Free Math Worksheets

- Did Paul add or remove any cubes when he rearranged them? - What does the number of unit cubes tell you about the volume? - Would rearranging \(18\) identical building blocks change the total amount of space occupied by the blocks?

1. The original shape contains \(18\) unit cubes. Each cube has a volume of \(1\,\text{cm}^3\), so its total volume is \(18\times1\,\text{cm}^3=18\,\text{cm}^3\). 2. Paul uses the same \(18\) cubes to make the rectangular prism. He does not add or remove any cubes. 3. Therefore, both shapes have a volume of \(18\,\text{cm}^3\). Rearranging the cubes changes the shape, but not the volume.

No. Both shapes use the same \(18\) unit cubes, so each has a volume of \(18\,\text{cm}^3\).

- Which descriptions tell how much can fit inside something? - Think about the difference among length, area, and three-dimensional space. - What happens to water on a very shallow surface when it is moved?

1. The capacity of a milk carton and the space inside a backpack describe three-dimensional space, so they refer to volume or capacity. 2. A bucket is deep and has high sides, so it can hold a large volume of water without spilling easily. 3. A baking sheet is shallow. It holds much less water, and the water can spill over its low sides when the sheet is moved.

The capacity of a milk carton and the space inside a backpack refer to volume or capacity. A bucket is better for carrying water because it is deep, has high sides, and can hold more water without spilling easily.

- Read the stated capacity of each container carefully. - Does pouring a liquid into another container change the amount of liquid? - Which shape can be stacked with the least empty space between containers?

1. Each container has a capacity of \(1\,\text{L}\), so all three hold the same amount of juice. 2. Pouring the juice into a different container changes its shape but not its volume. The volume remains \(1\,\text{L}\). 3. Rectangular cartons can be placed next to and on top of one another with little unused space, so they use the space in a shipping box efficiently.

a) All three containers hold the same amount: \(1\,\text{L}\). b) The volume remains \(1\,\text{L}\); only the shape of the juice changes. c) The rectangular cartons can be packed closely with little unused space.

- Use the finished prism's three dimensions. - Add all the stack heights in the plan. - Subtract the cubes already present from the total needed.

1. The finished prism's volume is \(4\times3\times3=36\,\text{cm}^3\), so it needs \(36\) unit cubes. 2. Add the stack heights shown in the plan: \(2+1+0+0+3+3+2+1+3+2+1+0=18\). 3. The number still needed is \(36-18=18\).

a) The finished prism's volume is \(36\,\text{cm}^3\). b) \(18\) more unit cubes are needed.

- Count each structure separately, stack by stack. - Remember that a stack's height includes every cube below its top cube. - Compare each total with \(15\).

1. Structure A contains \(5 + 5 + 5 + 5 = 20\) cubes. 2. Structure B contains \(2 + 3 + 6 + 4 = 15\) cubes. 3. Structure C contains \(4 + 3 + 4 + 3 = 14\) cubes. 4. Therefore, Structure B contains exactly \(15\) unit cubes.

Structure B

- Add the stack heights in each row. - Then add the three row totals. - Use the symmetry to check that the front and back row totals are equal.

1. The front row contains \(1 + 2 + 1 = 4\) cubes. 2. The middle row contains \(2 + 3 + 2 = 7\) cubes. 3. The back row contains \(1 + 2 + 1 = 4\) cubes. 4. The total is \(4 + 7 + 4 = 15\) cubes.

\(15\) unit cubes

- Count each structure row by row or stack by stack. - Include the cubes below the top cube in every tall stack. - Record all three totals before comparing them.

1. Structure A contains \(1 + 1 + 3 + 1 + 1 = 7\) cubes. 2. Structure B contains \(1 + 1 + 2 + 2 + 2 = 8\) cubes. 3. Structure C contains \(1 + 3 + 3 = 7\) cubes. 4. Structure B has the greatest total, with \(8\) unit cubes.

Structure B, with \(8\) unit cubes

- Use the symmetry to group stacks with the same height. - Count the height-\(3\) stacks, then the height-\(2\) stacks, and finally the center stack.

1. The four corner stacks each have height \(3\), for \(4\times3=12\) cubes. 2. The four middle-edge stacks each have height \(2\), for \(4\times2=8\) cubes. 3. The center stack has height \(1\). Altogether, the building contains \(12+8+1=21\) cubes.

The building contains \(21\) unit cubes.

- Determine each box's width, depth, and height from the outlines. - Multiply the three dimensions. - Compare the results.

1. Box 1 measures \(4\times3\times5\), so it holds \(60\) unit cubes. 2. Box 2 measures \(2\times6\times5\), so it also holds \(60\) unit cubes. 3. The boxes have equal volumes.

