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Volume of prisms

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Difficulty: 1 – 5

5111937

Two rectangular prisms each have a volume of \(480\,\text{cm}^3\). Prism A is \(12\,\text{cm}\) long and \(8\,\text{cm}\) wide. Prism B is \(10\,\text{cm}\) long and \(6\,\text{cm}\) high. Which prism is taller? Justify your answer with calculations.

Hints

- Find the missing height of Prism A. - Use \(V=Bh\). - Compare the calculated height with Prism B's given height.

Solution

1. Prism A has base area \(12\cdot8=96\,\text{cm}^2\). 2. Its height is \(h=V\div B=480\div96=5\,\text{cm}\). 3. Prism B's height is given as \(6\,\text{cm}\). 4. Since \(6>5\), Prism B is taller.

Answer

Prism B is taller. Prism A is \(5\,\text{cm}\) high, while Prism B is \(6\,\text{cm}\) high.

5112427

A rectangular rainwater collection basin in a school garden is \(4\,\text{m}\) long, \(2.5\,\text{m}\) wide, and \(1.5\,\text{m}\) deep. a) Find its maximum capacity in cubic meters. b) Find its maximum capacity in liters.

Hints

- Use the volume formula for a rectangular prism. - Convert cubic meters to liters.

Solution

1. The basin's volume is \(4\cdot2.5\cdot1.5=15\,\text{m}^3\). 2. Since \(1\,\text{m}^3=1000\,\text{L}\), \(15\,\text{m}^3=15{,}000\,\text{L}\).

Answer

a) The maximum capacity is \(15\,\text{m}^3\). b) The maximum capacity is \(15{,}000\,\text{L}\).

5113777

A rectangular sandbox is \(2.50\,\text{m}\) long and \(1.80\,\text{m}\) wide. Exactly \(1.35\,\text{m}^3\) of sand is spread evenly in the box. How deep is the sand, in centimeters?

Hints

- Find the area of the rectangular base. - Divide the sand volume by the base area. - Convert meters to centimeters.

Solution

1. The sandbox base area is \(2.50\cdot1.80=4.5\,\text{m}^2\). 2. Using \(V=Bh\), the depth is \(h=1.35\div4.5=0.3\,\text{m}\). 3. Convert the depth: \(0.3\,\text{m}=30\,\text{cm}\).

Answer

The sand is \(30\,\text{cm}\) deep.

5114047

A rectangular aquarium has a \(50\,\text{cm}\times30\,\text{cm}\) base. An additional \(12\,\text{L}\) of water is poured into it. How many centimeters does the water level rise?

Hints

- Convert liters to cubic centimeters. - Find the aquarium's base area. - Divide the added volume by the base area.

Solution

1. Convert the added volume: \(12\,\text{L}=12{,}000\,\text{cm}^3\). 2. The base area is \(50\cdot30=1500\,\text{cm}^2\). 3. The rise is \(12{,}000\div1500=8\,\text{cm}\).

Answer

The water level rises \(8\,\text{cm}\).

5117887

One aquarium is \(60\,\text{cm}\) long, \(40\,\text{cm}\) wide, and \(30\,\text{cm}\) high. A second aquarium has the same length and width but is \(50\,\text{cm}\) high. How many more liters can the second aquarium hold?

Hints

- Find each rectangular-prism volume. - Convert cubic centimeters to liters. - Subtract the smaller capacity from the larger one.

Solution

1. The first aquarium's volume is \(60\cdot40\cdot30=72{,}000\,\text{cm}^3=72\,\text{L}\). 2. The second aquarium's volume is \(60\cdot40\cdot50=120{,}000\,\text{cm}^3=120\,\text{L}\). 3. The difference is \(120-72=48\,\text{L}\).

Answer

The second aquarium holds \(48\,\text{L}\) more.

5118307

A rectangular road-sand container has a base area of \(12\,\text{m}^2\) and a total height of \(1.50\,\text{m}\). It contains \(15\,\text{m}^3\) of sand spread evenly. How far is the sand surface below the top edge?

Hints

- Divide the sand volume by the base area to find its depth. - Subtract the sand depth from the container's total height.

Solution

1. The sand depth is \(15\div12=1.25\,\text{m}\). 2. The distance below the top is \(1.50-1.25=0.25\,\text{m}\), or \(25\,\text{cm}\).

Answer

The sand surface is \(0.25\,\text{m}\), or \(25\,\text{cm}\), below the top.

5315597

The triangular prism shown has a triangular base with base \(8\,\text{cm}\) and corresponding height \(5\,\text{cm}\). The prism is \(15\,\text{cm}\) long. Find its volume in cubic centimeters.

Figure for problem 531559

Hints

- Find the area of the triangular base first. - A prism's volume is its base area multiplied by its length. - Substitute the given measurements into the formulas.

Solution

1. The area of the triangular base is \(B=\frac{1}{2}\cdot8\cdot5=20\,\text{cm}^2\). 2. The prism volume is \(V=Bh=20\cdot15=300\,\text{cm}^3\).

Answer

\(300\,\text{cm}^3\)

5315817

The aluminum part shown is a prism with an L-shaped base. The base has overall width \(5\,\text{cm}\), overall height \(6\,\text{cm}\), horizontal arm thickness \(2\,\text{cm}\), and vertical arm thickness \(2\,\text{cm}\). The prism is \(10\,\text{cm}\) long. Find its volume.

Figure for problem 531581

Hints

- Decompose the L-shaped base into rectangles. - Add the rectangle areas to find the base area. - Multiply the base area by the prism length.

Solution

1. Split the L-shaped base into a \(2\,\text{cm}\times6\,\text{cm}\) rectangle and a \(3\,\text{cm}\times2\,\text{cm}\) rectangle. 2. The base area is \(B=2\cdot6+3\cdot2=18\,\text{cm}^2\). 3. The volume is \(V=18\cdot10=180\,\text{cm}^3\).

Answer

\(180\,\text{cm}^3\)

5355527

A rectangular block of cheese measures \(12\,\text{cm}\times8\,\text{cm}\times5\,\text{cm}\). It is cut vertically along a diagonal of its \(12\,\text{cm}\times8\,\text{cm}\) base. Find the volume of one of the two triangular prisms formed.

Figure for problem 535552

Hints

- A diagonal divides a rectangle into two congruent triangles. - The two resulting prisms have equal volumes. - Find the original rectangular prism's volume, then divide by \(2\).

Solution

1. The cut divides the rectangular base into two congruent triangles, so the original rectangular prism is divided into two equal-volume triangular prisms. 2. The original volume is \(V=12\cdot8\cdot5=480\,\text{cm}^3\). 3. One triangular prism has volume \(480\div2=240\,\text{cm}^3\).

Answer

\(240\,\text{cm}^3\)

5355987

A structural channel has a U-shaped cross section. Its outside width is \(5\,\text{cm}\), each side is \(4\,\text{cm}\) tall, and the material is uniformly \(1\,\text{cm}\) thick. The channel is \(12\,\text{cm}\) long. Find its volume.

Figure for problem 535598

Hints

- Find the inner width and height from the uniform thickness. - Subtract the rectangular opening from the outer rectangle. - Multiply the cross-sectional area by the length.

Solution

1. The U-shaped cross-sectional area is the outer rectangle minus the opening: \(B=5\cdot4-3\cdot3=11\,\text{cm}^2\). 2. The volume is \(V=11\cdot12=132\,\text{cm}^3\).

Answer

\(132\,\text{cm}^3\)

5356007

A solid concrete block is a prism with a trapezoidal base. The trapezoid has parallel sides \(10\,\text{cm}\) and \(6\,\text{cm}\), and height \(4\,\text{cm}\). The block is \(15\,\text{cm}\) long. Find its volume.

Figure for problem 535600

Hints

- Use the trapezoid area formula. - Multiply the base area by the prism length. - Keep the area and volume units distinct.

Solution

1. The trapezoidal base area is \(B=\frac{10+6}{2}\cdot4=32\,\text{cm}^2\). 2. The volume is \(V=32\cdot15=480\,\text{cm}^3\).

