Free math worksheets

- Read the side length from the diagram. - A square has the same length and width. - Multiply the side length by itself and use square meters.

1. The diagram shows a side length of \(15\,\text{m}\). 2. The area of a square is side length times side length: \(15 \times 15=225\). 3. The garden's area is \(225\,\text{m}^2\).

The area of the garden is \(225\,\text{m}^2\).

- Focus on where the edging will be placed. - One measurement describes surface coverage, while the other describes the boundary. - Match the wording “go around the outside” to the appropriate measurement.

1. Edging goes around the outside boundary of the garden bed. 2. Perimeter measures the total distance around that boundary, so perimeter is the appropriate measurement.

Use the perimeter because it measures the distance around the garden bed's outside boundary.

- Think about what carpet covers: a boundary or a surface. - Recall what area measures and what perimeter measures. - No side lengths are needed to make this decision.

1. Carpet covers the surface inside the room. 2. Area measures the amount of surface inside a shape, while perimeter measures distance around its boundary.

Area, because carpet covers the surface inside the room.

- Area can be found by counting the unit squares that cover the region. - Look for equal rows and equal numbers of cells in each row. - A multiplication expression can represent the rows of unit squares.

1. The rectangle has \(3\) rows with \(5\) square units in each row. 2. Its area is \(3 \times 5 = 15\) square units.

\(15\) square units

- Find the distance around the entire rectangle. - No fencing is needed across the gate opening. - Subtract the opening from the full perimeter.

1. The garden perimeter is \(50+30+50+30=160\,\text{ft}\). 2. Subtract the gate opening: \(160-6=154\,\text{ft}\).

\(154\,\text{ft}\)

- Think of the playground as a rectangular array of square units. - Multiply the length by the width. - Break \(125\) into easier parts if needed.

1. Use the rectangle area formula: \(A=\ell \times w\). 2. Compute \(125 \times 80=(100 \times 80)+(20 \times 80)+(5 \times 80)=8000+1600+400=10{,}000\). 3. The playground area is \(10{,}000\,\text{ft}^2\).

\(10{,}000\,\text{ft}^2\)

- A square has four equal sides. - Multiply one side length by \(4\) for perimeter. - Multiply the side length by itself for area.

1. The perimeter is \(4\times 14=56\,\text{in.}\). 2. The area is \(14\times 14=196\,\text{in.}^2\).

The perimeter is \(56\,\text{in.}\), and the area is \(196\,\text{in.}^2\).

- Find the area of each rectangle separately. - Add the nonoverlapping areas.

1. The first rectangle has area \(5\times 2=10\,\text{ft}^2\). 2. The second rectangle has area \(3\times 1=3\,\text{ft}^2\). 3. The total area is \(10+3=13\,\text{ft}^2\).

The total floor area is \(13\,\text{ft}^2\).

- A rectangle has two sides of each length. - Subtract the two known sides from the perimeter. - Divide the remaining length equally between the other two sides.

1. The two known sides total \(2\times 8=16\,\text{ft}\). 2. The remaining two sides total \(26-16=10\,\text{ft}\). 3. Each remaining side is \(10\div 2=5\,\text{ft}\).

The other side is \(5\,\text{ft}\) long.

- Divide each figure into smaller rectangles or squares. - Count the complete unit squares inside each figure. - Add the areas of the parts.

1. Figure a) can be split into a \(4\times 1\) rectangle and two \(1\times 2\) rectangles. Its area is \(4\times 1+2\times(1\times 2)=4+4=8\) square units. 2. Figure b) is made of five \(1\times 1\) squares, so its area is \(5\) square units.

a) \(8\) square units b) \(5\) square units

- Count the unit squares in each figure. - Split a complicated figure into rectangles. - Use the shaded square as one square unit.

1. Figure 1 is made of \(5\) unit squares, so its area is \(5\) square units. 2. Figure 2 is a \(2 \times 3\) rectangle, so its area is \(6\) square units. 3. Figure 3 can be split into a \(3 \times 1\) rectangle and a \(1 \times 2\) rectangle, so its area is \(3+2=5\) square units. 4. Figures 1 and 3 have the same area.

Figures 1 and 3; \(5\) square units

- Divide the figure horizontally into two rectangles. - Use the labeled width and height of each rectangle. - Add the two rectangle areas.

1. The lower rectangle measures \(24\,\text{ft}\times 9\,\text{ft}\), so its area is \(216\,\text{ft}^2\). 2. The upper rectangle measures \(6\,\text{ft}\times 12\,\text{ft}\), so its area is \(72\,\text{ft}^2\). 3. The total area is \(216+72=288\,\text{ft}^2\).

\(288\,\text{ft}^2\)

- Read both side lengths from the diagram. - Express both side lengths in centimeters before multiplying. - Use square centimeters for the area unit.

1. The diagram shows side lengths \(14\,\text{cm}\) and \(50\,\text{mm}\). Convert \(50\,\text{mm}=5\,\text{cm}\). 2. The area is \(14 \times 5=70\,\text{cm}^2\).

The area is \(70\,\text{cm}^2\).

- Read the width value and the length expression from the diagram. - Substitute the width value into the length expression. - Use both side lengths to find the distance around the rectangle.

1. The diagram shows \(w=8\,\text{ft}\) and the length as \(w+4\,\text{ft}\). 2. The length is \(8+4=12\,\text{ft}\). 3. The perimeter is \(2 \times (12+8)=40\,\text{ft}\).

The perimeter is \(40\,\text{ft}\).

- Read both side lengths from the diagram. - Multiply the two side lengths. - Use square centimeters for the area unit.

1. The diagram shows side lengths \(25\,\text{cm}\) and \(14\,\text{cm}\). 2. Multiply the length by the width: \(25 \times 14=350\). 3. The poster's area is \(350\,\text{cm}^2\).

The poster's area is \(350\,\text{cm}^2\).

- Follow the entire outside boundary once. - Combine horizontal segments and vertical segments separately if that helps. - Check that every boundary segment is counted exactly once.

1. The three horizontal step segments total \(2 + 2 + 2 = 6\,\text{cm}\), and the three vertical step segments also total \(6\,\text{cm}\). 2. The bottom side and left side are each \(6\,\text{cm}\). 3. Add the boundary lengths: \(6 + 6 + 6 + 6 = 24\,\text{cm}\).

The perimeter is \(24\,\text{cm}\).

- Find the area of each field using length times width. - Compare the two areas. - Subtract to find how much greater one area is.

1. Field a) has area \(60\,\text{m} \times 40\,\text{m} = 2400\,\text{m}^2\). 2. Field b) has area \(50\,\text{m} \times 50\,\text{m} = 2500\,\text{m}^2\). 3. Field b) is greater, and the difference is \(2500 - 2400 = 100\,\text{m}^2\).

Field b) is \(100\,\text{m}^2\) greater than Field a).

- Divide the floor plan into a rectangle and a square. - Read each part's dimensions. - Add the two areas.

1. The left rectangle measures \(12\,\text{ft}\times 15\,\text{ft}\), so its area is \(180\,\text{ft}^2\). 2. The attached square measures \(6\,\text{ft}\times 6\,\text{ft}\), so its area is \(36\,\text{ft}^2\). 3. The total area is \(180+36=216\,\text{ft}^2\).

\(216\,\text{ft}^2\)

- Use the diagram to distinguish outside edges from shared interior edges. - Count how many small-triangle sides make up the outside boundary. - Equilateral means every small-triangle side has the same length.

1. Count only the small-triangle sides that lie on the outside boundary. There are \(6\) such sides. 2. Each outside side is \(8\,\text{cm}\), so the perimeter is \(6 \times 8=48\,\text{cm}\).

The perimeter of the figure is \(48\,\text{cm}\).

- Use the known area and one side to recover the missing side. - Check the missing side by multiplying the two side lengths. - Perimeter uses all four sides of the rectangle.

1. Since \(48 \div 6 = 8\), the length is \(8\,\text{ft}\). 2. The perimeter is \(2 \times (8 + 6) = 28\,\text{ft}\).

Length: \(8\,\text{ft}\) Perimeter: \(28\,\text{ft}\)

- Use side length times side length for the area of a square. - Think of the number of square feet as a missing factor in a multiplication equation with \(144\,\text{in.}^2\). - Check that the numerical value becomes smaller when the same area is expressed in the larger unit.

1. The area is \(24 \times 24=576\,\text{in.}^2\). 2. Since \(4 \times 144\,\text{in.}^2=576\,\text{in.}^2\), the area is \(4\,\text{ft}^2\).

The area is \(576\,\text{in.}^2\), or \(4\,\text{ft}^2\).

- Find the distance around the schoolyard once. - Multiply by the number of laps in one recess. - Multiply the daily distance by the number of school days.

1. One lap is the perimeter of the rectangle: \(100+60+100+60=320\,\text{ft}\). 2. Two laps cover \(2 \times 320=640\,\text{ft}\). 3. In \(5\) days, Felix runs \(5 \times 640=3200\,\text{ft}\).

a) \(640\,\text{ft}\) b) \(3200\,\text{ft}\)

- A full lap follows the entire boundary of the rectangle. - Use the corrected lap length for the four-lap distance. - Express the kilometer comparison in the same unit.

1. A lap follows all four sides: \(20\,\text{m} + 10\,\text{m} + 20\,\text{m} + 10\,\text{m} = 60\,\text{m}\). Keisha added only one length and one width. 2. Four laps are \(4 \times 60\,\text{m} = 240\,\text{m}\). Since \(1\,\text{km} = 1000\,\text{m}\), \(240\,\text{m}\) is less than \(1\,\text{km}\).

1. One lap is \(60\,\text{m}\). Keisha forgot to include the opposite length and width. 2. Four laps are \(240\,\text{m}\), which is less than \(1\,\text{km}\).