Box 1 holds \(60\) unit cubes, and Box 2 holds \(60\) unit cubes.

- Count the cubes in each stack, including the cubes below the top cube. - Work one row at a time so you do not skip a stack. - Compare the totals after you have counted both structures.

1. Structure A has \(1 + 0 + 1 = 2\) cubes in the front row and \(2 + 1 + 2 = 5\) cubes in the back row. Its total is \(2 + 5 = 7\) cubes. 2. Structure B has \(1 + 1 + 0 = 2\) cubes in the front row and \(3 + 1 + 1 = 5\) cubes in the back row. Its total is \(2 + 5 = 7\) cubes. 3. Both structures contain the same number of cubes.

Yes. Structure A has \(7\) cubes, and Structure B also has \(7\) cubes.

- Identify the height of the center stack. - Add the stack heights from one end to the center and then back down.

1. The stack heights from left to right are \(1, 2, 3, 4, 3, 2,\) and \(1\). 2. Add them: \(1 + 2 + 3 + 4 + 3 + 2 + 1 = 16\).

\(16\) unit cubes

- Count the structure one row at a time from front to back. - A stack of height \(2\) contains the top cube and the cube directly below it.

1. The front row contains \(1 + 1 + 0 = 2\) cubes. 2. The middle row contains \(1 + 1 + 1 = 3\) cubes. 3. The back row contains \(2 + 2 + 1 = 5\) cubes. 4. Altogether, the structure contains \(2 + 3 + 5 = 10\) cubes.

The structure contains \(10\) unit cubes.

- Add all the numbers in the plan to find the total number of cubes. - From the front, taller stacks hide shorter stacks directly behind them. - Find the greatest number in the middle column of the plan.

1. Add the stack heights: \(1+2+1+2+4+2+1+2+1=16\). The building contains \(16\) unit cubes. 2. For the front view, find the greatest height in each column of the plan. The middle column has heights \(2\), \(4\), and \(2\), so its visible height is \(4\).

a) \(16\) unit cubes b) \(4\) cubes high

- How many decimeters are in one meter? - Picture the cubes arranged in layers. How many cubes are in one layer, and how many layers are there?

1. Since \(1\,\text{m}=10\,\text{dm}\), exactly \(10\) small cubes fit in one row along each edge. 2. One layer contains \(10\times10=100\) small cubes. The large cube is \(10\,\text{dm}\) high, so it contains \(10\) layers. Therefore, the total number of small cubes is \(10\times10\times10=1000\).

a) \(10\) cubes b) \(1000\) cubes, because \(10\times10\times10=1000\).

- Determine how many cubes fit along each dimension. - For the larger cubes, check whether each box dimension is a multiple of \(2\,\text{cm}\). - Think about the \(5\,\text{cm}\) by \(4\,\text{cm}\) base one layer at a time.

1. For the \(1\,\text{cm}\) cubes, \(5\) fit along the length, \(4\) along the width, and \(3\) along the height. Therefore, \(5\times4\times3=60\) cubes fit. 2. For the \(2\,\text{cm}\) cubes, \(2\) fit along the \(5\,\text{cm}\) length, \(2\) fit along the \(4\,\text{cm}\) width, and \(1\) fits along the \(3\,\text{cm}\) height. 3. Therefore, \(2\times2\times1=4\) larger cubes fit completely inside. 4. The box dimensions \(5\,\text{cm}\) and \(3\,\text{cm}\) are not multiples of \(2\,\text{cm}\). The remaining strips of space are too narrow for another whole cube.

a) \(60\) cubes b) \(4\) cubes c) Some space remains empty because \(5\,\text{cm}\) and \(3\,\text{cm}\) are not divisible by \(2\,\text{cm}\) with no remainder.

- First find how many cubes are in one complete base layer. - Then determine how many equal layers make \(18\) cubes. - Think of the prism as a stack of equal floors.

1. One layer contains \(3\times3=9\) cubes. 2. Divide the total number of cubes by the number in each layer: \(18\div9=2\). 3. The prism is \(2\) cubes high.

The rectangular prism is \(2\) cubes high.

- Use the length and width to find the number of cubes in the bottom layer. - Compare the prism’s height with the edge length of one cube. - Multiply the number of cubes in one layer by the number of layers. - Imagine taking the prism apart and placing every cube in one straight row.