Answer

\(480\,\text{cm}^3\)

5360187

A stone block is a right prism with a parallelogram-shaped base. The parallelogram has base \(30\,\text{cm}\) and corresponding height \(18\,\text{cm}\). The block is \(50\,\text{cm}\) long. Find its volume.

Figure for problem 536018

Hints

- Use the parallelogram area formula. - Multiply the base area by the prism length.

Solution

1. The parallelogram base area is \(B=30\cdot18=540\,\text{cm}^2\). 2. The volume is \(V=540\cdot50=27{,}000\,\text{cm}^3\).

Answer

\(27{,}000\,\text{cm}^3\)

5360407

The interior of a storage shed is a prism with a trapezoidal side cross section. The front interior height is \(2.80\,\text{m}\), the rear interior height is \(2.20\,\text{m}\), and the interior depth is \(3.00\,\text{m}\). The shed is \(4.50\,\text{m}\) wide. Find the interior air volume.

Figure for problem 536040

Hints

- Treat the side cross section as the prism's base. - Use the trapezoid area formula. - Multiply the cross-sectional area by the shed width.

Solution

1. The trapezoidal side area is \(B=\frac{2.80+2.20}{2}\cdot3.00=7.50\,\text{m}^2\). 2. The volume is \(V=7.50\cdot4.50=33.75\,\text{m}^3\).

Answer

\(33.75\,\text{m}^3\)

5111397

A cube-shaped garden water tank has an inside edge length of \(1\,\text{m}\). It currently contains \(150\,\text{L}\) of water. a) How many cubic meters of water must be added to make the tank half full? b) How high is the water currently in the tank, in centimeters?

Hints

- Find half of the tank's total volume. - Convert liters to cubic meters. - Use \(V=Bh\) to find the water height.

Solution

1. The tank's total volume is \(1\cdot1\cdot1=1\,\text{m}^3\), so half its volume is \(0.5\,\text{m}^3\). 2. Convert the current amount: \(150\,\text{L}=0.15\,\text{m}^3\). 3. The amount to add is \(0.5-0.15=0.35\,\text{m}^3\). 4. The base area is \(1\,\text{m}^2\). Using \(V=Bh\), the current water height is \(h=0.15\div1=0.15\,\text{m}=15\,\text{cm}\).

Answer

a) Add \(0.35\,\text{m}^3\) of water. b) The water is \(15\,\text{cm}\) high.

5111737

One cube has an edge length of \(2\,\text{cm}\). A second cube has an edge length of \(4\,\text{cm}\), twice as long. a) Find the volume of each cube. b) A student claims, “Doubling a cube's edge length doubles its volume.” Test the claim and explain your conclusion.

Hints

- Use \(V=s^3\) for each cube. - Divide the larger volume by the smaller volume. - Think about how many dimensions are doubled.

Solution

1. The smaller cube's volume is \(2^3=8\,\text{cm}^3\). 2. The larger cube's volume is \(4^3=64\,\text{cm}^3\). 3. Compare the volumes: \(64\div8=8\). 4. The claim is false. Doubling all three dimensions multiplies the volume by \(2\cdot2\cdot2=8\), not by \(2\).

Answer

a) The volumes are \(8\,\text{cm}^3\) and \(64\,\text{cm}^3\). b) The claim is false. Doubling the edge length makes the volume \(8\) times as great.

5111947

A rectangular cistern has a base area of \(12\,\text{m}^2\). After a heavy rain, the water is \(45\,\text{cm}\) deep. How many liters of water are in the cistern?

Hints

- Convert the water depth to meters. - Use \(V=Bh\). - Convert cubic meters to liters.

Solution

1. Convert the water depth: \(45\,\text{cm}=0.45\,\text{m}\). 2. The water volume is \(V=Bh=12\cdot0.45=5.4\,\text{m}^3\). 3. Since \(1\,\text{m}^3=1000\,\text{L}\), \(5.4\,\text{m}^3=5400\,\text{L}\).

Answer

The cistern contains \(5400\,\text{L}\) of water.

5111957

An aquarium is \(80\,\text{cm}\) long and \(40\,\text{cm}\) wide. It contains \(96\,\text{L}\) of water. How deep is the water, in centimeters?

Hints

- Find the area of the aquarium's base. - Convert liters to cubic centimeters. - Divide the volume by the base area.

Solution

1. The aquarium's base area is \(80\cdot40=3200\,\text{cm}^2\). 2. Convert the water volume: \(96\,\text{L}=96{,}000\,\text{cm}^3\). 3. Use \(V=Bh\): \(h=96{,}000\div3200=30\,\text{cm}\).

Answer

The water is \(30\,\text{cm}\) deep.

5111967

Pool A has a square base with side length \(4\,\text{m}\). The water in it is \(1.5\,\text{m}\) deep. Pool B has a base area of \(20\,\text{m}^2\). All the water from Pool A is pumped into Pool B. How deep is the water in Pool B?

Hints

- First find the volume of water in Pool A. - The amount of water does not change when it is transferred. - Divide the volume by Pool B's base area.

Solution

1. Pool A's base area is \(4\cdot4=16\,\text{m}^2\). 2. Its water volume is \(16\cdot1.5=24\,\text{m}^3\). 3. The same volume goes into Pool B, so \(h=V\div B=24\div20=1.2\,\text{m}\).

Answer

The water in Pool B is \(1.2\,\text{m}\) deep.

5112017

A rectangular sandbox is \(2.4\,\text{m}\) long and \(1.5\,\text{m}\) wide. It will be filled with sand to a depth of \(20\,\text{cm}\). a) Find the volume of sand needed, in cubic meters. b) Sand costs \(\$25.00\) per cubic meter. Find the total cost. c) The same sand is also sold in \(20\,\text{L}\) bags. How many bags would be needed?

Hints

- Convert the sand depth to meters before finding volume. - Multiply the volume by the price per cubic meter. - Convert cubic meters to liters before finding the number of bags.

Solution

1. Convert the depth: \(20\,\text{cm}=0.2\,\text{m}\). 2. The sand volume is \(2.4\cdot1.5\cdot0.2=0.72\,\text{m}^3\). 3. The cost is \(0.72\cdot\$25.00=\$18.00\). 4. Convert the volume: \(0.72\,\text{m}^3=720\,\text{L}\). 5. The number of bags is \(720\div20=36\).

Answer

a) \(0.72\,\text{m}^3\) of sand is needed. b) The sand costs \(\$18.00\). c) \(36\) bags are needed.

5112057

A wooden cube has an edge length of \(4\,\text{cm}\). A wooden rectangular prism has twice the cube's length, three times its width, and half its height. a) Find the volume of each solid. b) By what factor is the rectangular prism's volume greater than the cube's volume? Explain how to find this factor directly from the dimension changes.

Hints

- Find the rectangular prism's three dimensions first. - Compare the volumes by division. - Multiply the three dimension-change factors.

Solution

1. The cube's volume is \(4^3=64\,\text{cm}^3\). 2. The rectangular prism's dimensions are \(8\,\text{cm}\), \(12\,\text{cm}\), and \(2\,\text{cm}\). 3. Its volume is \(8\cdot12\cdot2=192\,\text{cm}^3\). 4. The volume factor is \(192\div64=3\). 5. Directly, the product of the dimension factors is \(2\cdot3\cdot\frac{1}{2}=3\).

Answer

a) The cube's volume is \(64\,\text{cm}^3\), and the rectangular prism's volume is \(192\,\text{cm}^3\). b) The prism's volume is \(3\) times the cube's volume because \(2\cdot3\cdot\frac{1}{2}=3\).

5112067

An aquarium is \(50\,\text{cm}\) long, \(40\,\text{cm}\) wide, and \(30\,\text{cm}\) high. a) How many liters can it hold when filled to the top? b) The water is currently \(20\,\text{cm}\) deep. Find the current water volume in liters. c) Another \(10\,\text{L}\) of water is added. Find the new water depth.