- Think about the size of one kilometer compared with a classroom. - Decide which metric unit is reasonable for a room. - A walk around a rectangle follows how many sides?

1. The measurements are not reasonable. A classroom is not several kilometers long or wide; meters would be an appropriate unit. 2. The perimeter calculation is also incorrect. Using the stated measurements, the perimeter would be \(8 + 10 + 8 + 10 = 36\,\text{km}\), not \(18\,\text{km}\). The value \(18\) comes from adding only two sides.

1. Kilometers are not a reasonable unit for classroom dimensions; meters would be appropriate. 2. A perimeter includes all four sides. With the stated measurements, the perimeter would be \(36\,\text{km}\), not \(18\,\text{km}\).

- First find the width from the relationship in the problem. - Then use all four sides of the rectangle. - Check that opposite sides have equal lengths.

1. The width is \(10 \div 2=5\,\text{ft}\). 2. The perimeter is \(10+5+10+5=30\,\text{ft}\).

\(30\,\text{ft}\)

- A frame goes around all four sides. - Find the perimeter in inches first. - Use \(12\,\text{in}=1\,\text{ft}\) and a related multiplication fact to convert.

1. Find the perimeter: \(36+24+36+24=120\,\text{in}\). 2. Since \(10\times12\,\text{in}=120\,\text{in}\), \(120\,\text{in}=10\,\text{ft}\).

\(120\,\text{in}\), or \(10\,\text{ft}\)

- Find the area of each rectangle separately. - Multiply length by width. - Compare the two areas.

1. City Hall Plaza has area \(45 \times 60=2700\,\text{ft}^2\). 2. Theater Plaza has area \(55 \times 50=2750\,\text{ft}^2\). 3. Since \(2750>2700\), Theater Plaza has the greater area.

City Hall Plaza has area \(2700\,\text{ft}^2\). Theater Plaza has area \(2750\,\text{ft}^2\). Theater Plaza is larger.

- Find the area of the whole garden. - Find the area covered by the shed. - Subtract the shed area from the garden area. - Check that your answer is smaller than the total garden area.

1. The garden area is \(65 \times 40=2600\,\text{ft}^2\). 2. The shed area is \(12 \times 25=300\,\text{ft}^2\). 3. The uncovered area is \(2600-300=2300\,\text{ft}^2\).

\(2300\,\text{ft}^2\)

- Find the perimeter of the rectangle first. - A square has four equal sides. - Divide the square perimeter equally among its sides.

1. The rectangle perimeter is \(24+16+24+16=80\,\text{ft}\). 2. A square has four equal sides, so each side is \(80 \div 4=20\,\text{ft}\).

\(20\,\text{ft}\)

- Find the width first. - Calculate the full perimeter. - Compare the perimeter with the coverage from whole cans. - You must round up because a partial can cannot be purchased.

1. The width is \(120 \div 2=60\,\text{ft}\). 2. The perimeter is \(2 \times (120+60)=360\,\text{ft}\). 3. Two cans cover \(300\,\text{ft}\), which is not enough. Three cans cover \(450\,\text{ft}\), so \(3\) cans are needed.

\(3\) cans

- Picture the two crease lines after the paper is unfolded. - Find the side length of one small square. - A square has four equal sides.

1. Each crease halves a side of the large square, so a small square has side length \(16 \div 2=8\,\text{cm}\). 2. Its perimeter is \(4 \times 8=32\,\text{cm}\).

\(32\,\text{cm}\)

- Picture the square divided into three equal strips. - Find the length and width of one strip. - Calculate both perimeters. - Subtract to find the difference.

1. The square perimeter is \(4 \times 18=72\,\text{ft}\). 2. One strip is \(18\,\text{ft}\) by \(18 \div 3=6\,\text{ft}\). 3. The strip perimeter is \(2 \times (18+6)=48\,\text{ft}\). 4. The difference is \(72-48=24\,\text{ft}\).

a) \(48\,\text{ft}\) b) \(24\,\text{ft}\)

- Find how many \(1\)-foot-square pavers cover the walkway. - Compare the total number of pavers with whole packages of \(10\). - Choose the least whole number of packages that provides enough pavers.

1. The walkway needs \(12\) pavers along its length and \(4\) pavers across its width, so it needs \(12 \times 4=48\) pavers. 2. Four packages contain \(4 \times 10=40\) pavers, which is not enough. 3. Five packages contain \(5 \times 10=50\) pavers, which is enough. Therefore, \(5\) packages are needed.

\(5\) packages are needed.

- Each dimension gets a \(6\)-inch extension at both ends. - Combine the two \(6\)-inch extensions before converting them to feet. - Find the perimeter of the enlarged rectangle and compare it with \(31\,\text{ft}\).

1. Along each dimension, the border adds \(6\) inches at both ends. The two extensions total \(12\,\text{in.}=1\,\text{ft}\). 2. The outside length is \(8+1=9\,\text{ft}\), and the outside width is \(6+1=7\,\text{ft}\). 3. The outside perimeter is \(2 \times (9+7)=32\,\text{ft}\). 4. Since \(32\,\text{ft}>31\,\text{ft}\), the boards are not long enough.

No. The outside perimeter is \(32\,\text{ft}\), so the set is \(1\,\text{ft}\) too short.

- Find the dimensions after placing the squares side by side. - Convert all measurements to millimeters first. - Use the rectangle perimeter formula. - Convert the final result back to mixed units.

1. Convert the side length: \(9\,\text{cm}\,4\,\text{mm}=94\,\text{mm}\). 2. The rectangle is \(188\,\text{mm}\) long and \(94\,\text{mm}\) wide. 3. Its perimeter is \(2 \times (188+94)=564\,\text{mm}\). 4. Convert back: \(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\).

\(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\)

- Use the square perimeter to find one side. - Cutting through the middle halves one dimension. - Find the perimeter of the new rectangle. - Check that you used two lengths and two widths.

1. The square side length is \(32 \div 4=8\,\text{cm}\). 2. Each small rectangle is \(8\,\text{cm}\) by \(4\,\text{cm}\). 3. Its perimeter is \(2 \times (8+4)=24\,\text{cm}\).

\(24\,\text{cm}\)

- Try to find rectangles with the same area but different side lengths. - Look for more than one factor pair of \(25\). - Think about whether rearranging the same pieces changes how much surface they cover.

1. Statement a is false. A \(2\,\text{cm}\times 8\,\text{cm}\) rectangle and a \(4\,\text{cm}\times 4\,\text{cm}\) square both have area \(16\,\text{cm}^2\). Their perimeters are \(20\,\text{cm}\) and \(16\,\text{cm}\), respectively. 2. Statement b is false. A \(1\,\text{cm}\times 25\,\text{cm}\) rectangle also has area \(25\,\text{cm}^2\), but it is not a square. 3. Statement c is true. Cutting and rearranging the same pieces without gaps or overlaps does not change the sum of their areas.

a) False b) False c) True

- Treat each small square as a \(1\)-unit by \(1\)-unit square. - Count only the edges on the outside of each arrangement. - Compare the areas and perimeters of the two shapes.

1. The row forms a \(1\)-unit by \(4\)-unit rectangle, so its perimeter is \(2\times(1+4)=10\) units. 2. The \(2\times 2\) arrangement has perimeter \(2\times(2+2)=8\) units. 3. Both shapes have area \(4\) square units, but their perimeters are different.

a) \(10\) units b) \(8\) units c) Shapes can have the same area but different perimeters.

- Think of a body measurement or familiar object whose length you know approximately. - Estimate the wall's width and height separately. - Use the area formula for a rectangle.

1. Choose a reference length whose approximate size you know, such as your pace, arm span, height, or the width of a familiar object. 2. Estimate the wall's width by repeating the reference length across it. Estimate the height in the same way or compare it with your own height. 3. Express both estimates in the same unit. 4. Multiply the estimated width by the estimated height to find an approximate area in square units.

One possible method is to use a known reference length, such as an approximately \(1\)-foot pace. Estimate the wall's width and height with that reference, then multiply the two estimates. Answers will vary depending on the reference used.

- Use the two dimensions of one rectangular space before working with the number of spaces. - Find how many spaces the lot contains. - Keep track of square feet when you scale the area of one space to all spaces.

1. The area of one space is \(9\times18=162\,\text{ft}^2\). 2. The number of spaces is \(8\times15=120\). 3. The total area is \(120\times162=19{,}440\,\text{ft}^2\).

a) \(162\,\text{ft}^2\) b) \(19{,}440\,\text{ft}^2\)

- Find the area of the entire playground from its dimensions. - Find the area of each rectangular activity section separately. - Compare the combined activity area with the whole playground area.

1. The playground area is \(80\times75=6000\,\text{ft}^2\). 2. The activity areas are \(25\times40=1000\,\text{ft}^2\), \(20\times30=600\,\text{ft}^2\), and \(50\times40=2000\,\text{ft}^2\). 3. Their total area is \(1000+600+2000=3600\,\text{ft}^2\). 4. The open play space is \(6000-3600=2400\,\text{ft}^2\).

\(2400\,\text{ft}^2\)

- First express both side lengths in the same unit. - Add all four side lengths to find perimeter. - Multiply length by width to find area.

1. Convert the second side length: \(50\,\text{mm}=5\,\text{cm}\). 2. The perimeter is \(2\times(12+5)=34\,\text{cm}\). 3. The area is \(12\times 5=60\,\text{cm}^2\).

The perimeter is \(34\,\text{cm}\), and the area is \(60\,\text{cm}^2\).

- Find a number that multiplied by itself equals \(64\). - A square has four equal sides. - Add the four side lengths or multiply one side length by \(4\).

1. A square with area \(64\,\text{ft}^2\) has side length \(8\,\text{ft}\), because \(8\times 8=64\). 2. Its perimeter is \(4\times 8=32\,\text{ft}\).