1. The bottom layer contains \(5\times3=15\) cubes. 2. Because each cube is \(1\,\text{cm}\) high, a height of \(2\,\text{cm}\) gives \(2\) layers. 3. The entire prism contains \(15\times2=30\) cubes. 4. A row of \(30\) cubes with edge length \(1\,\text{cm}\) is \(30\times1\,\text{cm}=30\,\text{cm}\) long.

a) \(15\) cubes b) \(2\) layers c) \(30\) cubes d) \(30\,\text{cm}\)

- Examine the figure one layer at a time from bottom to top. - How many cubes are in a complete \(3\times3\) layer? - Use the layer descriptions to include cubes that may be hidden in the view. - How does the total number of \(1\,\text{cm}^3\) cubes relate to the volume?

1. The complete \(3\times3\) bottom layer contains \(3\times3=9\) cubes. 2. The middle layer contains \(9-1=8\) cubes because its center cube is missing. 3. The top layer contains \(4\) cubes. 4. Altogether, the figure contains \(9+8+4=21\) cubes. 5. Each cube has a volume of \(1\,\text{cm}^3\), so \(V=21\times1\,\text{cm}^3=21\,\text{cm}^3\).

The total volume is \(21\,\text{cm}^3\).

- Compare the stack heights one row at a time. - Keep track of which locations are blank. - After choosing the matching plan, add its stack heights to find the total number of cubes.

1. Read the structure row by row. The front row has two blank locations followed by a stack of height \(1\). 2. The middle row has a blank location followed by stacks of heights \(1\) and \(2\). 3. The back row has stacks of heights \(1\), \(2\), and \(3\). 4. Only plan C shows all three rows in those positions. 5. Add the stack heights in plan C: \(1 + 1 + 2 + 1 + 2 + 3 = 10\).

a) Plan C b) \(10\) unit cubes

- How many unit cubes fit in a complete \(3 \times 3 \times 3\) cube? - Count how many cubes are already shown. - Work row by row so that no cubes are missed. - Subtract the number present from the total capacity.

1. A completed \(3 \times 3 \times 3\) cube contains \(3\times3\times3=27\) unit cubes. 2. Count the cubes already present by rows: the back row has \(3+2+2=7\), the middle row has \(2+1+0=3\), and the front row has \(1+0+0=1\). Altogether, \(7+3+1=11\) cubes are present. 3. The number still needed is \(27-11=16\).

\(16\) unit cubes are still needed.

- For each view, use the tallest stack along each line of sight. - Keep the columns or rows in order. - To find the volume, add the heights of all the stacks.

1. From the front, use the tallest stack in each left-to-right column. The heights are \(3\), \(2\), and \(1\), so the front view is \((3, 2, 1)\). 2. From the right, use the tallest stack in each front-to-back row. The heights are \(3\), \(2\), and \(1\), so the right view is \((3, 2, 1)\). Therefore, Option A is correct. 3. Add the heights of all nine stacks: \(3+1+0+2+2+1+0+1+1=11\). The building contains \(11\) unit cubes, so its volume is \(11\) cubic units.

a) Option A b) \(11\) cubic units

- Use the outline to determine each box's length, width, and height. - Count the cubes already shown in each box. - Subtract the existing cubes from the full volume.

1. Box A measures \(3\times3\times4\), so its total volume is \(36\) unit cubes. 2. Box A contains \(4+1+1+4+4=14\) cubes, so \(36-14=22\) cubes are missing. 3. Box B measures \(5\times2\times3\), so its total volume is \(30\) unit cubes. 4. Box B contains \(3+1+1+1+1+3=10\) cubes, so \(30-10=20\) cubes are missing. 5. Since \(22>20\), Box A needs more cubes.

a) Box A has a total volume of \(36\) unit cubes. Box B has a total volume of \(30\) unit cubes. b) Box A needs more cubes; \(22\) cubes are missing.

- Find the tallest stack in each column and each row. - A side view turns the building's rows into visible columns. - Add every stack height to find the volume.

1. The front-view column heights are \((3, 2, 1)\), so the back view is \((1, 2, 3)\). 2. The tallest stacks in the three front-to-back rows have heights \((3, 2, 2)\). From the left, that order is reversed, giving \((2, 2, 3)\). Therefore, the view is from the left. 3. Add the heights of all nine stacks: \(3+1+1+1+2+1+2+0+1=12\). The volume is \(12\) cubic units.

a) From the left b) \(12\) cubic units

- Find the tallest stack and its location. - Match every occupied position and stack height. - Add the plan's stack heights to find the volume.

1. Read the stack heights by location. The back-left stack has height \(3\), the stack directly to its right has height \(1\), the center stack has height \(2\), and the front-right stack has height \(1\). 2. Plan A shows exactly those positions and heights. Plans B and C place one or more stacks differently, so Plan A is correct. 3. Add the stack heights: \(3+1+2+1=7\). The building has volume \(7\) cubic units.

a) Plan A b) \(7\) cubic units

- Mentally rotate each building. - Add the stack heights to find each volume. - A rotation does not change the number of cubes.