Hints

- Use the volume formula for a rectangular prism. - Relate \(1000\,\text{cm}^3\) to \(1\,\text{L}\). - For part c, find the new volume, then divide by the base area.

Solution

1. The full aquarium's volume is \(50\cdot40\cdot30=60{,}000\,\text{cm}^3=60\,\text{L}\). 2. At a depth of \(20\,\text{cm}\), the water volume is \(50\cdot40\cdot20=40{,}000\,\text{cm}^3=40\,\text{L}\). 3. After adding water, the volume is \(40+10=50\,\text{L}=50{,}000\,\text{cm}^3\). 4. The base area is \(50\cdot40=2000\,\text{cm}^2\), so the new depth is \(50{,}000\div2000=25\,\text{cm}\).

Answer

a) The aquarium holds \(60\,\text{L}\). b) It currently contains \(40\,\text{L}\). c) The new water depth is \(25\,\text{cm}\).

5112077

A rectangular storage box is \(40\,\text{cm}\) long, \(30\,\text{cm}\) wide, and \(20\,\text{cm}\) high. a) The box contains \(18{,}000\,\text{cm}^3\) of play sand. Find the volume of the empty space in the box. b) All the sand is poured into a narrower box with a \(20\,\text{cm}\times20\,\text{cm}\) base. How high will the sand be in the new box?

Hints

- Subtract the sand volume from the first box's total volume. - The amount of sand stays the same when it is transferred. - Divide the sand volume by the new base area.

Solution

1. The first box's volume is \(40\cdot30\cdot20=24{,}000\,\text{cm}^3\). 2. The empty volume is \(24{,}000-18{,}000=6000\,\text{cm}^3\). 3. The new box's base area is \(20\cdot20=400\,\text{cm}^2\). 4. The sand height is \(18{,}000\div400=45\,\text{cm}\).

Answer

a) The empty volume is \(6000\,\text{cm}^3\). b) The sand will be \(45\,\text{cm}\) high.

5112087

A rectangular garden water tank has a \(100\,\text{cm}\times80\,\text{cm}\) base. a) The tank contains \(240\,\text{L}\) of water. Find the water depth in centimeters. b) Water is removed at a constant rate of \(40\,\text{L}\) per hour. How many hours will it take to empty the tank? c) After a rainstorm, the water level rises by \(5\,\text{cm}\). How many liters of rainwater entered the tank?

Hints

- Convert liters to cubic centimeters for part a. - Divide total water by the hourly removal rate for part b. - Treat the \(5\,\text{cm}\) rise as a rectangular-prism layer.

Solution

1. Convert the initial volume: \(240\,\text{L}=240{,}000\,\text{cm}^3\). 2. The base area is \(100\cdot80=8000\,\text{cm}^2\), so the depth is \(240{,}000\div8000=30\,\text{cm}\). 3. The time to empty the tank is \(240\div40=6\) hours. 4. The added rainwater volume is \(100\cdot80\cdot5=40{,}000\,\text{cm}^3=40\,\text{L}\).

Answer

a) The water is \(30\,\text{cm}\) deep. b) The tank will be empty after \(6\) hours. c) \(40\,\text{L}\) of rainwater entered the tank.

5112117

Two rectangular wooden planters will be filled with soil. Planter A is \(80\,\text{cm}\) long, \(40\,\text{cm}\) wide, and \(30\,\text{cm}\) high. Planter B has a square base with side length \(50\,\text{cm}\) and is \(40\,\text{cm}\) high. a) Which planter has the greater capacity? b) Soil is sold in \(20\,\text{L}\) bags. What is the minimum number of bags needed to fill Planter A?

Hints

- Find both planter volumes in the same unit. - Relate cubic centimeters to liters. - Round the number of bags up to the next whole bag.

Solution

1. Planter A's volume is \(80\cdot40\cdot30=96{,}000\,\text{cm}^3=96\,\text{L}\). 2. Planter B's volume is \(50\cdot50\cdot40=100{,}000\,\text{cm}^3=100\,\text{L}\). 3. Since \(100>96\), Planter B has the greater capacity. 4. Planter A requires \(96\div20=4.8\) bags. Because only whole bags can be purchased, at least \(5\) bags are needed.

Answer

a) Planter B has the greater capacity: \(100\,\text{L}\), compared with \(96\,\text{L}\) for Planter A. b) At least \(5\) bags are needed.

5112127

Weather reports often say that \(1\,\text{mm}\) of rain equals \(1\,\text{L}\) of water per square meter. a) Verify this statement by finding the volume of a \(1\,\text{mm}\)-deep layer of water over \(1\,\text{m}^2\). Give the volume in cubic meters and liters. b) A storm drops \(15\,\text{mm}\) of rain on a flat roof that is \(8\,\text{m}\) long and \(5\,\text{m}\) wide. How many liters of rain fall on the roof?

Hints

- Convert millimeters to meters. - Model the rainfall as a very thin rectangular prism. - Convert cubic meters to liters.

Solution

1. Convert the rainfall depth: \(1\,\text{mm}=0.001\,\text{m}\). 2. Over \(1\,\text{m}^2\), the volume is \(1\cdot0.001=0.001\,\text{m}^3\). 3. Since \(1\,\text{m}^3=1000\,\text{L}\), \(0.001\,\text{m}^3=1\,\text{L}\). The statement is correct. 4. The roof area is \(8\cdot5=40\,\text{m}^2\), and \(15\,\text{mm}=0.015\,\text{m}\). 5. The rainfall volume is \(40\cdot0.015=0.6\,\text{m}^3=600\,\text{L}\).

Answer

a) The volume is \(0.001\,\text{m}^3\), or \(1\,\text{L}\), so the statement is correct. b) \(600\,\text{L}\) of rain fall on the roof.

5112137

A rectangular property is \(20\,\text{m}\) long and \(15\,\text{m}\) wide. A pool on the property has a water-surface area of \(12\,\text{m}^2\). During one night, \(2.4\,\text{cm}\) of rain falls. a) How many liters of rain fall on the entire property, including the pool? b) By how many centimeters does the pool's water level rise, assuming no water evaporates or drains away? Explain. c) How many liters of rain fall directly into the pool?

Hints

- Convert the rainfall depth to meters. - Treat the rain as a thin layer over each area. - The depth of the rain layer does not depend on the area's shape.

Solution

1. The property area is \(20\cdot15=300\,\text{m}^2\), and the rain depth is \(2.4\,\text{cm}=0.024\,\text{m}\). 2. The total rain volume is \(300\cdot0.024=7.2\,\text{m}^3=7200\,\text{L}\). 3. The pool level rises by exactly \(2.4\,\text{cm}\) because rain adds a layer with that same uniform depth directly to the pool. 4. The rain volume entering the pool is \(12\cdot0.024=0.288\,\text{m}^3=288\,\text{L}\).

Answer

a) \(7200\,\text{L}\) of rain fall on the property. b) The pool level rises \(2.4\,\text{cm}\), the same as the rainfall depth. c) \(288\,\text{L}\) fall directly into the pool.

5112157

A rectangular grain storage area is \(12\,\text{m}\) long and \(8\,\text{m}\) wide. Grain is piled evenly to a height of \(2.5\,\text{m}\). a) Find the grain's volume in cubic meters. b) One cubic meter of this grain has a mass of about \(750\,\text{kg}\). Find the total mass in metric tons.

Hints

- Use the volume formula for a rectangular prism. - Multiply the volume by the mass per cubic meter. - Convert kilograms to metric tons.

Solution

1. The grain volume is \(12\cdot8\cdot2.5=240\,\text{m}^3\). 2. Its mass is \(240\cdot750=180{,}000\,\text{kg}\). 3. Since \(1000\,\text{kg}=1\) metric ton, the total mass is \(180\) metric tons.

Answer

a) The volume is \(240\,\text{m}^3\). b) The grain has a mass of about \(180\) metric tons.