The perimeter is \(32\,\text{ft}\).

- A rectangle has two sides of each length. - Divide the perimeter by \(2\) to find the sum of two adjacent sides. - Find the missing side before calculating area.

1. Half of the perimeter is the sum of one length and one width: \(28\div 2=14\,\text{ft}\). 2. The missing side length is \(14-6=8\,\text{ft}\). 3. The area is \(6\times 8=48\,\text{ft}^2\).

The area is \(48\,\text{ft}^2\).

- Divide the perimeter equally among the four sides. - Multiply the side length by itself to find the area. - Label length in feet and area in square feet.

1. A square has four equal sides, so one side is \(48\div 4=12\,\text{ft}\). 2. The area is \(12\times 12=144\,\text{ft}^2\).

a) \(12\,\text{ft}\) b) \(144\,\text{ft}^2\)

- Convert the mixed measurement to inches first. - A square has four equal sides. - Multiply the side length by itself to find area.

1. Convert the side length: \(2\,\text{ft}\ 1\,\text{in.}=25\,\text{in.}\). 2. The perimeter is \(4\times 25=100\,\text{in.}\). 3. The area is \(25\times 25=625\,\text{in.}^2\).

a) \(100\,\text{in.}\) b) \(625\,\text{in.}^2\)

- Use the stated rectangular model to estimate the area of one sheet first. - Break \(32\) into tens and ones when multiplying by the one-sheet area. - Use \(144\,\text{in.}^2=1\,\text{ft}^2\) to convert the estimated total area.

1. Using the stated estimation model, one sheet has area \(9\times12=108\,\text{in.}^2\). 2. For \(32\) sheets, \(108\times32=(108\times30)+(108\times2)=3240+216=3456\,\text{in.}^2\). 3. Since \(24\times144\,\text{in.}^2=3456\,\text{in.}^2\), the estimated total area is \(24\,\text{ft}^2\). 4. \(24\,\text{ft}^2\) is closer to \(21\,\text{ft}^2\) than to \(8\,\text{ft}^2\), so the sheets cover about the area of an interior door.

The sheets would cover an area closer to an interior door. Their estimated total area is \(24\,\text{ft}^2\).

- Find the area of one square note first. - Multiply by the number of notes. - Build the square-foot equivalent with convenient multiples of \(144\,\text{in.}^2\) instead of using a large divisor.

1. One note has area \(3 \times 3=9\,\text{in.}^2\). 2. The \(400\) notes have total area \(400 \times 9=3600\,\text{in.}^2\). 3. Twenty square feet is \(20 \times 144=2880\,\text{in.}^2\), and another \(5\) square feet is \(5 \times 144=720\,\text{in.}^2\). Since \(2880+720=3600\), the notes cover \(25\,\text{ft}^2\). 4. \(25\,\text{ft}^2\) is much closer to \(24\,\text{ft}^2\) than to \(2\,\text{ft}^2\).

The total area is closer to the area of the large wall mirror. The notes have a combined area of \(25\,\text{ft}^2\).

- First find the side length of the second square. - Use four equal sides for perimeter and side times side for area. - Compare the two areas with a multiplication equation rather than a division equation.

1. For the smaller square, the perimeter is \(4 \times 5=20\,\text{in.}\), and the area is \(5 \times 5=25\,\text{in.}^2\). 2. The larger square has side length \(2 \times 5=10\,\text{in.}\). Its perimeter is \(4 \times 10=40\,\text{in.}\), and its area is \(10 \times 10=100\,\text{in.}^2\). 3. Since \(4 \times 25\,\text{in.}^2=100\,\text{in.}^2\), the larger area is \(4\) times as great.

1) Smaller square: perimeter \(20\,\text{in.}\), area \(25\,\text{in.}^2\); larger square: perimeter \(40\,\text{in.}\), area \(100\,\text{in.}^2\) 2) The larger area is \(4\) times the smaller area.

- Divide a square's perimeter by \(4\) to find its side length. - Find a number that multiplied by itself equals \(100\). - Compare the two side lengths.

1. Square A has side length \(20\div 4=5\,\text{ft}\), so its area is \(5\times 5=25\,\text{ft}^2\). 2. Square B has side length \(10\,\text{ft}\), because \(10\times 10=100\). Its perimeter is \(4\times 10=40\,\text{ft}\). 3. Square B has the longer side because \(10\,\text{ft}>5\,\text{ft}\).

Square A has area \(25\,\text{ft}^2\). Square B has perimeter \(40\,\text{ft}\). Square B has the longer side length.

- Half the perimeter equals one length plus one width. - Check whether each unit measures length. - Convert millimeters to centimeters before comparing.

1. One length plus one width must equal half the perimeter, or \(15\,\text{cm}\). A proposed length of \(22\,\text{cm}\) is already greater than \(15\,\text{cm}\), so it cannot work. 2. Square centimeters measure area, not length, so \(14\,\text{cm}^2\) cannot be a side length. 3. Since \(40\,\text{mm}=4\,\text{cm}\), the proposed length plus the width would be only \(4+8=12\,\text{cm}\), not the required \(15\,\text{cm}\).

1) It is greater than half the perimeter. 2) It uses an area unit instead of a length unit. 3) \(40\,\text{mm}=4\,\text{cm}\), which is too short.

- Divide the perimeter by the number of equal sides. - Check whether each unit measures length or area. - Convert all proposed lengths to centimeters.

1. A square has four equal sides, so the side length is \(200\div 4=50\,\text{cm}\). Choice 1 is correct. 2. Square centimeters measure area, not length. Choice 2 is incorrect. 3. Since \(2\,\text{m}=200\,\text{cm}\), this proposed side equals the entire perimeter. Choice 3 is incorrect. 4. Since \(500\,\text{mm}=50\,\text{cm}\), choice 4 is correct.

Choices 2 and 3 are incorrect. Choice 2 uses an area unit, and choice 3 equals the entire perimeter rather than one-fourth of it.

- Half the perimeter equals one length plus one width. - Subtract the known length from half the perimeter. - For part b, consider only the two equal width sides.

1. Half the perimeter is one length plus one width: \(22\div 2=11\,\text{ft}\). 2. The width is \(11-6=5\,\text{ft}\). 3. If the width were \(16\,\text{ft}\), the two width sides alone would total \(32\,\text{ft}\), which is already greater than the entire \(22\,\text{ft}\) perimeter.

a) \(5\,\text{ft}\) b) Two \(16\)-foot widths would already total \(32\,\text{ft}\), more than the entire perimeter.

- Divide the area by the known side length. - Once both side lengths are known, add all four sides. - Length units and area units are different.

1. The missing side length is \(72\div 9=8\,\text{cm}\). 2. The perimeter is \(2\times(9+8)=34\,\text{cm}\).

The missing side is \(8\,\text{cm}\), and the perimeter is \(34\,\text{cm}\).

- Divide the area by the known side length to find each missing side. - Find each perimeter separately. - Subtract the smaller perimeter from the larger one.

1. Rug A has width \(12\div 4=3\,\text{ft}\), so its perimeter is \(2\times(4+3)=14\,\text{ft}\). 2. Rug B has width \(12\div 6=2\,\text{ft}\), so its perimeter is \(2\times(6+2)=16\,\text{ft}\). 3. Rug B has the greater perimeter, and the difference is \(16-14=2\,\text{ft}\).

Rug B has the greater perimeter. The difference is \(2\,\text{ft}\).

- Convert the perimeter to centimeters first. - Half the perimeter equals one length plus one width. - Multiply the two side lengths to find area.

1. Convert the perimeter: \(400\,\text{mm}=40\,\text{cm}\). 2. Half the perimeter is one length plus one width: \(40\div 2=20\,\text{cm}\). 3. The width is \(20-12=8\,\text{cm}\). 4. The area is \(12\times 8=96\,\text{cm}^2\).

The width is \(8\,\text{cm}\), and the area is \(96\,\text{cm}^2\).

- Convert the entire perimeter to inches. - Divide by \(4\) to find one side of the square. - Multiply the side length by itself.

1. Convert the perimeter: \(10\,\text{ft}\ 8\,\text{in.}=128\,\text{in.}\). 2. One side is \(128\div 4=32\,\text{in.}\). 3. The area is \(32\times 32=1024\,\text{in.}^2\).

The area is \(1024\,\text{in.}^2\).

- Convert the perimeter to centimeters. - Half the perimeter equals the sum of two adjacent sides. - Multiply the side lengths to find area.

1. Convert the perimeter: \(1\,\text{m}=100\,\text{cm}\). 2. Half the perimeter is \(100\div 2=50\,\text{cm}\). 3. The other side is \(50-35=15\,\text{cm}\). 4. The area is \(35\times 15=525\,\text{cm}^2\).

The other side is \(15\,\text{cm}\), and the area is \(525\,\text{cm}^2\).

- Treat the shorter side as the missing factor in the area equation. - Check which number multiplied by \(30\) gives \(600\). - Use both side lengths to find the distance around the lot.

1. Since \(20 \times 30=600\), the shorter side is \(20\,\text{ft}\). 2. The perimeter is \(2 \times (30+20)=100\,\text{ft}\).

The shorter side is \(20\,\text{ft}\), and the perimeter is \(100\,\text{ft}\).

- Use the place value in \(200\) to simplify the area multiplication. - For part b), think of the number of plots as a missing factor. - Check your plot count by multiplying it by \(500\,\text{ft}^2\).

1. For the total area, \(150 \times 2=300\). Since \(200=2 \times 100\), \(150 \times 200=30{,}000\,\text{ft}^2\). 2. Since \(60 \times 500\,\text{ft}^2=30{,}000\,\text{ft}^2\), the area can be divided into \(60\) study plots.

a) \(30{,}000\,\text{ft}^2\) b) \(60\) study plots

- Treat the width as a missing factor in the area equation. - Use the place values in \(400\) and \(80{,}000\) to test a likely factor. - Use both dimensions to find the distance around the field.

