that the water level is 55 cccc below the top? If the aquarium is 6666 cccc deep, then she wants the water to be 5555 cccc deep. Water takes on the shape of its container, so the water will form a right rectangular prism with a length of 9999 cccc, a width of 4444 cccc, and a height of 5555 cccc. = (llll)hh VV = (9999 cccc 4444 cccc) 5555 cccc VV = 44444444 ccmm 5555 cccc VV = ccmm 33 ccmm 33 =. 66 LL The volume of water needed is. 66 LL. Lesson 24: 332
Lesson 24 3. The inside space of two different water tanks are shown below. Which tank has a greater capacity? Justify your answer. VV 11 = BBBB = (llll)hh VV 11 = 66 iiii. 11 11 iiii. 33 iiii. Tank 2 VV 11 = (66 iinn + 33 iinn ) 33 iinn VV 11 = 99 iinn 33 iinn Tank 1 VV 11 = iinn 33 VV = BBBB = (llll)hh VV = 11 11 iiii. iiii. 99 iiii. VV = ( iinn + 11 iinn ) 99 iiii. VV = 33 iinn 99 iiii. VV = iinn 33 The tanks have the same volume, iinn 33. Each prism has a face with an area of 1111 iinn (base) and a height that is 11 11 iiii. 4. The inside of a tank is in the shape of a right rectangular prism. The base of that prism is 8888 cccc by 6666 cccc. What is the minimum height inside the tank if the volume of the liquid in the tank is LL? = (llll)hh 9999999999 ccmm 33 = (8888 cccc 6666 cccc) hh 9999999999 ccmm 33 = 55555555 ccmm hh 11 9999999999 ccmm 33 55555555 ccmm = 55555555 11 ccmm hh 55555555 ccmm 9999999999 cccc = 11 hh 55555555 1111 3333 cccc = hh 3333 The minimum height of the inside of the tank is 1111 3333 3333 cccc. Lesson 24: 333
Lesson 24 5. An oil tank is the shape of a right rectangular prism. The inside of the tank is 3333. 55 cm long, 5555 cm wide, and cm high. If 4444 liters of oil have been removed from the tank since it was full, what is the current depth of oil left in the tank? = (llll)hh VV = (3333. 55 cccc 5555 cccc) cccc VV = 11111111 ccmm cccc VV = 5555555555 ccmm 33 The tank has a capacity of 5555555555 ccmm 33, or 5555. 000000 LL. If 4444 LL of oil have been removed from the tank, then 5555. 000000 LL 4444 LL = 1111. 000000 LL are left in the tank. = (llll)hh 1111111111 ccmm 33 = (3333. 55 cccc 5555 cccc) hh 1111111111 ccmm 33 = 11111111 ccmm hh 11 1111111111 ccmm 33 11111111 ccmm = 11111111 11 ccmm hh 11111111 ccmm 1111111111 cccc = 11 hh 11111111 55. cccc hh The depth of oil left in the tank is approximately 55. cccc. 6. The inside of a right rectangular prism-shaped tank has a base that is 1111 cccc by cccc and a height of 6666 cccc. The tank is filled to its capacity with water, and then 1111. 9999 LL of water is removed. How far did the water level drop? = (llll)hh VV = (1111 cccc cccc) 6666 cccc VV = 333333 ccmm 6666 cccc VV = ccmm 33 The capacity of the tank is ccmm 33 or 00. 1111 LL. When 1111. 9999 LL or 1111111111 ccmm 33 of water is removed from the tank, there remains ccmm 33 1111111111 ccmm 33 = 99999999 ccmm 33 of water in the tank. = (llll)hh 99999999 ccmm 33 = (1111 cccc cccc) hh 99999999 ccmm 33 = 333333 ccmm hh 11 99999999 ccmm 33 333333 ccmm = 333333 11 ccmm hh 333333 ccmm 99999999 cccc = 11 hh 333333 11 cccc = hh The depth of the water left in the tank is 11 cccc. This means that the water level has dropped 6666 cccc 11 cccc = 3333 11 cccc. Lesson 24: 334
Lesson 24 7. A right rectangular prism-shaped container has inside dimensions of 77 11 cccc long and 44 33 55 cccc wide. The tank is 33 55 full of vegetable oil. It contains 00. 444444 LL of oil. Find the height of the container. = (llll)hh 444444 ccmm 33 = 77 11 cccc 44 33 cccc hh 55 444444 ccmm 33 = 3333 11 ccmm hh 444444 ccmm 33 = 6666 ccmm hh 6666 444444 ccmm 33 = 6666 ccmm ccmm hh 666666mm 888888 cccc = 11 hh 6666 1111 cccc = hh The vegetable oil in the container is 1111 cccc deep, but this is only 33 of the container s depth. Let dd represent the 55 depth of the container in centimeters. 1111 cccc = 33 55 dd 1111 cccc 55 33 = 33 55 55 33 dd 6666 cccc = 11 dd 33 cccc = dd The depth of the container is cccc. Lesson 24: 335
Lesson 24 8. A right rectangular prism with length of 1111 iiii., width of 1111 iiii., and height of 1111 iiii. is filled with water. If the 33 water is emptied into another right rectangular prism with a length of 1111 iiii., a width of 1111 iiii., and height of 99 iiii., will the second container hold all of the water? Explain why or why not. Determine how far (above or below) the water level would be from the top of the container. 1111 iiii. = iiii. = 88 iiii. The height of the water in the first prism is 88 iiii. 33 33 = (llll)hh VV = (1111 iiii. 1111 iiii. ) 88 iiii. VV = 111111 iinn 88 iiii. VV = 11111111 iinn 33 The volume of water is 11111111 iinn 33. = (llll)hh VV = (1111 iiii. 1111 iiii. ) 99 iiii. VV = 111111 iinn 99 iiii. VV = 11111111 iinn 33 The capacity of the second prism is 11111111 iinn 33, which is greater than the volume of water, so the water will fit in the second prism. = (llll)hh Let hh represent the depth of the water in the second prism in inches. 11111111 iinn 33 = (1111 iiii. 1111 iiii. ) hh 11111111 iinn 33 = (111111 iinn ) hh 11 118888 iinn 33 111111 iinn = 111111 11 iinn hh 111111 iinn 11111111 iiii. = 11 hh 111111 88 111111 iiii. = hh 111111 88 88 iiii. = hh 99 The depth of the water in the second prism is 88 88 99 iiii. The water level will be 99 iiii. 88 88 99 iiii. = 11 iiii. from the top of the second prism. 99 Lesson 24: 336