5112167

An aquarium has an \(80\,\text{cm}\times40\,\text{cm}\) base and is \(50\,\text{cm}\) high. a) Find its maximum capacity in liters. b) The water is currently \(10\,\text{cm}\) below the top. How many liters of water are in the aquarium? c) Decorative stones with a total volume of \(1600\,\text{cm}^3\) are completely submerged. How many centimeters does the water level rise?

Hints

- Use the full height for part a and the actual water depth for part b. - A fully submerged object displaces its own volume. - Divide the displaced volume by the aquarium's base area.

Solution

1. The aquarium's full volume is \(80\cdot40\cdot50=160{,}000\,\text{cm}^3=160\,\text{L}\). 2. The current water depth is \(50-10=40\,\text{cm}\), so the water volume is \(80\cdot40\cdot40=128{,}000\,\text{cm}^3=128\,\text{L}\). 3. The stones displace \(1600\,\text{cm}^3\). The base area is \(80\cdot40=3200\,\text{cm}^2\). 4. The water-level rise is \(1600\div3200=0.5\,\text{cm}\).

Answer

a) The aquarium holds \(160\,\text{L}\). b) It contains \(128\,\text{L}\) of water. c) The water level rises \(0.5\,\text{cm}\).

5112197

A flat rectangular garage roof measures \(3\,\text{m}\times6\,\text{m}\). During a storm, \(1.5\,\text{cm}\) of rain falls. All the water is directed into a rain barrel that holds \(200\,\text{L}\). Find the total rainwater volume in liters and determine whether one barrel is enough.

Hints

- Find the roof area first. - Convert the rainfall depth to meters. - Convert the resulting cubic meters to liters and compare.

Solution

1. The roof area is \(3\cdot6=18\,\text{m}^2\). 2. Convert the rainfall depth: \(1.5\,\text{cm}=0.015\,\text{m}\). 3. The water volume is \(18\cdot0.015=0.27\,\text{m}^3=270\,\text{L}\). 4. Since \(270>200\), one barrel is not enough.

Answer

The roof receives \(270\,\text{L}\) of rainwater. One \(200\,\text{L}\) barrel is not enough.

5112227

A rectangular aquarium has a volume of \(48{,}000\,\text{cm}^3\), a height of \(30\,\text{cm}\), and a square base. a) Find the side length of the square base. b) By how many liters does the water volume decrease if the water level is lowered by \(5\,\text{cm}\)?

Hints

- Divide the volume by the height to find the base area. - Find the side length of a square with that area. - Model the removed water as a rectangular-prism layer.

Solution

1. The base area is \(48{,}000\div30=1600\,\text{cm}^2\). 2. Since \(40\cdot40=1600\), the square base has side length \(40\,\text{cm}\). 3. Lowering the level by \(5\,\text{cm}\) removes a layer with volume \(1600\cdot5=8000\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{L}\), the decrease is \(8\,\text{L}\).

Answer

a) Each side of the base is \(40\,\text{cm}\). b) The water volume decreases by \(8\,\text{L}\).

5112357

A rectangular container measures \(5\,\text{cm}\times4\,\text{cm}\times10\,\text{cm}\). a) Find its volume in cubic centimeters and liters. b) How many full containers must be emptied into a \(10\,\text{L}\) bucket to fill the bucket exactly?

Hints

- Multiply the three dimensions. - Convert cubic centimeters to liters. - Divide the bucket's capacity by one container's capacity.

Solution

1. The container's volume is \(5\cdot4\cdot10=200\,\text{cm}^3\). 2. Since \(1000\,\text{cm}^3=1\,\text{L}\), \(200\,\text{cm}^3=0.2\,\text{L}\). 3. The number of containers is \(10\div0.2=50\).

Answer

a) The volume is \(200\,\text{cm}^3\), or \(0.2\,\text{L}\). b) \(50\) full containers are needed.

5112447

A rectangular water tank is \(2\,\text{m}\) long, \(1.5\,\text{m}\) wide, and \(2\,\text{m}\) high. It is initially half full. A pump then adds water at \(150\,\text{L}\) per minute for exactly \(10\) minutes. a) How many centimeters does the water level rise? b) What is the final water depth, in meters?

Hints

- Find the volume added in \(10\) minutes. - Divide that volume by the tank's base area. - Add the rise to the initial half-full depth.

Solution

1. The tank's base area is \(2\cdot1.5=3\,\text{m}^2\). 2. The pump adds \(10\cdot150=1500\,\text{L}=1.5\,\text{m}^3\). 3. The water-level rise is \(1.5\div3=0.5\,\text{m}=50\,\text{cm}\). 4. Half of the tank's \(2\,\text{m}\) height is \(1\,\text{m}\). The final depth is \(1+0.5=1.5\,\text{m}\).

Answer

a) The water level rises \(50\,\text{cm}\). b) The final water depth is \(1.5\,\text{m}\).

5112527

All three dimensions of a rectangular prism are tripled. a) By what factor does the area of each corresponding face change? b) By what factor does the volume change? Explain why the two factors are different.

Hints

- A face uses two dimensions. - Volume uses three dimensions. - Multiply the scale factor once for each dimension involved.

Solution

1. A face area is the product of two dimensions. Tripling both gives \((3l)(3w)=9lw\), so each face area is multiplied by \(9\). 2. Volume is the product of three dimensions. Tripling all three gives \((3l)(3w)(3h)=27lwh\), so volume is multiplied by \(27\). 3. Area uses two length factors, so the scale factor is squared. Volume uses three length factors, so the scale factor is cubed.

Answer

a) Each corresponding face area is multiplied by \(9\). b) The volume is multiplied by \(27\). Area scales by \(3^2\), while volume scales by \(3^3\).

5118277

Aquarium A is \(50\,\text{cm}\) long, \(30\,\text{cm}\) wide, and \(40\,\text{cm}\) high. Aquarium B is \(60\,\text{cm}\) long, \(25\,\text{cm}\) wide, and also \(40\,\text{cm}\) high. a) Which aquarium has the greater volume, or are the volumes equal? Show your calculations. b) Aquarium A is filled with \(45\,\text{L}\) of water. How deep is the water?

Hints

- Find both rectangular-prism volumes. - Convert liters to cubic centimeters. - Divide the water volume by Aquarium A's base area.

Solution

1. Aquarium A's volume is \(50\cdot30\cdot40=60{,}000\,\text{cm}^3\). 2. Aquarium B's volume is \(60\cdot25\cdot40=60{,}000\,\text{cm}^3\). The volumes are equal. 3. Convert the water volume: \(45\,\text{L}=45{,}000\,\text{cm}^3\). 4. Aquarium A's base area is \(50\cdot30=1500\,\text{cm}^2\), so the water depth is \(45{,}000\div1500=30\,\text{cm}\).

Answer

a) The aquariums have equal volumes of \(60{,}000\,\text{cm}^3\), or \(60\,\text{L}\). b) The water is \(30\,\text{cm}\) deep.

5118327

A cube-shaped container with edge length \(40\,\text{cm}\) is completely full of water. All the water is poured into an empty rectangular tank with a \(50\,\text{cm}\times32\,\text{cm}\) base and a height of \(50\,\text{cm}\). How deep is the water in the second tank?

Hints

- The water volume stays the same when it is poured. - Find the second tank's base area. - Divide the water volume by the base area.

Solution

1. The cube contains \(40^3=64{,}000\,\text{cm}^3\) of water. 2. The second tank's base area is \(50\cdot32=1600\,\text{cm}^2\). 3. The water depth is \(64{,}000\div1600=40\,\text{cm}\).

Answer

The water is \(40\,\text{cm}\) deep in the second tank.

5138727

A prism has volume \(720\,\text{cm}^3\) and height \(15\,\text{cm}\). Its base can be divided exactly into three congruent rectangles. a) Find the total base area \(B\). b) Find the area of one of the three rectangles. c) One rectangle is \(4\,\text{cm}\) wide. Find its length.

Hints

- Rearrange the volume formula to find the base area. - Divide the base area into three equal parts. - Use rectangle area to find the missing side length.