1. Since \(200 \times 400=80{,}000\), the width is \(200\,\text{ft}\). 2. The perimeter is \(2 \times (400+200)=1200\,\text{ft}\).

a) \(200\,\text{ft}\) b) \(1200\,\text{ft}\) of fencing

- Break \(30\) into \(3 \times 10\) when finding the cleared area. - Compare the two areas in the same unit. - Subtract to find how much larger the replanted area is.

1. First, \(4000 \times 3=12{,}000\). Since \(30=3 \times 10\), the cleared area is \(120{,}000\,\text{ft}^2\). 2. Since \(150{,}000>120{,}000\), the replanted area is large enough. 3. The difference is \(150{,}000-120{,}000=30{,}000\,\text{ft}^2\).

a) \(120{,}000\,\text{ft}^2\) b) Yes. The replanted area is \(30{,}000\,\text{ft}^2\) larger.

- A square has four equal sides. - Multiply the side length by itself to find area. - Divide a square's perimeter by \(4\) to find its side length.

1. For Square A, \(P=4\times 12=48\,\text{cm}\), and \(A=12\times 12=144\,\text{cm}^2\). 2. For Square B, the side length is \(36\div 4=9\,\text{m}\), and the area is \(9\times 9=81\,\text{m}^2\).

<table> <tr><td>Square</td><td>Side length \(s\)</td><td>Perimeter \(P\)</td><td>Area \(A\)</td></tr> <tr><td>A</td><td>\(12\,\text{cm}\)</td><td>\(48\,\text{cm}\)</td><td>\(144\,\text{cm}^2\)</td></tr> <tr><td>B</td><td>\(9\,\text{m}\)</td><td>\(36\,\text{m}\)</td><td>\(81\,\text{m}^2\)</td></tr> </table>

- The wire length becomes the perimeter of the first square. - Find a number that multiplied by itself equals \(64\). - Compare the two side lengths.

1. The wire length is the first square's perimeter, so its side length is \(28\div 4=7\,\text{in.}\). 2. The second square has side length \(8\,\text{in.}\), because \(8\times 8=64\). 3. Since \(8\,\text{in.}>7\,\text{in.}\), the second square has the longer side.

The second square has the longer side. Its side is \(8\,\text{in.}\), compared with \(7\,\text{in.}\) for the wire square.

- Find the first square's side length from its area. - Double that side length. - Use the new side length to find perimeter and area.

1. The first poster has side length \(7\,\text{in.}\), because \(7\times 7=49\). 2. The second poster has side length \(2\times 7=14\,\text{in.}\). 3. Its perimeter is \(4\times 14=56\,\text{in.}\). 4. Its area is \(14\times 14=196\,\text{in.}^2\).

The second poster has perimeter \(56\,\text{in.}\) and area \(196\,\text{in.}^2\).

- Find the area of one square photo first. - Multiply by the number of photos. - Treat the number of square feet as the missing factor in a multiplication equation with \(144\,\text{in.}^2\).

1. One photo has area \(6 \times 6=36\,\text{in.}^2\). 2. All \(40\) photos have area \(40 \times 36=1440\,\text{in.}^2\). 3. Since \(10 \times 144\,\text{in.}^2=1440\,\text{in.}^2\), the total area is \(10\,\text{ft}^2\).

a) \(36\,\text{in.}^2\) b) \(1440\,\text{in.}^2\), or \(10\,\text{ft}^2\)

- Split \(2.5\) feet into \(2\) feet and one-half foot. - Use the banner width to find the total wall length. - Find the area of one banner, then multiply by the number of banners.

1. Split the width as \(2.5\,\text{ft}=2\,\text{ft}+0.5\,\text{ft}\). The banners cover \(12\times 2=24\,\text{ft}\) plus \(12\times 0.5=6\,\text{ft}\), for a total of \(30\,\text{ft}\). 2. One banner has area \(8\times 2+8\times 0.5=16+4=20\,\text{ft}^2\). 3. All \(12\) banners have area \(12\times 20=240\,\text{ft}^2\).

a) \(30\,\text{ft}\) b) \(240\,\text{ft}^2\)

- When strips are end to end, their lengths add. - Repeated equal lengths can be found with multiplication. - For part b, regroup the inch total into feet and leftover inches.

1. Fifteen strips have total length \(15 \times 22\,\text{in} = 330\,\text{in}\). 2. Eighty strips have total length \(80 \times 22\,\text{in} = 1760\,\text{in}\). 3. Since \(1760 = 146 \times 12 + 8\), \(1760\,\text{in} = 146\,\text{ft}\ 8\,\text{in}\).

a) \(330\,\text{in}\) b) \(1760\,\text{in}\), or \(146\,\text{ft}\ 8\,\text{in}\)

- Find the full square area. - Subtract the border width from both ends of the side length. - Subtract the inner area from the total area. - Multiply the flower-bed area by the number of flowers per square foot.

1. The full lawn area is \(10\times 10=100\,\text{ft}^2\). 2. The inner lawn has side length \(10-1-1=8\,\text{ft}\), so its area is \(8\times 8=64\,\text{ft}^2\). 3. The flower-bed area is \(100-64=36\,\text{ft}^2\). 4. The number of flowers is \(36\times 8=288\).

\(288\) flowers are needed.

- Add the lawn and pond areas. - Subtract their combined area from the total garden area.

1. The lawn and pond use \(3500+450=3950\,\text{ft}^2\). 2. The vegetable beds use \(8000-3950=4050\,\text{ft}^2\).

\(4050\,\text{ft}^2\)

- Find the area of one sticker for part a). - For part b), first determine how many \(4\)-inch sticker sides fit along one \(24\)-inch poster side. - Use the same number of stickers in each row and column.

1. One sticker has area \(4 \times 4=16\,\text{in.}^2\). 2. Along one side of the poster, \(6\) stickers fit because \(6 \times 4\,\text{in.}=24\,\text{in.}\). 3. The square arrangement has \(6\) rows of \(6\) stickers, so \(6 \times 6=36\) stickers are needed.

a) \(16\,\text{in.}^2\) b) \(36\) stickers

- Express both dimensions in feet. - Multiply length by width. - Label the result in square feet.

1. Convert the width: \(96\,\text{in.}=8\,\text{ft}\). 2. The area is \(15\times 8=120\,\text{ft}^2\).

The area is \(120\,\text{ft}^2\).

- Find a number that multiplied by itself equals \(10{,}000\). - A square has four equal sides.

1. The side length is \(100\,\text{ft}\), because \(100\times 100=10{,}000\). 2. The perimeter is \(4\times 100=400\,\text{ft}\).

The perimeter is \(400\,\text{ft}\).

- Find the first rectangle's area. - Use the same area for the second rectangle. - Treat the unknown width as the missing factor in \(10 \times ? = 60\).

1. The first rectangle's area is \(15 \times 4=60\,\text{cm}^2\). 2. The second rectangle also has area \(60\,\text{cm}^2\). Since \(10 \times 6=60\), its width is \(6\,\text{cm}\).

The width is \(6\,\text{cm}\).

- Find the square's perimeter. - Use that same perimeter for the rectangle. - Half the rectangle's perimeter is one length plus one width.

1. The square's perimeter is \(4\times 8=32\,\text{ft}\). 2. For the rectangle, one length plus one width is half the perimeter: \(32\div 2=16\,\text{ft}\). 3. The length is \(16-6=10\,\text{ft}\).

The rectangular play area is \(10\,\text{ft}\) long.

- Find the rectangle's area first. - Find a square side length whose product with itself equals that area. - Multiply the square's side length by \(4\).

1. The rectangle's area is \(16\times 9=144\,\text{in.}^2\). 2. The square has side length \(12\,\text{in.}\), because \(12\times 12=144\). 3. Its perimeter is \(4\times 12=48\,\text{in.}\).

a) \(144\,\text{in.}^2\) b) \(48\,\text{in.}\)

- Find the original area and then the new total area. - Use the unchanged length to find the new width. - Compare the original and new perimeters.

1. The original area is \(5 \times 3 = 15\,\text{ft}^2\). 2. The new area is \(15 + 10 = 25\,\text{ft}^2\). 3. The new width is \(25 \div 5 = 5\,\text{ft}\). 4. The original perimeter is \(2 \times (5 + 3) = 16\,\text{ft}\), and the new perimeter is \(2 \times (5 + 5) = 20\,\text{ft}\). 5. The perimeter increases by \(20 - 16 = 4\,\text{ft}\).

a) \(5\,\text{ft}\) b) The perimeter increases by \(4\,\text{ft}\).

- Find the area of each section separately. - Add the two areas. - Divide the total area by the area covered by one tile.

1. The rectangular area is \(12\times 5=60\,\text{ft}^2\). 2. The square area is \(6\times 6=36\,\text{ft}^2\). 3. The total area is \(60+36=96\,\text{ft}^2\). 4. The number of tiles is \(96\div 4=24\).

\(24\) tiles are needed.

- Split \(8.5\) feet into \(8\) feet and one-half foot. - Find the area of each rug. - Compare the two areas and subtract to find the difference.

1. Split \(8.5\,\text{ft}\) into \(8\,\text{ft}+0.5\,\text{ft}\). The rectangular rug has area \(8\times 6+0.5\times 6=48+3=51\,\text{ft}^2\). 2. The square rug has area \(7\times 7=49\,\text{ft}^2\). 3. The rectangular rug is larger by \(51-49=2\,\text{ft}^2\).

a) Rectangular rug: \(51\,\text{ft}^2\); square rug: \(49\,\text{ft}^2\) b) The rectangular rug is larger by \(2\,\text{ft}^2\).