1. Building A is an L-shaped structure with a stack of height \(3\) at the corner and two arms. Building B is Building A after a rotation, so the two buildings have the same stack heights and connections. 2. Building A has volume \(3+1+1+1+1=7\) cubic units. Building B also has volume \(1+1+3+1+1=7\) cubic units. 3. Building C has volume \(3+1+1+1=6\) cubic units because one arm is shorter. 4. Building C is different. Its smaller volume confirms that it cannot be a rotation of A or B.

a) Building C b) Building A: \(7\) cubic units; Building B: \(7\) cubic units; Building C: \(6\) cubic units c) A and B have equal volume and matching arrangements after rotation, while C has one fewer cube.

- Compare the occupied positions before comparing heights. - Match the cross pattern and the four-corner pattern. - Add the stack heights to find each volume.

1. Building A has stacks of height \(2\) in the center and at the four side-middle positions. This matches Plan 1. 2. Building B has four stacks of height \(2\), one at each corner, with all other positions empty. This matches Plan 2. Plan 3 shows a complete ring around an empty center, so it matches neither building. 3. Building A has \(5\) stacks of height \(2\), so its volume is \(5 \times 2=10\) cubic units. 4. Building B has \(4\) stacks of height \(2\), so its volume is \(4 \times 2=8\) cubic units.

a) Building A matches Plan 1. Building B matches Plan 2. Plan 3 is not used. b) Building A: \(10\) cubic units; Building B: \(8\) cubic units

- Begin with the front row and compare it with the bottom table row. - Each plan entry gives the height of one stack. - After completing the plan, add all six stack heights.

1. In the front row, the stack heights from left to right are \((2, 1, 2)\). Therefore, \(b = 2\). 2. In the back row, the stack heights are \((3, 2, 3)\). Therefore, \(a = 2\). 3. Add all six stack heights: \(2 + 1 + 2 + 3 + 2 + 3 = 13\).

1) \(a = 2\) and \(b = 2\) 2) \(13\) unit cubes

- Read the front row from the bottom row of Plan P. - Compare the stack heights in both structures. - After choosing the match, add its stack heights.

1. Plan P has front-row stack heights \((2, 1)\) and back-row stack heights \((1, 1)\). 2. Structure A has those heights in the same locations. Structure B switches the two front-row heights. 3. Add the stack heights in structure A: \(2 + 1 + 1 + 1 = 5\).

a) Structure A b) \(5\) unit cubes

- For the front view, use the tallest stack in each left-to-right column. - Add every stack height to find the volume. - Look for cubes hidden behind cubes at the same height.

1. In the left column, the tallest stack is \(2\) cubes high, so \(2\) square faces are visible. In the right column, the tallest stack is \(3\) cubes high, so \(3\) square faces are visible. Altogether, \(2+3=5\) square faces are visible. 2. Add all stack heights: \(2+1+0+3=6\). The volume is \(6\) cubic units. 3. One cube is directly behind another cube at the same height, so its front face is hidden. That is why the front view shows \(5\) squares even though the building contains \(6\) cubes.

a) \(5\) square faces b) \(6\) cubic units c) One cube's front face is hidden behind another cube.

- Use the tallest stack in each left-to-right column for the front view. - Add all stack heights to find each volume. - Matching views and volumes do not always mean the buildings are identical.

1. Building A has visible front heights \((3, 1)\): the left column is \(3\) cubes high and the right column is \(1\) cube high. Building B has the same visible front heights, so the front views match. 2. Building A has volume \(3+1+1+1=6\) cubic units. Building B has volume \(3+1+2+0=6\) cubic units. 3. The buildings are not identical. They have the same front view and volume, but the stack heights behind the front row are arranged differently.

a) Yes. Both front views have column heights \((3, 1)\). b) Each building has volume \(6\) cubic units. c) No. Their hidden stack arrangements are different.

- For the front view, find the tallest stack in each column. - For the right-side view, find the tallest stack in each row. - Add every stack height to find the volume.

1. From the front, the visible column heights are \(3\), \(1\), and \(2\). Therefore, \(3+1+2=6\) square faces are visible. 2. From the right, the visible row heights are \(1\), \(2\), and \(3\). Therefore, \(1+2+3=6\) square faces are visible. 3. Add all stack heights: \(1+0+1+2+0+1+3+1+2=11\). The volume is \(11\) cubic units.

a) \(6\) square faces b) \(6\) square faces c) \(11\) cubic units

- Compare each stack with the corresponding plan entry. - Add the corrected stack heights to find the volume. - Compare the correct total with the total from the incorrect plan.