Solution

1. From \(V=Bh\), \(B=\frac{V}{h}=\frac{720\,\text{cm}^3}{15\,\text{cm}}=48\,\text{cm}^2\). 2. Each of the three congruent rectangles has area \(\frac{48\,\text{cm}^2}{3}=16\,\text{cm}^2\). 3. The rectangle's length is \(\frac{16\,\text{cm}^2}{4\,\text{cm}}=4\,\text{cm}\).

Answer

a) \(B=48\,\text{cm}^2\) b) \(16\,\text{cm}^2\) c) \(4\,\text{cm}\)

5140027

A paint manufacturer states that \(1\,\text{L}\) of paint covers exactly \(10\,\text{m}^2\). a) Find the theoretical thickness of the paint layer in millimeters if it is spread uniformly. b) A wall is \(4\,\text{m}\) wide and \(2.5\,\text{m}\) high. A painter applies a layer twice as thick as the manufacturer's stated coverage. How many liters of paint are needed?

Hints

- Convert volume and area to compatible units. - Use \(V=At\) for a uniform layer. - At the same area, how does doubling thickness affect volume?

Solution

1. Convert the quantities: \(1\,\text{L}=1000\,\text{cm}^3\), and \(10\,\text{m}^2=100{,}000\,\text{cm}^2\). 2. The layer thickness is \(t=\frac{V}{A}=\frac{1000\,\text{cm}^3}{100{,}000\,\text{cm}^2}=0.01\,\text{cm}=0.1\,\text{mm}\). 3. The wall area is \(4\,\text{m}\cdot2.5\,\text{m}=10\,\text{m}^2\), so a standard layer would use \(1\,\text{L}\). 4. Doubling the thickness doubles the volume, so \(2\,\text{L}\) are needed.

Answer

a) \(0.1\,\text{mm}\) b) \(2\,\text{L}\)

5141387

A rectangular aquarium has a base measuring \(80\,\text{cm}\times40\,\text{cm}\). It contains \(48\,\text{L}\) of water and is exactly half full. 1) How deep is the water? 2) What is the total height of the aquarium?

Hints

- Convert liters to cubic centimeters before calculating. - Treat the water as a rectangular prism. - If the aquarium is half full, compare the water depth with the total height.

Solution

1. Convert the water volume: \(48\,\text{L}=48{,}000\,\text{cm}^3\). 2. The base area is \(B=80\cdot40=3200\,\text{cm}^2\). 3. The water depth is \(h=\frac{48{,}000}{3200}\,\text{cm}=15\,\text{cm}\). 4. Because the aquarium is half full, its total height is \(2\cdot15\,\text{cm}=30\,\text{cm}\).

Answer

1) The water is \(15\,\text{cm}\) deep. 2) The aquarium is \(30\,\text{cm}\) tall.

5141397

Two prisms have equal volumes. Prism A is a rectangular prism with dimensions \(12\,\text{cm}\times8\,\text{cm}\times10\,\text{cm}\). Prism B has a triangular base with base length \(16\,\text{cm}\) and corresponding height \(6\,\text{cm}\). Find the height of Prism B.

Hints

- Equal-volume solids have the same numerical volume. - Find the volume of Prism A first. - Use the area formula for a triangle to find the base area of Prism B.

Solution

1. The volume of Prism A is \(V=12\cdot8\cdot10=960\,\text{cm}^3\). 2. The area of the triangular base of Prism B is \(B=\frac{1}{2}\cdot16\cdot6=48\,\text{cm}^2\). 3. Since the prisms have equal volumes, \(48h=960\). 4. Therefore, \(h=960\div48=20\,\text{cm}\).

Answer

The height of Prism B is \(20\,\text{cm}\).

5223127

A rectangular water basin has a base area of \(15\,\text{m}^2\) and is \(2\,\text{m}\) deep. a) Find its maximum capacity in cubic meters. b) The water is currently \(1.20\,\text{m}\) deep. How many liters are needed to fill the basin to the top?

Hints

- Use base area times depth. - Find the unfilled depth before finding the missing volume. - Convert cubic meters to liters.

Solution

1. The maximum volume is \(15\cdot2=30\,\text{m}^3\). 2. The unfilled depth is \(2-1.20=0.80\,\text{m}\). 3. The missing volume is \(15\cdot0.80=12\,\text{m}^3\). 4. Convert to liters: \(12\,\text{m}^3=12{,}000\,\text{L}\).

Answer

a) The maximum capacity is \(30\,\text{m}^3\). b) \(12{,}000\,\text{L}\) are needed.

5279347

A rectangular swimming pool is divided into three equal-width lanes. The pool is \(25\,\text{m}\) long, each lane is \(2.5\,\text{m}\) wide, and the pool is \(2\,\text{m}\) deep. a) Find the total water volume when the pool is full. b) For cleaning, the water level is lowered by \(40\,\text{cm}\) throughout the pool. How much water remains?

Hints

- Find the volume of one lane, then multiply by three. - Convert the lowered amount to meters. - Use the new water depth for part b.

Solution

1. The full volume of one lane is \(25\cdot2.5\cdot2=125\,\text{m}^3\). 2. The full pool volume is \(3\cdot125=375\,\text{m}^3\). 3. Lowering the level by \(40\,\text{cm}=0.4\,\text{m}\) leaves a depth of \(2-0.4=1.6\,\text{m}\). 4. One lane then contains \(25\cdot2.5\cdot1.6=100\,\text{m}^3\), so the pool contains \(3\cdot100=300\,\text{m}^3\).

Answer

a) The full pool contains \(375\,\text{m}^3\) of water. b) \(300\,\text{m}^3\) of water remains.

5315637

The concrete L-shaped retaining block shown has an L-shaped cross section. The vertical section is \(20\,\text{cm}\) wide and \(60\,\text{cm}\) high. The base extends to a total width of \(50\,\text{cm}\) and is \(15\,\text{cm}\) high. The block is \(40\,\text{cm}\) long. Find its volume in cubic decimeters.

Figure for problem 531563

Hints

- Split the L-shaped cross section into two rectangles. - Multiply the cross-sectional area by the block length. - Convert cubic centimeters to cubic decimeters.

Solution

1. Split the cross section into a \(20\,\text{cm}\times60\,\text{cm}\) rectangle and a \(30\,\text{cm}\times15\,\text{cm}\) rectangle. 2. The cross-sectional area is \(20\cdot60+30\cdot15=1650\,\text{cm}^2\). 3. The volume is \(1650\cdot40=66{,}000\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{dm}^3\), the volume is \(66\,\text{dm}^3\).

Answer

The block's volume is \(66\,\text{dm}^3\).

5315737

Two chocolate bars have the shapes shown. Bar A is a prism with an L-shaped base. Its base has overall width \(6\,\text{cm}\), overall height \(4\,\text{cm}\), horizontal arm thickness \(2\,\text{cm}\), and vertical arm width \(3\,\text{cm}\). The bar is \(10\,\text{cm}\) long. Bar B is a rectangular prism measuring \(6\,\text{cm}\times10\,\text{cm}\times3\,\text{cm}\). Do the bars have equal volumes? Calculate both volumes to justify your answer.

Figure for problem 531573

Hints

- Decompose the L-shaped base into rectangles. - Find each prism's base area. - Multiply each base area by the prism length.

Solution

1. Split the L-shaped base of bar A into a \(6\,\text{cm}\times2\,\text{cm}\) rectangle and a \(3\,\text{cm}\times2\,\text{cm}\) rectangle. 2. Its base area is \(B_A=6\cdot2+3\cdot2=18\,\text{cm}^2\). 3. The volume of bar A is \(V_A=18\cdot10=180\,\text{cm}^3\). 4. The volume of bar B is \(V_B=6\cdot10\cdot3=180\,\text{cm}^3\). 5. Therefore, the bars have equal volumes.

Answer

Yes. Each bar has volume \(180\,\text{cm}^3\).

5316657

A new L-shaped sandbox is shown in the diagram. It will be filled evenly with sand to a depth of \(25\,\text{cm}\). Each bag of play sand holds \(25\,\text{L}\). How many bags of sand are needed?