- Find the area of one poster first. - Use \(50=5 \times 10\) to simplify the multiplication by \(50\). - Build the square-foot equivalent with convenient multiples of \(144\,\text{in.}^2\).

1. One poster has area \(12 \times 18=216\,\text{in.}^2\). 2. First, \(216 \times 5=1080\). Since \(50=5 \times 10\), all \(50\) posters have area \(10{,}800\,\text{in.}^2\). 3. To convert, \(144 \times 70=10{,}080\) and \(144 \times 5=720\). Since \(10{,}080+720=10{,}800\), the total area is \(75\,\text{ft}^2\).

The posters cover \(75\,\text{ft}^2\).

- Use the array to determine the patio's side length. - Put the patio dimensions in feet before comparing areas. - Calculate both areas; matching one side length is not enough to decide the comparison.

1. Four pavers lie along each patio side, so the side length is \(4\times18=72\,\text{in}\). 2. Convert the patio side: \(72\,\text{in}=6\,\text{ft}\). 3. The patio area is \(6\times6=36\,\text{ft}^2\). 4. The shed area is \(5\times7=35\,\text{ft}^2\). 5. The patio is larger by \(36-35=1\,\text{ft}^2\).

The patio is larger by \(1\,\text{ft}^2\).

- Find the original perimeter first. - Write the doubled side lengths before finding the new perimeter. - Compare the two perimeters with a multiplication relationship.

1. The original perimeter is \(2 \times (6+4)=20\,\text{cm}\). 2. The new side lengths are \(12\,\text{cm}\) and \(8\,\text{cm}\). The new perimeter is \(2 \times (12+8)=40\,\text{cm}\). 3. Since \(2 \times 20=40\), doubling both side lengths doubles the perimeter.

a) \(20\,\text{cm}\) b) \(40\,\text{cm}\) c) The perimeter doubles when both side lengths are doubled.

- Write an area expression for each garden bed. - Subtract the smaller area from the larger area. - Use the shared width as the variable. - Solve the resulting equation.

1. Let \(x\) feet be the common width. 2. The difference in area is \(60x - 45x = 375\). 3. Combine like terms: \(15x = 375\). 4. Divide by \(15\): \(x = 25\).

The common width is \(25\,\text{ft}\).

- Enclose the L-shape in one large rectangle. - Find the dimensions of the missing corner. - Subtract the missing area from the large rectangle.

1. Enclose the L-shape in a \(10\,\text{ft}\times 8\,\text{ft}\) rectangle, which has area \(10\times 8=80\,\text{ft}^2\). 2. The missing corner measures \((10-4)\,\text{ft}\times(8-4)\,\text{ft}=6\,\text{ft}\times 4\,\text{ft}\), so its area is \(24\,\text{ft}^2\). 3. The patio area is \(80-24=56\,\text{ft}^2\).

\(56\,\text{ft}^2\)

- Use the labeled segments in the diagram to find the full rectangle dimensions. - Find the full perimeter, then account for the open gate segment shown in the diagram. - Whole rolls must provide at least as much fencing as the required length.

1. From the diagram, the garden length is \(12 + 4 + 14 = 30\,\text{ft}\), and the width is \(18\,\text{ft}\). 2. The full perimeter is \(2 \times (30 + 18) = 96\,\text{ft}\). 3. Leave the marked \(4\)-foot gate segment unfenced: \(96 - 4 = 92\,\text{ft}\) of fencing is needed. 4. Nine rolls provide \(90\,\text{ft}\), which is not enough. Ten rolls provide \(100\,\text{ft}\), so \(10\) rolls are required. 5. The total cost is \(10 \times \$45.00 = \$450.00\).

a) \(92\,\text{ft}\) of fencing b) \(10\) rolls, costing \(\$450.00\)

- Use the total height and the inner vertical length to find the lower height. - Split the L-shape into two rectangles. - Find the missing horizontal side by subtraction, then add every outside edge for the perimeter.

1. The lower section is \(7-4=3\,\text{ft}\) high. Split the figure into a bottom rectangle measuring \(8\,\text{ft}\times 3\,\text{ft}\) and an upper-left rectangle measuring \(3\,\text{ft}\times 4\,\text{ft}\). 2. The area is \(8\times 3+3\times 4=24+12=36\,\text{ft}^2\). 3. The unlabeled horizontal side is \(8-3=5\,\text{ft}\). Add the six outside edges: \(8+3+5+4+3+7=30\,\text{ft}\).

a) \(36\,\text{ft}^2\) b) \(30\,\text{ft}\)

- Trace and add every outer and inner boundary segment. - Use subtraction to find the unlabeled arm width and bottom-strip height. - Split the U-shape into three nonoverlapping rectangles and add their areas.

1. Add all eight boundary lengths: \(7+5+2+3+3+3+2+5=30\,\text{ft}\). 2. The left side width is \(7-2-3=2\,\text{ft}\), and the bottom strip height is \(5-3=2\,\text{ft}\). Split the shape into two \(2\,\text{ft}\times 5\,\text{ft}\) side rectangles and one \(3\,\text{ft}\times 2\,\text{ft}\) bottom rectangle. 3. The area is \(2\times(2\times 5)+3\times 2=20+6=26\,\text{ft}^2\).

a) \(30\,\text{ft}\) b) \(26\,\text{ft}^2\)

- Read the outer patio dimensions and inner pool dimensions from the diagram. - Subtract the pool area from the full patio area. - Use the tiled area, not the full patio area, when finding the cost.

1. From the diagram, the full patio measures \(12\,\text{ft}\times8\,\text{ft}\), so its area is \(12 \times 8=96\,\text{ft}^2\). 2. The pool measures \(6\,\text{ft}\times4\,\text{ft}\), so its area is \(6 \times 4=24\,\text{ft}^2\). 3. The tiled area is \(96-24=72\,\text{ft}^2\). 4. The total tile cost is \(72 \times \$35.00=\$2520.00\).

a) \(72\,\text{ft}^2\) b) \(\$2520.00\)

- Split the T-shape into two nonoverlapping rectangles. - Find each rectangle's dimensions. - Add the two areas.

1. Split the T-shape into a top rectangle measuring \(12\,\text{ft}\times 4\,\text{ft}\) and a lower rectangle measuring \(4\,\text{ft}\times 6\,\text{ft}\). 2. Their areas are \(12\times 4=48\,\text{ft}^2\) and \(4\times 6=24\,\text{ft}^2\). 3. The total area is \(48+24=72\,\text{ft}^2\).

The patio area is \(72\,\text{ft}^2\).

- Imagine vertical and horizontal lines that create nonoverlapping rectangles. - Use the grid spacing to determine each rectangle's dimensions. - Add the component areas.

1. Figure A can be split into a \(5\times 2\) bottom rectangle and two \(2\times 2\) upper rectangles. Its area is \(5\times 2+2\times 2+2\times 2=10+4+4=18\) square units. 2. Figure B can be split into a \(5\times 2\) middle rectangle and two \(3\times 1\) rectangles. Its area is \(5\times 2+3\times 1+3\times 1=10+3+3=16\) square units.

a) \(18\) square units b) \(16\) square units

- Perimeter is the total distance around the outside boundary. - Count peg-to-peg segments once around each figure. - Do not skip or double-count an edge. - Each adjacent-peg segment is \(1\,\text{cm}\).

1. Figure A has side lengths \(4,3,1,2,2,2,1,3\), so its perimeter is \(4+3+1+2+2+2+1+3=18\,\text{cm}\). 2. Figure B has side lengths \(4,1,1,1,1,1,1,1,1,4\), so its perimeter is \(4+1+1+1+1+1+1+1+1+4=16\,\text{cm}\).

Figure A: \(18\,\text{cm}\) Figure B: \(16\,\text{cm}\)

- Trace the outside edge to find each perimeter. - Pay special attention to inward and outward corners. - Count unit squares or split each figure into rectangles. - Compare both perimeter and area.

1. Figure A is a \(4 \times 2\) rectangle. Its perimeter is \(4+2+4+2=12\) units, and its area is \(4 \times 2=8\) square units. 2. Figure B has perimeter \(3+2+2+2+1+4=14\) units. It can be split into a \(1 \times 4\) rectangle and a \(2 \times 2\) rectangle, so its area is \(4+4=8\) square units. 3. Figure B has the greater perimeter, but the two figures have equal areas.

a) Figure A: \(12\) units; Figure B: \(14\) units b) Figure B c) Each figure has area \(8\) square units. Equal-area figures can have different perimeters.

- For area, count unit squares or split the figure into rectangles. - For perimeter, trace the outside edge once. - Each grid segment is \(1\,\text{cm}\) long.

1. Split the figure into vertical strips containing \(3\), \(2\), and \(1\) unit squares. The area is \(3+2+1=6\,\text{cm}^2\). 2. Add the lengths around the outside boundary: \(3+1+1+1+1+1+1+3=12\,\text{cm}\).

a) \(6\,\text{cm}^2\) b) \(12\,\text{cm}\)

- Imagine a line between the vertical stem and the horizontal bar. - Find each rectangle's area and add. - For part b, find factor pairs of the area from part a.

1. The vertical stem is a \(2\times 4\) rectangle, so its area is \(2\times 4=8\) square units. 2. The horizontal bar is an \(8\times 2\) rectangle, so its area is \(8\times 2=16\) square units. 3. The total area is \(8+16=24\) square units. 4. Two rectangles with area \(24\) square units can have side lengths \(3\) and \(8\), or \(4\) and \(6\).

a) \(24\) square units b) Possible side-length pairs are \(3\) and \(8\), and \(4\) and \(6\).

- To find perimeter, count the grid-unit lengths along the boundary. - To find area, count the unit squares inside each figure. - Organize the perimeter and area values so you can compare them. - Think about how changing a rectangle from compact to long and narrow affects its area.