1. The front row of building G has heights \((1, 2, 1)\), which matches the front row of plan P. The back row of the building has heights \((2, 2, 2)\). 2. The middle entry of the plan's back row is \(3\), but the corresponding stack is only \(2\) cubes high. The \(3\) should be replaced with \(2\). 3. The actual volume is \(1+2+1+2+2+2=10\) cubic units. 4. The incorrect plan totals \(11\) cubic units, so it gives a volume that is \(1\) cubic unit too large.

a) The \(3\) in the middle of the back row is wrong. It should be \(2\). b) \(10\) cubic units c) \(1\) cubic unit too large

- For each building, record the height of every stack. - Add the stack heights for each building separately.

1. Building A contains \(1+(2+1)+(3+2+1)=10\) cubes. 2. Building B contains \((2+1)+(2+2+1)+(3+2+2)=15\) cubes. 3. Building C contains \((2+2)+(4+2+2)=12\) cubes. 4. Therefore, Building B has volume \(15\) cubic units.

Building B has volume \(15\) cubic units.

- Find the number of cubes in the full block. - Find the number of cubes removed for the shaft. - Subtract the removed cubes from the original total.

1. The full cube contains \(5\times5\times5=125\) unit cubes. 2. The shaft contains \(3\times3\times5=45\) unit cubes. 3. The number remaining is \(125-45=80\).

\(80\) unit cubes remain.

- Count each structure row by row or stack by stack. - Include every cube below the top cube in a tall stack. - Subtract the smaller total from the larger total.

1. Structure A contains \(1 + 1 = 2\) cubes in the front row and \(2 + 1 = 3\) cubes in the back row, for a total of \(2 + 3 = 5\) cubes. 2. Structure B contains \(0 + 1 + 0 = 1\) cube in the front row and \(2 + 2 + 1 = 5\) cubes in the back row, for a total of \(1 + 5 = 6\) cubes. 3. Since \(6 - 5 = 1\), Structure B contains one more cube.

1) Structure B 2) \(1\) unit cube

- For the bottom layer, count one cube at every occupied position. - For the total, count every cube in each stack. - Check whether the position at the far right of the back row is occupied.

1. The bottom layer has one cube at each occupied position. There are \(2\) occupied positions in the front row and \(3\) in the back row, so the bottom layer contains \(2 + 3 = 5\) cubes. 2. The front row contains \(2 + 1 + 0 = 3\) cubes, and the back row contains \(2 + 2 + 1 = 5\) cubes. The entire structure contains \(3 + 5 = 8\) cubes.

The bottom layer contains \(5\) unit cubes. The entire structure contains \(8\) unit cubes.

- Record the height of the stack at each occupied position. - List the back row first and the front row last. - Add all the stack heights to find the total number of cubes.

1. The back row has stack heights \((2, 2, 2)\), the middle row has heights \((1, \text{empty}, 1)\), and the front row has heights \((1, 1, 1)\). 2. Add the stack heights: \(2+2+2+1+1+1+1+1=11\) cubes.

a) Back row: \((2, 2, 2)\); middle row: \((1, \text{empty}, 1)\); front row: \((1, 1, 1)\) b) The building contains \(11\) unit cubes.

- Match the front row of the 3D building to a row in the plan. - Use the compass orientation to name that side. - Add all plan entries to find the volume.

1. The front row of the 3D building has stack heights \((2, 3, 1)\). 2. In Plan P, this pattern is the top row, which is the north row. Because the north row appears at the front of the 3D view, the building is viewed from the north. 3. Add all stack heights in the plan: \(1+1+1+1+2+1+2+3+1=13\). The volume is \(13\) cubic units.

a) From the north b) \(13\) cubic units

- Match the front row of the building with the bottom row of each plan. - Add all stack heights in the correct plan. - Find the number of unit cubes in a solid \(3 \times 3 \times 3\) cube.

1. Plan A matches every stack height and position. 2. Add the stack heights: \(3+2+7 \times 1=12\) cubes. 3. A solid \(3 \times 3 \times 3\) cube contains \(27\) unit cubes. The number missing is \(27-12=15\).

a) Plan A b) \(12\) cubes c) \(15\) cubes

- Add all the stack heights in the diagram. - Find the volume of one cube before finding the structure's volume. - Compare the number of cubes shown with the number in a \(4\times3\times2\) prism.

1. Add the stack heights: \(1+0+0+1+0+0+1+1+0+2+2+1=9\). The structure contains \(9\) cubes. 2. One cube has volume \(2\times2\times2=8\,\text{cm}^3\), so the structure has volume \(9\times8=72\,\text{cm}^3\). 3. A \(4\times3\times2\) rectangular prism contains \(4\times3\times2=24\) cubes. 4. The number of additional cubes needed is \(24-9=15\).

a) The volume is \(72\,\text{cm}^3\). b) \(15\) more cubes are needed.