Figure for problem 531665

Hints

- How can you split the L-shaped base into two rectangles? - Convert the fill depth to meters before finding the volume. - Remember that \(1\,\text{m}^3=1000\,\text{L}\).

Solution

1. Split the L-shaped base into two rectangles. Their total area is \(4.0\cdot1.5+1.5\cdot2.0=6.0+3.0=9.0\,\text{m}^2\). 2. Convert the depth: \(25\,\text{cm}=0.25\,\text{m}\). 3. The required volume is \(9.0\cdot0.25=2.25\,\text{m}^3\). 4. Since \(1\,\text{m}^3=1000\,\text{L}\), \(2.25\,\text{m}^3=2250\,\text{L}\). 5. The number of bags is \(2250\div25=90\).

Answer

The sandbox requires \(90\) bags of sand.

5356287

A feed trough is a right prism with a trapezoidal cross section. The trapezoid is \(25\,\text{cm}\) wide at the top, \(15\,\text{cm}\) wide at the bottom, and \(12\,\text{cm}\) deep. The trough is \(80\,\text{cm}\) long. Find its capacity in liters.

Figure for problem 535628

Hints

- Find the area of the trapezoidal cross section. - Multiply the cross-sectional area by the trough length. - Convert cubic centimeters to liters.

Solution

1. The trapezoidal cross-sectional area is \(B=\frac{25+15}{2}\cdot12=240\,\text{cm}^2\). 2. The volume is \(V=240\cdot80=19{,}200\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), the capacity is \(19.2\,\text{L}\).

Answer

\(19.2\,\text{L}\)

5356327

A steel part is a prism with the grooved cross-section shown. The cross-section is \(15\,\text{mm}\) wide and \(10\,\text{mm}\) high. A rectangular groove \(5\,\text{mm}\) wide and \(6\,\text{mm}\) deep has been removed from the center. Find the volume of a part that is \(25\,\text{mm}\) long.

Figure for problem 535632

Hints

- Find the area of the cross-section by subtracting the groove from the full rectangle. - Multiply the cross-sectional area by the prism's length.

Solution

1. The full rectangular cross-section has area \(15\cdot10=150\,\text{mm}^2\). 2. The groove has area \(5\cdot6=30\,\text{mm}^2\). 3. The actual cross-sectional area is \(150-30=120\,\text{mm}^2\). 4. The volume is \(120\cdot25=3000\,\text{mm}^3\).

Answer

The part has volume \(3000\,\text{mm}^3\).

5357527

An architect designs a house with the L-shaped floor plan shown. a) Find the area of the floor plan. b) The house will have a uniform interior height of \(2.8\,\text{m}\). Find its interior volume. c) Construction is estimated to cost \(\$310\) per cubic meter of interior volume. Find the estimated total cost.

Figure for problem 535752

Hints

- Divide the L-shaped floor plan into two rectangles. - Multiply the floor area by the interior height. - Multiply the volume by the cost per cubic meter.

Solution

1. Divide the floor plan into rectangles with areas \(15\cdot10=150\,\text{m}^2\) and \(7\cdot8=56\,\text{m}^2\). The total area is \(150+56=206\,\text{m}^2\). 2. The interior volume is \(206\cdot2.8=576.8\,\text{m}^3\). 3. The estimated cost is \(576.8\cdot\$310=\$178{,}808\).

Answer

a) The floor-plan area is \(206\,\text{m}^2\). b) The interior volume is \(576.8\,\text{m}^3\). c) The estimated cost is \(\$178{,}808\).

5357557

A warehouse has the floor plan shown. The two rectangular cutouts along the lower side are loading entrances. a) Find the area of the warehouse floor. b) The warehouse is \(7\,\text{m}\) high. Find the enclosed volume.

Figure for problem 535755

Hints

- Subtract the two rectangular cutouts from the large rectangle. - Multiply the floor area by the height.

Solution

1. Begin with a \(25\,\text{m}\times12\,\text{m}\) rectangle, which has area \(25\cdot12=300\,\text{m}^2\). 2. Each cutout measures \(4\,\text{m}\times3\,\text{m}\), so their combined area is \(2(4\cdot3)=24\,\text{m}^2\). 3. The floor area is \(300-24=276\,\text{m}^2\). 4. The enclosed volume is \(276\cdot7=1932\,\text{m}^3\).

Answer

a) The floor area is \(276\,\text{m}^2\). b) The enclosed volume is \(1932\,\text{m}^3\).

5357587

A long concrete planter has the U-shaped end shown. a) Find the area of the end face. b) The planter is \(40\,\text{cm}\) long. Find the volume of concrete in cubic centimeters and in liters.

Figure for problem 535758

Hints

- Subtract the inner rectangle from the outer rectangle. - Multiply the end-face area by the planter's length. - Use \(1000\,\text{cm}^3=1\,\text{L}\).

Solution

1. The outer rectangle has area \(120\cdot50=6000\,\text{cm}^2\), and the rectangular opening has area \(100\cdot40=4000\,\text{cm}^2\). 2. The end-face area is \(6000-4000=2000\,\text{cm}^2\). 3. The volume is \(2000\cdot40=80{,}000\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{L}\), the volume is \(80\,\text{L}\).

Answer

a) The end-face area is \(2000\,\text{cm}^2\). b) The concrete volume is \(80{,}000\,\text{cm}^3\), or \(80\,\text{L}\).

5357857

A concrete part is a prism with a trapezoidal end face. The parallel sides of the trapezoid are \(8.0\,\text{dm}\) and \(4.0\,\text{dm}\), and its height is \(3.0\,\text{dm}\). The prism is \(20.0\,\text{dm}\) long. Find the volume in cubic decimeters.

Figure for problem 535785

Hints

- Find the area of the trapezoidal end face. - Multiply that area by the prism's length.

Solution

1. The trapezoidal end has area \(\frac{8.0+4.0}{2}\cdot3.0=18.0\,\text{dm}^2\). 2. The volume is \(18.0\cdot20.0=360.0\,\text{dm}^3\).

Answer

The volume is \(360.0\,\text{dm}^3\).

5357887

A garden shed is \(3.0\,\text{m}\) wide and \(3.5\,\text{m}\) high overall. Its vertical side walls are \(2.5\,\text{m}\) high, and the shed is \(4.0\,\text{m}\) long. Find the volume of air inside the shed.

Figure for problem 535788

Hints

- Divide the end face into a rectangle and a triangle. - Find the height of the triangular roof section. - Multiply the end-face area by the shed's length.

Solution

1. The rectangular part of the end face has area \(3.0\cdot2.5=7.5\,\text{m}^2\). 2. The triangular roof section has height \(3.5-2.5=1.0\,\text{m}\) and area \(\frac12\cdot3.0\cdot1.0=1.5\,\text{m}^2\). 3. The total cross-sectional area is \(7.5+1.5=9.0\,\text{m}^2\). 4. The interior volume is \(9.0\cdot4.0=36.0\,\text{m}^3\).

Answer

The shed contains \(36.0\,\text{m}^3\) of air.

5357947

A concrete stair step is a prism with an L-shaped cross section. The cross section has overall width \(40\,\text{cm}\), overall height \(30\,\text{cm}\), lower section height \(15\,\text{cm}\), and upper section width \(20\,\text{cm}\). The step is \(80\,\text{cm}\) long. Find the volume of concrete in liters.

Figure for problem 535794

Hints

- Decompose the L-shaped cross section into rectangles. - Multiply the cross-sectional area by the step length. - Convert cubic centimeters to liters.

Solution

1. Split the L-shaped cross section into a \(40\,\text{cm}\times15\,\text{cm}\) rectangle and a \(20\,\text{cm}\times15\,\text{cm}\) rectangle. 2. The cross-sectional area is \(B=40\cdot15+20\cdot15=900\,\text{cm}^2\). 3. The volume is \(V=900\cdot80=72{,}000\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{L}\), the concrete volume is \(72\,\text{L}\).