1. Figure 1 is a square. Its perimeter is \(3 + 3 + 3 + 3 = 12\) units, and its area is \(3 \times 3 = 9\) square units. 2. Figure 2 is a rectangle. Its perimeter is \(4 + 2 + 4 + 2 = 12\) units, and its area is \(4 \times 2 = 8\) square units. 3. Figure 3 is a rectangle. Its perimeter is \(5 + 1 + 5 + 1 = 12\) units, and its area is \(5 \times 1 = 5\) square units. 4. All three figures have the same perimeter, but their areas are different. 5. Figure 1 has the greatest area for the same perimeter.

a) Figure 1: perimeter \(12\) units, area \(9\) square units. Figure 2: perimeter \(12\) units, area \(8\) square units. Figure 3: perimeter \(12\) units, area \(5\) square units. b) The perimeters are equal, but the areas are different. c) Figure 1 encloses the greatest area.

- Find each area by counting or decomposing unit squares. - For perimeter, count only the outside boundary segments. - Compare whether the same area can be arranged with different amounts of boundary.

1. Figure 1 is a \(4\,\text{cm}\times 2\,\text{cm}\) rectangle. Its area is \(4\times 2=8\,\text{cm}^2\), and its perimeter is \(2\times(4+2)=12\,\text{cm}\). 2. Figure 2 can be split into a \(5\,\text{cm}\times 1\,\text{cm}\) rectangle and a \(3\,\text{cm}\times 1\,\text{cm}\) rectangle. Its area is \(5+3=8\,\text{cm}^2\). Its perimeter is \(5+1+2+1+3+2=14\,\text{cm}\). 3. Jordan is not correct. Equal areas do not require equal perimeters.

Figure 1: area \(8\,\text{cm}^2\), perimeter \(12\,\text{cm}\) Figure 2: area \(8\,\text{cm}^2\), perimeter \(14\,\text{cm}\) Jordan is not correct.

- Find the area and perimeter of each figure separately. - Split each L-shape into rectangles. - For perimeter, count only the outside edges.

1. Figure a) can be split into a \(4 \times 1\) rectangle and a \(1 \times 3\) rectangle. Its area is \(4+3=7\) square units. Its perimeter is \(4+1+3+3+1+4=16\) units. 2. Figure b) can be split into a \(3 \times 2\) rectangle and a \(2 \times 1\) rectangle. Its area is \(6+2=8\) square units. Its perimeter is \(3+2+1+1+2+3=12\) units. 3. Since \(8>7\), Figure b) has the greater area.

Figure b) has the greater area: \(8\) square units compared with \(7\) square units for Figure a). The perimeters are \(16\) units for Figure a) and \(12\) units for Figure b).

- Identify the two vertical rectangles and the bottom rectangle. - For perimeter, include the edges inside the U-shaped indentation. - Check the area against the surrounding \(6\times 3\) rectangle.

1. Decompose the figure into two \(1\times 3\) vertical rectangles and one \(4\times 1\) horizontal rectangle. The area is \(2\times(1\times 3)+4\times 1=6+4=10\) square units. 2. Add the lengths of the entire boundary, including the indentation. The perimeter is \(6+3+1+2+4+2+1+3=22\) length units.

Area: \(10\) square units Perimeter: \(22\) length units

- First count the boundary in grid segments and the area in grid squares. - Use the given peg spacing for boundary lengths. - Find the area of one grid square before scaling to the whole figure.

1. The boundary contains \(12\) grid segments, so the perimeter is \(12 \times 3\,\text{in} = 36\,\text{in}\). 2. The figure contains \(5\) grid squares. Each grid square has area \(3\,\text{in} \times 3\,\text{in} = 9\,\text{in}^2\). 3. The total area is \(5 \times 9\,\text{in}^2 = 45\,\text{in}^2\).

Perimeter: \(36\,\text{in}\) Area: \(45\,\text{in}^2\)

- Use subtraction to find the unlabeled lower height and lower-right width. - Imagine one straight segment that divides the L-shape into two rectangles. - Find each rectangle's area and add.

1. The lower section is \(18-12=6\,\text{ft}\) high, and the lower-right section is \(15-6=9\,\text{ft}\) wide. 2. Using a vertical decomposition, the left rectangle measures \(6\,\text{ft}\times 18\,\text{ft}\), so its area is \(108\,\text{ft}^2\). The lower-right rectangle measures \(9\,\text{ft}\times 6\,\text{ft}\), so its area is \(54\,\text{ft}^2\). 3. The kitchen's area is \(108+54=162\,\text{ft}^2\). 4. A horizontal decomposition also works: \(15\times 6+6\times 12=90+72=162\,\text{ft}^2\).

\(162\,\text{ft}^2\)

- Use the total height and upper height to find the lower height. - Divide the L-shape into two rectangles. - Find and add the two rectangle areas.

1. The lower section is \(15-6=9\,\text{ft}\) high. 2. Decompose the patio into a lower rectangle measuring \(15\,\text{ft}\times 9\,\text{ft}\) and an upper-left rectangle measuring \(6\,\text{ft}\times 6\,\text{ft}\). 3. Their areas are \(15\times 9=135\,\text{ft}^2\) and \(6\times 6=36\,\text{ft}^2\). 4. The patio's area is \(135+36=171\,\text{ft}^2\).

\(171\,\text{ft}^2\)

- Divide the figure into three equal-width rectangles. - Use the labeled step lengths to determine each rectangle's height. - Add the three rectangle areas.

1. Each step is \(8\,\text{in}\) wide and \(8\,\text{in}\) high. Decompose the panel into three vertical rectangles. 2. Their widths are each \(8\,\text{in}\), and their heights are \(24\,\text{in}\), \(16\,\text{in}\), and \(8\,\text{in}\). 3. The total area is \(8 \times 24 + 8 \times 16 + 8 \times 8 = 384\,\text{in}^2\).

\(384\,\text{in}^2\)

- Perimeter is the total distance around the figure. - Find the two unlabeled side lengths from the full dimensions. - Include all six boundary sides.

1. The missing side lengths are \(5-3=2\,\text{ft}\) and \(6-2=4\,\text{ft}\). 2. Add all six boundary sides: \(6+2+4+3+2+5=22\,\text{ft}\).

\(22\,\text{ft}\)

- Use subtraction to find the unlabeled lower height and lower-right width. - Split the L-shape into two nonoverlapping rectangles. - Add the two rectangle areas.

1. The lower section is \(8 - 5 = 3\,\text{cm}\) high, and the lower-right section is \(8 - 3 = 5\,\text{cm}\) wide. 2. Decompose the figure into a left rectangle measuring \(3\,\text{cm} \times 8\,\text{cm}\) and a lower-right rectangle measuring \(5\,\text{cm} \times 3\,\text{cm}\). 3. Their areas are \(24\,\text{cm}^2\) and \(15\,\text{cm}^2\). 4. The total area is \(39\,\text{cm}^2\).

\(39\,\text{cm}^2\)

- Find the area of rectangle A first. - Split Area B into two rectangles. - Use the total height to find the missing height of the upper rectangle. - Compare the two total areas.

1. Area A is \(8\,\text{m} \times 6\,\text{m} = 48\,\text{m}^2\). 2. Split Area B into two rectangles. The lower rectangle has area \(10\,\text{m} \times 4\,\text{m} = 40\,\text{m}^2\). The upper rectangle has height \(7 - 4 = 3\,\text{m}\), so its area is \(4\,\text{m} \times 3\,\text{m} = 12\,\text{m}^2\). 3. Area B is \(40 + 12 = 52\,\text{m}^2\). 4. Since \(52 > 48\), Area B is greater and will need more grass seed.

Area B is greater. It has an area of \(52\,\text{m}^2\), compared with \(48\,\text{m}^2\) for Area A.

- Imagine completing the shape to make one large rectangle. - Find the area of the missing planter rectangle. - Subtract the missing area from the large rectangle.

1. The surrounding rectangle measures \(18\,\text{ft}\times 12\,\text{ft}\), so its area is \(216\,\text{ft}^2\). 2. The planter measures \(6\,\text{ft}\times 6\,\text{ft}\), so its area is \(36\,\text{ft}^2\). 3. The balcony floor area is \(216-36=180\,\text{ft}^2\).

\(180\,\text{ft}^2\)

- Use subtraction to find the missing main-width and nook-height measures. - Divide the floor plan into two rectangles. - Add their areas.

1. The main rectangle is \(18 - 6 = 12\,\text{ft}\) wide, and the reading nook is \(15 - 6 = 9\,\text{ft}\) high. 2. Decompose the floor plan into a \(12\,\text{ft} \times 15\,\text{ft}\) rectangle and a \(6\,\text{ft} \times 9\,\text{ft}\) rectangle. 3. Their areas are \(180\,\text{ft}^2\) and \(54\,\text{ft}^2\). 4. The total floor area is \(234\,\text{ft}^2\).

\(234\,\text{ft}^2\)

- Use subtraction to find the unlabeled left width. - Add the labeled vertical steps to find each section's height. - Divide the stepped shape into three rectangles and add their areas.

1. The left section is \(36-12-9=15\,\text{ft}\) wide. The three section heights are \(6+6+12=24\,\text{ft}\), \(6+6=12\,\text{ft}\), and \(6\,\text{ft}\). 2. Divide the floor plan into three vertical rectangles measuring \(15\,\text{ft}\times 24\,\text{ft}\), \(12\,\text{ft}\times 12\,\text{ft}\), and \(9\,\text{ft}\times 6\,\text{ft}\). 3. Their areas are \(360\,\text{ft}^2\), \(144\,\text{ft}^2\), and \(54\,\text{ft}^2\). 4. The total area is \(360+144+54=558\,\text{ft}^2\).

\(558\,\text{ft}^2\)

- First find the area as though there were no opening. - Find the area of the square opening. - Subtract the opening from the full slab.