- Count the stacks row by row from front to back. - A stack's height equals the number of cubes at that plan location. - Include zeros for locations with no stack.

1. Add the stack heights: \(1 + 2 + 1 + 2 + 3 + 3 + 3 = 15\). 2. The front row is \((1, 2, 1)\), the middle row is \((0, 2, 0)\), and the back row is \((3, 3, 3)\).

a) \(15\) unit cubes b) Front: \((1, 2, 1)\) Middle: \((0, 2, 0)\) Back: \((3, 3, 3)\)

- Match each building stack with its plan cell. - The front row of the building is the bottom row of the table. - Compare the current total with four stacks of height \(4\).

1. The front-left stack is \(3\) cubes high, so \(x=3\). 2. The back-right stack is \(2\) cubes high, so \(y=2\). 3. The building contains \(3+1+2+2=8\) cubes. Four stacks of height \(4\) contain \(4 \times 4=16\) cubes. Therefore, \(16-8=8\) cubes must be added.

a) \(x=3\) and \(y=2\) b) \(8\) cubes

- Track where each stack moves during the rotation. - Keep each stack height unchanged. - Add all stack heights to find the volume.

1. Before the rotation, the front row is \((2, 1)\) and the back row is \((3, 0)\). 2. After a \(90^\circ\) clockwise rotation, the front row is \((1, 0)\) and the back row is \((2, 3)\). Plan B shows this arrangement. 3. The building contains \(2+1+3+0=6\) unit cubes, so its volume is \(6\) cubic units both before and after the rotation. 4. A rotation changes only the positions of the stacks; it does not add or remove cubes.

a) Plan B b) \(6\) cubic units before and \(6\) cubic units after c) The same cubes are only rearranged by the rotation.

- Break the staircase into sections with the same depth. - Find the capacity of one full layer of the box. - Remember to include cubes hidden below the upper cubes.

1. The four steps are each \(4\) cubes deep. They contain \(4\times1=4\), \(4\times2=8\), \(4\times3=12\), and \(4\times4=16\) cubes. The building contains \(4+8+12+16=40\) cubes. 2. The \(4 \times 4 \times 4\) box holds \(4\times4\times4=64\) cubes. 3. The number still needed is \(64-40=24\).

1. \(40\) unit cubes 2. \(64\) unit cubes 3. \(24\) unit cubes

- Add every number in the plan to find the total number of cubes. - For the front view, compare the stacks that line up behind one another. - In each visible column, only the tallest aligned stack determines the height. - For the left-side view, compare the stacks in each row from front to back.

1. Add all the stack heights: \(1+2+0+3+1+2=9\). The building contains \(9\) unit cubes. 2. For the front view, use the greatest height in each column: left \(\max(1,3)=3\), middle \(\max(2,1)=2\), and right \(\max(0,2)=2\). 3. For the left-side view, use the greatest height in each row: front \(\max(1,2,0)=2\) and back \(\max(3,1,2)=3\).

a) \(9\) unit cubes b) Left: \(3\); middle: \(2\); right: \(2\) c) Front: \(2\); back: \(3\)

- For the front view, find the tallest stack in each column. - For the left-side view, find the tallest stack in each row. - Add every stack height to find the total number of cubes.

1. For the front view, take the greatest stack height in each column: \(\max(2,2,2)=2\), \(\max(2,1,0)=2\), and \(\max(0,0,2)=2\). The front-view heights are \((2, 2, 2)\). 2. For the left-side view, take the greatest height in each row. The back, middle, and front rows each have greatest height \(2\), so the left-side-view heights are \((2, 2, 2)\). 3. Add the stack heights: \((2+2+0)+(2+1+0)+(2+0+2)=4+3+4=11\).

a) \((2, 2, 2)\) b) \((2, 2, 2)\) c) \(11\) unit cubes

- Each number in the plan gives the height of one stack. - First find the total number of cubes needed for the building. - Subtract that number from the starting supply.

1. Add the occupied stack heights by rows: \(1+2+3+2=8\), \(2+1+2+1=6\), and \(1+1=2\). 2. The building uses \(8+6+2=16\) cubes. 3. Jonah has \(20-16=4\) cubes left.

Jonah has \(4\) cubes left.

- Count each structure carefully, including cubes below the top cubes. - Find the total for each structure. - Subtract the smaller total from the larger total.

1. Structure A contains \(1 + 1 + 0 = 2\) cubes in the front row and \(2 + 1 + 1 = 4\) cubes in the back row, for a total of \(2 + 4 = 6\) cubes. 2. Structure B contains \(1 + 1 + 1 = 3\) cubes in the front row and \(2 + 2 + 1 = 5\) cubes in the back row, for a total of \(3 + 5 = 8\) cubes. 3. Maria must remove \(8 - 6 = 2\) cubes from Structure B.