Answer

\(72\,\text{L}\)

5358097

An L-shaped concrete block is made from a base measuring \(80\,\text{cm}\times40\,\text{cm}\times20\,\text{cm}\) and an upper prism measuring \(20\,\text{cm}\times40\,\text{cm}\times40\,\text{cm}\). Find the volume in liters in two different ways: by adding the two prism volumes and by subtracting a missing prism from a larger rectangular prism.

Figure for problem 535809

Hints

- Add the volumes of the two visible prisms. - For a second method, subtract the missing section from the enclosing prism. - Use \(1000\,\text{cm}^3=1\,\text{L}\).

Solution

1. By decomposition, the volume is \(80\cdot40\cdot20+20\cdot40\cdot40=64{,}000+32{,}000=96{,}000\,\text{cm}^3\). 2. By subtraction, the enclosing prism has volume \(80\cdot40\cdot60=192{,}000\,\text{cm}^3\), and the missing prism has volume \(60\cdot40\cdot40=96{,}000\,\text{cm}^3\). The difference is \(96{,}000\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), the volume is \(96\,\text{L}\).

Answer

The block has volume \(96\,\text{L}\).

5358637

The interior of a doghouse is a prism with a house-shaped pentagonal cross section. The inside width is \(6\,\text{dm}\), the vertical wall height is \(4\,\text{dm}\), the total height to the roof peak is \(7\,\text{dm}\), and the interior is \(10\,\text{dm}\) deep. Find the interior volume.

Figure for problem 535863

Hints

- Split the house-shaped cross section into a rectangle and a triangle. - Find the roof triangle's height from the total height and wall height. - Multiply the cross-sectional area by the depth.

Solution

1. The rectangular part of the cross section has area \(6\cdot4=24\,\text{dm}^2\). 2. The roof triangle has height \(7-4=3\,\text{dm}\) and area \(\frac{1}{2}\cdot6\cdot3=9\,\text{dm}^2\). 3. The total cross-sectional area is \(B=24+9=33\,\text{dm}^2\). 4. The interior volume is \(V=33\cdot10=330\,\text{dm}^3\).

Answer

\(330\,\text{dm}^3\)

5359767

A concrete ramp connects directly to a rectangular platform. In the side cross section, the entire structure is \(10\,\text{cm}\) long, the level platform is \(4\,\text{cm}\) long, and the maximum height is \(3\,\text{cm}\). The structure is \(8\,\text{cm}\) wide. Find its volume.

Figure for problem 535976

Hints

- Split the side cross section into a rectangle and a right triangle. - Subtract the platform length from the total length to find the ramp length. - Multiply the cross-sectional area by the width.

Solution

1. The ramp portion has horizontal length \(10-4=6\,\text{cm}\). 2. The rectangular part of the cross section has area \(4\cdot3=12\,\text{cm}^2\). 3. The triangular part has area \(\frac{1}{2}\cdot6\cdot3=9\,\text{cm}^2\). 4. The total cross-sectional area is \(B=12+9=21\,\text{cm}^2\). 5. The volume is \(V=21\cdot8=168\,\text{cm}^3\).

Answer

\(168\,\text{cm}^3\)

5360207

A \(100\,\text{m}\)-long section of an earthen levee has a trapezoidal cross section. The bottom is \(18\,\text{m}\) wide, the top is \(4\,\text{m}\) wide, and the levee is \(5\,\text{m}\) high. Find the volume of soil used to build this section.

Figure for problem 536020

Hints

- Model the levee as a prism whose base is the trapezoidal cross section. - Which measurements determine the area of the trapezoid? - Multiply the cross-sectional area by the levee length.

Solution

1. Find the area of the trapezoidal cross section: \(B=\frac{18+4}{2}\cdot5=11\cdot5=55\,\text{m}^2\). 2. Treat the levee section as a prism. Its volume is \(V=B\cdot l=55\cdot100=5{,}500\,\text{m}^3\).

Answer

The levee section contains \(5{,}500\,\text{m}^3\) of soil.

5111997

A rectangular grain storage room is \(8\,\text{m}\) long, \(5\,\text{m}\) wide, and \(4\,\text{m}\) high. a) Find the room's maximum volume. b) The room currently contains \(120\,\text{m}^3\) of grain spread evenly across the floor. How high is the grain? c) The grain will be moved into rectangular containers measuring \(1\,\text{m}\times1\,\text{m}\times0.5\,\text{m}\). How many containers are needed to hold all \(120\,\text{m}^3\) of grain?

Hints

- Use the rectangular-prism volume formula for part a. - For part b, divide the grain volume by the floor area. - For part c, find one container's volume before dividing.

Solution

1. The room's maximum volume is \(8\cdot5\cdot4=160\,\text{m}^3\). 2. Its floor area is \(8\cdot5=40\,\text{m}^2\). The grain height is \(120\div40=3\,\text{m}\). 3. One container's volume is \(1\cdot1\cdot0.5=0.5\,\text{m}^3\). 4. The number of containers needed is \(120\div0.5=240\).

Answer

a) The maximum volume is \(160\,\text{m}^3\). b) The grain is \(3\,\text{m}\) high. c) \(240\) containers are needed.

5112027

Two rectangular rain barrels are compared. Barrel A has a \(60\,\text{cm}\times60\,\text{cm}\) base and is \(1\,\text{m}\) high. Barrel B has an \(80\,\text{cm}\times50\,\text{cm}\) base and is also \(1\,\text{m}\) high. a) Find each barrel's maximum capacity in liters. b) Barrel A is half full. Barrel B contains exactly \(170\,\text{L}\). Which barrel contains more water? Justify your answer. c) If the amount of water in the half-full Barrel A were poured into Barrel B, how high would the water be, in centimeters?

Hints

- Express all dimensions in centimeters before finding capacity. - Half full means half of the total volume. - For part c, divide the transferred volume by Barrel B's base area.

Solution

1. Convert the height to \(100\,\text{cm}\). Barrel A's volume is \(60\cdot60\cdot100=360{,}000\,\text{cm}^3=360\,\text{L}\). 2. Barrel B's volume is \(80\cdot50\cdot100=400{,}000\,\text{cm}^3=400\,\text{L}\). 3. Half of Barrel A's capacity is \(360\div2=180\,\text{L}\). Since \(180>170\), Barrel A contains more water. 4. Barrel B's base area is \(80\cdot50=4000\,\text{cm}^2\). The transferred volume is \(180{,}000\,\text{cm}^3\), so the water height is \(180{,}000\div4000=45\,\text{cm}\).

Answer

a) Barrel A holds \(360\,\text{L}\), and Barrel B holds \(400\,\text{L}\). b) Barrel A contains more water because it has \(180\,\text{L}\), compared with \(170\,\text{L}\) in Barrel B. c) The water would be \(45\,\text{cm}\) high in Barrel B.

5112047

A toy company wants to make a rectangular block whose volume is exactly \(6\) times the volume of a block with edge lengths \(a\), \(b\), and \(c\). Give three different ways to multiply or divide the edge lengths so that the new volume is \(6abc\).

Hints

- Volume is the product of the three edge lengths. - The three change factors must have a product of \(6\). - Look for several different factor combinations.

Solution

1. The product of the three dimension-change factors must equal \(6\). 2. One option is to multiply one edge length by \(6\) and leave the other two unchanged: \((6a)bc=6abc\). 3. A second option is to double one edge and triple another: \((2a)(3b)c=6abc\). 4. A third option is to multiply one edge by \(12\) and divide another by \(2\): \((12a)\left(\frac{b}{2}\right)c=6abc\).

Answer

Sample answers are: 1) Multiply one edge length by \(6\). 2) Double one edge length and triple another. 3) Multiply one edge length by \(12\) and divide another by \(2\). In each case, the product of the change factors is \(6\).

5112097

A rectangular school aquarium has a capacity of \(540\,\text{L}\) and a length of \(150\,\text{cm}\). a) Find the area of the rectangular end face, width times height, in square centimeters. b) Give one reasonable pair of dimensions for the width and height. Briefly explain why your dimensions make sense for an aquarium.