1. The full slab measures \(8\,\text{ft}\times 6\,\text{ft}\), so its area is \(48\,\text{ft}^2\). 2. The square opening measures \(1\,\text{ft}\times 1\,\text{ft}\), so its area is \(1\,\text{ft}^2\). 3. The concrete surface area is \(48-1=47\,\text{ft}^2\).

\(47\,\text{ft}^2\)

- Complete the shape to make one large rectangle. - Find the dimensions of the missing rectangle. - Subtract the missing area.

1. View the floor plan as a \(90\,\text{ft}\times 45\,\text{ft}\) rectangle with a \(30\,\text{ft}\times 15\,\text{ft}\) rectangle removed. 2. The large rectangle has area \(90\times 45=4050\,\text{ft}^2\). The missing rectangle has area \(30\times 15=450\,\text{ft}^2\). 3. The hall's area is \(4050-450=3600\,\text{ft}^2\).

\(3600\,\text{ft}^2\)

- Imagine a horizontal segment that makes two rectangles. - Use the labeled width and height of each rectangle. - Add the two areas.

1. The lower rectangle measures \(36\,\text{ft}\times 15\,\text{ft}\), so its area is \(540\,\text{ft}^2\). 2. The upper rectangle measures \(12\,\text{ft}\times 21\,\text{ft}\), so its area is \(252\,\text{ft}^2\). 3. The total area is \(540+252=792\,\text{ft}^2\).

a) The lower rectangle has area \(540\,\text{ft}^2\), and the upper rectangle has area \(252\,\text{ft}^2\). b) The total area is \(792\,\text{ft}^2\).

- Read the square's side length from the diagram. - Convert meters to centimeters before finding the area. - Use \(120=12 \times 10\) to organize the multiplication with place value.

1. The diagram shows a side length of \(1.2\,\text{m}\). Convert: \(1.2\,\text{m}=120\,\text{cm}\). 2. Write \(120=12 \times 10\). Then the area is \((12 \times 12) \times (10 \times 10)=144 \times 100=14{,}400\,\text{cm}^2\).

The area is \(14{,}400\,\text{cm}^2\).

- The two rectangles use different length units in their labels. - Put both rectangles' dimensions in one common unit before finding their areas. - Compare the resulting square-unit measures.

1. For Rectangle A, convert \(25\,\text{mm} = 2.5\,\text{cm}\), so its area is \(8\,\text{cm} \times 2.5\,\text{cm} = 20\,\text{cm}^2\). 2. Rectangle B has area \(4\,\text{cm} \times 6\,\text{cm} = 24\,\text{cm}^2\). 3. Since \(24\,\text{cm}^2 > 20\,\text{cm}^2\), Rectangle B has the greater area.

Rectangle B has the greater area: \(24\,\text{cm}^2\), compared with \(20\,\text{cm}^2\) for Rectangle A.

- Decide whether the problem needs area or boundary length. - Use the overall width and height together with the labeled shorter sides to infer the two unlabeled boundary sides. - Add all six boundary sides, then use the price per foot.

1. The unlabeled right vertical side is \(6-3=3\,\text{ft}\). 2. The unlabeled horizontal step is \(8-5=3\,\text{ft}\). 3. The perimeter is \(8+3+3+3+5+6=28\,\text{ft}\). 4. The total cost is \(28\times\$12.50=\$350.00\).

\(\$350.00\)

- Divide the floor plan into two nonoverlapping rectangles. - Add their areas. - Interpret \(\$0.50\) as one-half dollar and find half of the total area.

1. Decompose the floor plan into a \(15\,\text{ft}\times 12\,\text{ft}\) rectangle and a \(9\,\text{ft}\times 6\,\text{ft}\) rectangle. 2. The areas are \(15\times 12=180\,\text{ft}^2\) and \(9\times 6=54\,\text{ft}^2\), for a total of \(234\,\text{ft}^2\). 3. A rate of \(\$0.50\) per square foot is one-half dollar per square foot. Half of \(234\) is \(117\), so the cost is \(\$117.00\).

\(\$117.00\)

- Subtract the cutout's area from the original rectangle's area. - Add every visible boundary segment for the perimeter. - Compare the removed outer edges with the new inner edges.

1. The original panel measures \(10\,\text{cm}\times 8\,\text{cm}\), so its area is \(80\,\text{cm}^2\). The removed rectangle measures \(3\,\text{cm}\times 3\,\text{cm}\), so its area is \(9\,\text{cm}^2\). The remaining area is \(80-9=71\,\text{cm}^2\). 2. Adding the boundary segments gives \(10+5+3+3+7+8=36\,\text{cm}\). The corner cut replaces two outer segments with two equal-length inner segments, so the perimeter remains the same as the original rectangle's perimeter.

Perimeter: \(36\,\text{cm}\) Area: \(71\,\text{cm}^2\)

- Count every outside side. - Use the total width to find the unlabeled horizontal side. - Use the total height to find the unlabeled vertical side.

1. The missing horizontal side is \(10-4=6\,\text{ft}\). 2. The missing vertical side is \(8-5=3\,\text{ft}\). 3. The perimeter is \(10+5+6+3+4+8=36\,\text{ft}\).

\(36\,\text{ft}\)

- Use subtraction to find the addition's unlabeled width. - Divide the floor plan into two rectangles and add their areas. - Add only the outer sides for the perimeter.

1. The addition is \(18-12=6\,\text{ft}\) wide. Decompose the floor plan into an \(18\,\text{ft}\times 12\,\text{ft}\) rectangle and a \(6\,\text{ft}\times 9\,\text{ft}\) rectangle. 2. The total area is \(18\times 12+6\times 9=216+54=270\,\text{ft}^2\). 3. Add the outside sides: \(18+21+6+9+12+12=78\,\text{ft}\).

Area: \(270\,\text{ft}^2\) Perimeter: \(78\,\text{ft}\)

- Subtract the missing corner's area from the large rectangle. - Count every side of the remaining shape. - Compare the removed outer sides with the new inner sides.

1. The original rectangle measures \(9\,\text{cm}\times 6\,\text{cm}\), so its area is \(54\,\text{cm}^2\). The removed corner measures \(3\,\text{cm}\times 2\,\text{cm}\), so its area is \(6\,\text{cm}^2\). The logo's area is \(54-6=48\,\text{cm}^2\). 2. The perimeter is \(9+4+3+2+6+6=30\,\text{cm}\). The removed outer segments are replaced by equal-length inner segments.

Area: \(48\,\text{cm}^2\) Perimeter: \(30\,\text{cm}\)

- View the figure as a large rectangle with a smaller rectangle removed. - Include the inward-facing sides when finding the perimeter.

1. The surrounding rectangle measures \(8\,\text{cm}\times 6\,\text{cm}\), with area \(48\,\text{cm}^2\). The rectangular opening measures \(4\,\text{cm}\times 4\,\text{cm}\), with area \(16\,\text{cm}^2\). The figure's area is \(48-16=32\,\text{cm}^2\). 2. Adding the outer and inner boundary segments gives \(8+6+2+4+4+4+2+6=36\,\text{cm}\).

Area: \(32\,\text{cm}^2\) Perimeter: \(36\,\text{cm}\)

- Read the bedroom dimensions and rug side length from the diagram. - Find the room's area and the rug's area separately. - Subtract the rug's area from the room's area.

1. From the diagram, the bedroom measures \(24\,\text{ft}\times15\,\text{ft}\), so its area is \(24 \times 15=360\,\text{ft}^2\). 2. The rug has side length \(12\,\text{ft}\), so its area is \(12 \times 12=144\,\text{ft}^2\). 3. The uncovered area is \(360-144=216\,\text{ft}^2\).

\(216\,\text{ft}^2\)

- Imagine one straight segment that makes two rectangles. - Find the area of each rectangle. - Add the two areas.

1. One decomposition uses a lower \(10\,\text{cm}\times 4\,\text{cm}\) rectangle and an upper \(4\,\text{cm}\times 8\,\text{cm}\) rectangle. 2. The total area is \(10\times 4+4\times 8=40+32=72\,\text{cm}^2\). 3. A vertical decomposition also gives \(4\times 12+6\times 4=48+24=72\,\text{cm}^2\).

\(72\,\text{cm}^2\)

- Use subtraction to find any missing side length. - Find area by multiplying side lengths or subtracting a missing square. - For perimeter, add every outside edge. - Equal areas do not necessarily mean equal perimeters.

1. Figure 1 is a rectangle, so its area is \(4\,\text{cm} \times 3\,\text{cm} = 12\,\text{cm}^2\). 2. In Figure 2, the missing right-side length is \(4 - 2 = 2\,\text{cm}\), and the missing inward horizontal length is also \(4 - 2 = 2\,\text{cm}\). 3. Figure 2 is a \(4\,\text{cm} \times 4\,\text{cm}\) square with a \(2\,\text{cm} \times 2\,\text{cm}\) square removed. Its area is \(16 - 4 = 12\,\text{cm}^2\). 4. The figures have the same area. 5. The perimeter of Figure 1 is \(2 \times (4 + 3) = 14\,\text{cm}\). 6. The perimeter of Figure 2 is \(4 + 2 + 2 + 2 + 2 + 4 = 16\,\text{cm}\). 7. Figure 2 has the greater perimeter.

a) Both figures have an area of \(12\,\text{cm}^2\). b) Figure 1 has a perimeter of \(14\,\text{cm}\), and Figure 2 has a perimeter of \(16\,\text{cm}\). Figure 2 has the greater perimeter.

- Use the diagram to identify which side becomes an interior shared edge. - Interior segments are not part of the perimeter. - Add only the labeled side lengths that remain on the outside boundary.