Maria must remove \(2\) unit cubes.

- For each plan, find the tallest stack in each column. - Add all plan entries to find each volume. - Compare the positions of stacks, not only the totals.

1. In Plan A, the greatest height in the left column is \(3\), and the greatest height in the right column is \(2\). Its front view is \((3, 2)\). 2. In Plan B, the greatest heights are \(2\) and \(3\). Its front view is \((2, 3)\). Therefore, Plan A matches the given front view. 3. Plan A has volume \(3+1+2+2=8\) cubic units. Plan B has volume \(1+3+2+2=8\) cubic units. 4. The plans have equal volumes but place their stacks differently, so their front views are different.

a) Plan A b) Each plan represents \(8\) cubic units. c) Equal volume does not determine where the stacks are located.

- How many cubes are represented by the known numbers? - How many more cubes are needed to reach the total of \(15\)?

1. Add the known entries in the plan: \(3 + 2 + 1 + 2 + 1 + 2 + 2 + 0 = 13\). 2. The structure contains \(15\) cubes, so the missing number is \(15 - 13 = 2\).

\(2\)

- For each view, identify which stacks line up behind one another. - The tallest stack in each line determines the visible column height. - To find the volume, add the heights of all stacks in the plan.

1. Looking from the front, the greatest heights in the four plan columns are \(3\), \(2\), \(3\), and \(2\). Looking from the back reverses their order, so the back-view heights are \((2, 3, 2, 3)\). 2. For the right-side view, find the greatest stack height in each row from front to back. The row maxima are \(3\), \(2\), and \(3\), so the right-side heights are \((3, 2, 3)\). 3. Add all stack heights in the plan: \((3+1+2)+(1+2+1)+(1+3)=6+4+4=14\). Therefore, the volume is \(14\) cubic units.

a) \((2, 3, 2, 3)\) b) \((3, 2, 3)\) c) \(14\) cubic units

- Compare stack heights at matching positions. - Adding one cube makes exactly one stack one cube taller and leaves every other stack unchanged. - Use the total number of cubes as a final check.

1. Structure S has stack heights \((1, 2)\) in the front row and \((2, 3)\) in the back row. 2. Adding one cube must leave every stack unchanged except one stack, which becomes exactly one cube taller. 3. Structure A matches S except that the back-right stack increases from \(3\) cubes to \(4\) cubes. Structure B changes two stacks, and Structure C has one shorter stack. 4. Therefore, Structure A could be the result of adding one cube.

Structure A

- At each base position, the two stack heights must add to \(2\). - A position with height \(2\) in piece A needs height \(0\) in the matching piece. - Compare the required heights with pieces B, C, and D.

1. At each position in the \(2\times2\) base, the two stack heights must add to \(2\). 2. Piece A has stack heights \((2, 1)\) in one row and \((1, 0)\) in the other row. 3. The matching piece must therefore have heights \((0, 1)\) and \((1, 2)\). 4. Piece B has exactly these stack heights.

Piece B fits piece A exactly.

- Determine how many cubes fit along each inside dimension. - How many cubes are in one layer, and how many complete layers fit? - What happens when the box height is not a multiple of the cube edge length? - Can part of a cube be packed as an additional whole cube?

1. In a standard box, \(50\div5=10\) cubes fit along the length, \(30\div5=6\) fit along the width, and \(20\div5=4\) fit along the height. 2. Therefore, \(10\times6\times4=240\) cubes fit in the standard box. 3. In the shorter box, only \(3\) complete cubes fit along the height because \(3\times5=15\) and \(4\times5=20>18\). 4. Therefore, \(10\times6\times3=180\) cubes fit in the shorter box. 5. Dividing the two volumes would treat all empty space as though it could be combined into space for additional cubes. The unused \(3\,\text{cm}\) of height is not tall enough for another complete cube.

a) \(240\) cubes b) \(180\) cubes. Only whole cubes can be packed, and the remaining \(3\,\text{cm}\) of height cannot hold another cube.

- First express all dimensions in the same unit. - How many cubes fit along each side of the crate? - How many cubes cover the bottom in one layer, and how many layers fit? - Which metric volume unit is equal to one liter?