Hints

- Convert the capacity to cubic centimeters. - Divide the volume by the aquarium's length. - Choose two reasonable dimensions whose product is the end-face area.

Solution

1. Convert the volume: \(540\,\text{L}=540{,}000\,\text{cm}^3\). 2. Divide by the length to find the end-face area: \(540{,}000\div150=3600\,\text{cm}^2\). 3. The width and height must have a product of \(3600\). One reasonable choice is \(90\,\text{cm}\times40\,\text{cm}\). 4. These dimensions make a tank that is wider than it is tall and are practical for viewing and access.

Answer

a) The end-face area is \(3600\,\text{cm}^2\). b) One possible pair is a width of \(90\,\text{cm}\) and a height of \(40\,\text{cm}\). Answers will vary.

5112107

A rectangular garden water container is \(2\,\text{m}\) long and \(1.5\,\text{m}\) wide. a) Twelve stone blocks, each with a volume of \(15\,\text{dm}^3\), are placed completely underwater. No water spills. How many centimeters does the water level rise? b) How many liters of water would fill the empty container to a depth of \(10\,\text{cm}\), without the blocks?

Hints

- A fully submerged object displaces its own volume of water. - Divide the displaced volume by the container's base area. - For part b, use the base area and the target depth.

Solution

1. The blocks displace their total volume: \(12\cdot15=180\,\text{dm}^3=0.18\,\text{m}^3\). 2. The container's base area is \(2\cdot1.5=3\,\text{m}^2\). 3. The rise in water level is \(0.18\div3=0.06\,\text{m}=6\,\text{cm}\). 4. A depth of \(10\,\text{cm}=0.1\,\text{m}\) gives volume \(3\cdot0.1=0.3\,\text{m}^3=300\,\text{L}\).

Answer

a) The water level rises \(6\,\text{cm}\). b) The container needs \(300\,\text{L}\) of water.

5112147

Two weather stations report rainfall in nearby areas. Station A reports \(12\,\text{mm}\) of rain over a \(4\,\text{ha}\) forest. Station B reports \(0.8\,\text{cm}\) of rain over a \(0.05\,\text{km}^2\) field. Which area received the greater total volume of water, in liters? Find the difference.

Hints

- Convert both areas to square meters. - Convert both rainfall depths to meters. - Find each thin-layer volume, then convert to liters and compare.

Solution

1. For Area A, \(4\,\text{ha}=40{,}000\,\text{m}^2\) and \(12\,\text{mm}=0.012\,\text{m}\). 2. Its rain volume is \(40{,}000\cdot0.012=480\,\text{m}^3=480{,}000\,\text{L}\). 3. For Area B, \(0.05\,\text{km}^2=50{,}000\,\text{m}^2\) and \(0.8\,\text{cm}=0.008\,\text{m}\). 4. Its rain volume is \(50{,}000\cdot0.008=400\,\text{m}^3=400{,}000\,\text{L}\). 5. Area A received more water, by \(480{,}000-400{,}000=80{,}000\,\text{L}\).

Answer

The forest in Area A received more water: \(480{,}000\,\text{L}\), compared with \(400{,}000\,\text{L}\) in Area B. The difference is \(80{,}000\,\text{L}\).

5112177

A reservoir with vertical sides has a water-surface area of \(0.4\,\text{km}^2\). During a hot summer, evaporation lowers the water level by \(12\,\text{cm}\). a) Find the volume of water lost, in cubic meters. b) Water from a rectangular reserve basin measuring \(100\,\text{m}\) by \(40\,\text{m}\) will replace the loss. How far would the water level in the reserve basin have to drop to supply the entire amount? c) The reserve basin is \(5\,\text{m}\) deep and initially full. Determine whether it contains enough water to replace the loss.

Hints

- Convert square kilometers to square meters and centimeters to meters. - Treat the lost water as a rectangular-prism layer. - Compare the required volume with the reserve basin's full capacity.

Solution

1. Convert the area and depth: \(0.4\,\text{km}^2=400{,}000\,\text{m}^2\) and \(12\,\text{cm}=0.12\,\text{m}\). 2. The evaporated volume is \(400{,}000\cdot0.12=48{,}000\,\text{m}^3\). 3. The reserve basin's base area is \(100\cdot40=4000\,\text{m}^2\). 4. To supply \(48{,}000\,\text{m}^3\), its level would have to drop \(48{,}000\div4000=12\,\text{m}\). 5. A full \(5\,\text{m}\)-deep reserve basin contains \(4000\cdot5=20{,}000\,\text{m}^3\), which is less than \(48{,}000\,\text{m}^3\).

Answer

a) \(48{,}000\,\text{m}^3\) of water evaporated. b) The reserve level would have to drop \(12\,\text{m}\). c) The reserve basin does not contain enough water; it holds only \(20{,}000\,\text{m}^3\).

5125047

A large cube with edge length \(k\) units is built from unit cubes, where \(k\ge3\) is a whole number. The outside of the large cube is painted red, and then the cube is taken apart. a) Write an expression for the number of unit cubes that have no red paint. b) Find this number when \(k=5\). c) How many unit cubes have at least one red face when \(k=10\)?

Hints

- Imagine removing one layer of unit cubes from every face. - How does removing the outer layer change each edge length? - Use the volume formula for a cube. - For part c), subtract the number of unpainted cubes from the total number of cubes.

Solution

1. Removing the painted outer layer leaves an inner cube with edge length \(k-2\). 2. Therefore, the number of unpainted unit cubes is \((k-2)^3\). 3. When \(k=5\), \((5-2)^3=3^3=27\). 4. When \(k=10\), the large cube contains \(10^3=1000\) unit cubes, and the unpainted inner cube contains \((10-2)^3=8^3=512\). 5. Thus, \(1000-512=488\) unit cubes have at least one red face.

Answer

a) \((k-2)^3\) b) \(27\) unit cubes c) \(488\) unit cubes

5315847

An open-top concrete planter has outside dimensions \(8\,\text{dm}\times5\,\text{dm}\times4\,\text{dm}\). Its bottom and walls are uniformly \(1\,\text{dm}\) thick. Find the volume of concrete used, in cubic decimeters and liters.

Figure for problem 531584

Hints

- Find the volume of the outside rectangular prism. - Determine the inside dimensions after accounting for wall and bottom thickness. - Subtract the hollow volume from the outside volume.

Solution

1. The outside volume is \(8\cdot5\cdot4=160\,\text{dm}^3\). 2. The hollow inside measures \((8-2)\,\text{dm}\times(5-2)\,\text{dm}\times(4-1)\,\text{dm}=6\,\text{dm}\times3\,\text{dm}\times3\,\text{dm}\). 3. The hollow volume is \(6\cdot3\cdot3=54\,\text{dm}^3\). 4. The concrete volume is \(160-54=106\,\text{dm}^3=106\,\text{L}\).

Answer

The planter uses \(106\,\text{dm}^3\), or \(106\,\text{L}\), of concrete.

5357417

A solid concrete foundation has the stepped cross-section shown. a) Find the area of the cross-section in square meters. b) Find the volume of the foundation if it is \(15\,\text{m}\) long.

Figure for problem 535741

Hints

- Divide the cross-section into three horizontal rectangles. - Convert square centimeters to square meters carefully. - Multiply the cross-sectional area by the length.

Solution

1. Divide the cross-section into three horizontal rectangles. Their areas are \(120\cdot30=3600\,\text{cm}^2\), \(80\cdot30=2400\,\text{cm}^2\), and \(40\cdot30=1200\,\text{cm}^2\). 2. The total cross-sectional area is \(3600+2400+1200=7200\,\text{cm}^2\). 3. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), the area is \(7200\div10{,}000=0.72\,\text{m}^2\). 4. The volume is \(0.72\cdot15=10.8\,\text{m}^3\).

Answer

a) The cross-sectional area is \(0.72\,\text{m}^2\). b) The volume is \(10.8\,\text{m}^3\).

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