1. Convert the crate dimensions to centimeters: \(1.2\,\text{m}=120\,\text{cm}\), \(8\,\text{dm}=80\,\text{cm}\), and the height is \(60\,\text{cm}\). 2. Along the three dimensions, \(120\div20=6\), \(80\div20=4\), and \(60\div20=3\) cubes fit. 3. One layer has \(6\times4=24\) cubes. With \(3\) layers, the crate holds \(24\times3=72\) cubes. 4. In decimeters, the crate measures \(12\,\text{dm}\times8\,\text{dm}\times6\,\text{dm}\). Its volume is \(12\times8\times6=576\,\text{dm}^3\). 5. Since \(1\,\text{dm}^3=1\,\text{L}\), the volume is \(576\,\text{L}\).

a) \(72\) cubes b) Each layer has \(6\times4=24\) cubes, and \(3\) layers give \(24\times3=72\) cubes. c) \(576\,\text{L}\)

- Determine how many times the bottom layer fits into the total number of cubes. - Find two whole numbers whose product is \(15\). - Picture how the cubes could be arranged on the bottom of the box.

1. Each layer contains \(15\) cubes, so the number of layers is \(45\div15=3\). 2. The length and width of the bottom layer must be whole-number factors of \(15\). One possible pair is \(5\,\text{cm}\) and \(3\,\text{cm}\), since \(5\times3=15\).

a) \(3\) layers b) One possible answer is \(5\,\text{cm}\) by \(3\,\text{cm}\).

- Use the tallest stack in each row for the side view. - Add all stack heights to find each volume. - A side view can hide cubes behind taller stacks.

1. For Building A, the tallest stack in each row is \(2\), so the left-side view shows \(2+2+2=6\) square faces. 2. For Building B, the tallest stack heights by row are \(2\), \(3\), and \(2\), so the left-side view shows \(2+3+2=7\) square faces. Therefore, Building B has more visible squares. 3. Building A has volume \(1+2+1+1+2+1+1+2+1=12\) cubic units. Building B has volume \(2+1+1+0+3+0+1+1+2=11\) cubic units. 4. Building B has more squares visible from the left, but Building A has the greater volume. A side view records only the tallest stack in each row, not every cube.

a) Building B: \(7\) visible square faces compared with \(6\) for Building A b) Building A: \(12\) cubic units; Building B: \(11\) cubic units c) No. Building B has more visible squares, but Building A has greater volume.

- Find the total number of cubes in the completed large cube. - Divide the current building into horizontal layers and count each layer. - Look for a pattern in the layer counts. - Subtract the number present from the completed total.

1. A complete \(4 \times 4 \times 4\) cube contains \(4\times4\times4=64\) unit cubes. 2. Count the building by horizontal layers from top to bottom: \(1,3,6,10\). It contains \(1+3+6+10=20\) cubes. 3. The number still needed is \(64-20=44\).

\(44\) unit cubes must be added.

- Mentally rotate each building from above. - Compare where the one-cube stack is attached to the two taller stacks. - Count the cubes in every building before deciding what volume can show.

1. In Building 1, two adjacent stacks of height \(2\) form the corner of an L-shape, and a one-cube stack is attached at the third position. Building 2 has the same relative arrangement after a \(90^\circ\) rotation. 2. In Building 3, the one-cube stack is attached at the other end of the L-shape, so it cannot be obtained from Building 1 or 2 by a rotation. 3. Each building has volume \(2+2+1=5\) cubic units. 4. Equal volume tells only the number of cubes. Buildings can use the same number of cubes in different arrangements, as Building 3 shows.

a) Buildings 1 and 2 b) Each building has volume \(5\) cubic units. c) Equal volume does not determine the arrangement of the cubes.

- Count the unit cubes in each piece. - Find the number of cubes in a \(2 \times 2 \times 2\) cube. - Which four piece counts add to that total? - Check that those four shapes can fit within the large cube.

1. Count the unit cubes in each piece: A has \(1\), B has \(2\), C has \(2\), D has \(3\), and E has \(4\). 2. All five pieces contain \(1+2+2+3+4=12\) cubes. A \(2 \times 2 \times 2\) cube contains \(8\) cubes, so the unused piece must contain \(12-8=4\) cubes. Only Piece E has \(4\) cubes. 3. Pieces A, B, C, and D can fill the large cube: place B and C as two vertical stacks in the front row; place D in the back row with its two-cube stack on the left and its single cube at the upper right; then A fills the lower-right space in the back row.

Piece E

- Find the total number of cubes in the building. - Two identical stacked pieces must contain the same number of cubes. - Locate the empty position in the front or back row. - The bottom plan row represents the front of the building.

1. The building has five stacks of height \(2\) and one empty position, so it contains \(5 \times 2=10\) cubes. 2. Because the two pieces are identical and stacked directly, each piece contains one cube at each occupied position. Each piece therefore contains \(5\) cubes. 3. The empty position is in the middle of the front row. Plan A shows exactly that arrangement. Plan B has only \(4\) cubes, and Plan C places the empty position in the back row.

Plan A