1. The diagram shows that the two sides labeled \(b\) are joined and lie inside the combined figure, so they are not part of the perimeter. 2. Each tile contributes sides \(a\), \(c\), and \(d\) to the outside boundary. 3. The perimeter is \(2 \times (5+7+3)=30\,\text{cm}\).

The perimeter of the combined figure is \(30\,\text{cm}\).

- Read the full garden dimensions and the three-way partitions from the diagram. - Use the equal partitions to find one small bed's dimensions. - Use those dimensions for both the area and the center-bed perimeter.

1. From the diagram, the garden is \(12\,\text{ft}\) by \(9\,\text{ft}\), with each side split into \(3\) equal parts. The small-bed dimensions are \(12 \div 3=4\,\text{ft}\) and \(9 \div 3=3\,\text{ft}\). 2. One bed has area \(4 \times 3=12\,\text{ft}^2\). 3. The center bed's perimeter is \(2 \times (4+3)=14\,\text{ft}\).

a) \(4\,\text{ft}\) by \(3\,\text{ft}\) b) \(12\,\text{ft}^2\) c) \(14\,\text{ft}\)

- Read all five side lengths from the diagram. - Add the side lengths before converting units. - Use \(10\,\text{mm}=1\,\text{cm}\) to express the total in centimeters.

1. Add the five side lengths shown: \(50+40+35+50+35=210\,\text{mm}\). 2. Since \(21 \times 10\,\text{mm}=210\,\text{mm}\), the perimeter is \(21\,\text{cm}\).

The perimeter is \(21\,\text{cm}\).

- Find area and perimeter separately for each rectangle. - Compare like quantities: area with area and perimeter with perimeter. - Use the two results to decide whether equal perimeter determines area.

1. Rectangle A has perimeter \(2 \times (6 + 4) = 20\,\text{ft}\) and area \(6 \times 4 = 24\,\text{ft}^2\). 2. Rectangle B has perimeter \(2 \times (8 + 2) = 20\,\text{ft}\) and area \(8 \times 2 = 16\,\text{ft}^2\). 3. Rectangle A has \(24 - 16 = 8\,\text{ft}^2\) more area. 4. Equal perimeters do not force rectangles to have equal areas.

Rectangle A has the greater area by \(8\,\text{ft}^2\). Rectangles can have the same perimeter but different areas.

- The wire forms the entire perimeter. - A square has four equal sides. - For a rectangle, the length plus width is half the perimeter. - In part c), first consider the two opposite long sides.

1. For the square, divide the perimeter among four equal sides: \(60 \div 4=15\,\text{cm}\). 2. For the rectangle, the length and width together equal half the perimeter: \(60 \div 2=30\,\text{cm}\). The width is \(30-20=10\,\text{cm}\). 3. Two sides of length \(35\,\text{cm}\) would require \(2 \times 35=70\,\text{cm}\), which is already longer than the wire.

a) \(15\,\text{cm}\) b) \(10\,\text{cm}\) c) Two \(35\,\text{cm}\) sides require \(70\,\text{cm}\), but the wire is only \(60\,\text{cm}\) long.

- Use \(P=4s\) and \(A=s \times s\) to complete the table. - Compare the first and third columns with multiplication factors. - Check whether the perimeter factor and area factor are the same or different.

1. For \(s=3\,\text{in.}\), \(P=4 \times 3=12\,\text{in.}\) and \(A=3 \times 3=9\,\text{in.}^2\). 2. For \(s=6\,\text{in.}\), \(P=4 \times 6=24\,\text{in.}\) and \(A=6 \times 6=36\,\text{in.}^2\). 3. For \(s=9\,\text{in.}\), \(P=4 \times 9=36\,\text{in.}\) and \(A=9 \times 9=81\,\text{in.}^2\). 4. The side length is multiplied by \(3\). The perimeter is also multiplied by \(3\), because \(3 \times 12=36\). The area is multiplied by \(9\), because \(9 \times 9=81\).

<table> <tr><td>Side length \(s\)</td><td>\(3\,\text{in.}\)</td><td>\(6\,\text{in.}\)</td><td>\(9\,\text{in.}\)</td></tr> <tr><td>Perimeter \(P\)</td><td>\(12\,\text{in.}\)</td><td>\(24\,\text{in.}\)</td><td>\(36\,\text{in.}\)</td></tr> <tr><td>Area \(A\)</td><td>\(9\,\text{in.}^2\)</td><td>\(36\,\text{in.}^2\)</td><td>\(81\,\text{in.}^2\)</td></tr> </table> The perimeter is multiplied by \(3\), and the area is multiplied by \(9\).

- Area is the product of length and width. - Think about how each side-length change affects one factor in that product. - In each case, decide whether the two side changes reinforce each other, offset each other, or act only once before calculating.

1. Tripling one side while keeping the other side unchanged triples the area: \(48\times3=144\,\text{cm}^2\). 2. Doubling the length first doubles the area to \(96\,\text{cm}^2\). Halving the width then halves the area back to \(48\,\text{cm}^2\). 3. Halving one side halves the area from \(48\,\text{cm}^2\) to \(24\,\text{cm}^2\). Halving the other side halves it again: \(24\div2=12\,\text{cm}^2\).

1) \(144\,\text{cm}^2\) 2) \(48\,\text{cm}^2\) 3) \(12\,\text{cm}^2\)

- Find the original area and then the target area. - Look for different factor pairs of the target area. - Check each proposed pair by multiplying its dimensions.

1. The original area is \(4 \times 3 = 12\,\text{m}^2\), so the new area must be \(4 \times 12 = 48\,\text{m}^2\). 2. One option is \(16\,\text{m} \times 3\,\text{m}\), which has area \(48\,\text{m}^2\). 3. Another option is \(8\,\text{m} \times 6\,\text{m}\), which also has area \(48\,\text{m}^2\).

Possible answers include \(16\,\text{m}\) by \(3\,\text{m}\), \(4\,\text{m}\) by \(12\,\text{m}\), and \(8\,\text{m}\) by \(6\,\text{m}\). Any two valid pairs are acceptable.

- Express all side lengths in inches first. - Find each area and compare. - Find each perimeter and compare.

1. Convert Rectangle B's longer side: \(2\,\text{ft}\ 6\,\text{in.}=30\,\text{in.}\). 2. Rectangle A has area \(15\times 20=300\,\text{in.}^2\). Rectangle B has area \(30\times 10=300\,\text{in.}^2\). Their areas are equal. 3. Rectangle A has perimeter \(2\times(15+20)=70\,\text{in.}\). Rectangle B has perimeter \(2\times(30+10)=80\,\text{in.}\). 4. Rectangle B has the greater perimeter.

The rectangles have equal areas of \(300\,\text{in.}^2\). Rectangle B has the greater perimeter: \(80\,\text{in.}\) compared with \(70\,\text{in.}\).

- Find the square's area first. - Use the same area and the rectangle's width to find its length. - Find and compare both perimeters.

1. The square's area is \(20\times 20=400\,\text{in.}^2\). 2. The rectangle's length is \(400\div 4=100\,\text{in.}\). 3. The square's perimeter is \(4\times 20=80\,\text{in.}\). 4. The rectangle's perimeter is \(2\times(100+4)=208\,\text{in.}\). 5. The rectangular piece has the greater perimeter.

The rectangle is \(100\,\text{in.}\) long. Its perimeter is \(208\,\text{in.}\), compared with \(80\,\text{in.}\) for the square, so the rectangle has the greater perimeter.

- For a missing side from an area, think of the side as a missing factor. - Half the perimeter equals one length plus one width. - Convert kilometers to meters before using the perimeter.

1. Rectangle 1: Since \(8 \times 12=96\), \(w=8\,\text{cm}\). Then \(P=2 \times (12+8)=40\,\text{cm}\). 2. Rectangle 2: Half of \(140\,\text{m}\) is \(70\,\text{m}\), so \(w=70-25=45\,\text{m}\). The area is \(25 \times 45=1125\,\text{m}^2\). 3. Rectangle 3: Since \(30 \times 50=1500\), \(w=30\,\text{m}\). Then \(P=2 \times (50+30)=160\,\text{m}\). 4. Rectangle 4: \(1\,\text{km}=1000\,\text{m}\), so half the perimeter is \(500\,\text{m}\). Thus \(w=500-200=300\,\text{m}\). For the area, \(200 \times 3=600\), then multiplying by \(100\) gives \(200 \times 300=60{,}000\,\text{m}^2\).

<table> <tr><th>Rectangle</th><th>Length \(l\)</th><th>Width \(w\)</th><th>Perimeter \(P\)</th><th>Area \(A\)</th></tr> <tr><td>1</td><td>\(12\,\text{cm}\)</td><td>\(8\,\text{cm}\)</td><td>\(40\,\text{cm}\)</td><td>\(96\,\text{cm}^2\)</td></tr> <tr><td>2</td><td>\(25\,\text{m}\)</td><td>\(45\,\text{m}\)</td><td>\(140\,\text{m}\)</td><td>\(1125\,\text{m}^2\)</td></tr> <tr><td>3</td><td>\(50\,\text{m}\)</td><td>\(30\,\text{m}\)</td><td>\(160\,\text{m}\)</td><td>\(1500\,\text{m}^2\)</td></tr> <tr><td>4</td><td>\(200\,\text{m}\)</td><td>\(300\,\text{m}\)</td><td>\(1\,\text{km}\)</td><td>\(60{,}000\,\text{m}^2\)</td></tr> </table>

- For a given perimeter, find the sum of one length and one width. - Convert Rectangle B's length to centimeters. - Compare the three areas after calculating them.

1. Rectangle A: Half the perimeter is \(22\div 2=11\,\text{cm}\), so the width is \(11-6=5\,\text{cm}\). Its area is \(6\times 5=30\,\text{cm}^2\). 2. Rectangle B: Convert \(80\,\text{mm}=8\,\text{cm}\). Its area is \(8\times 2=16\,\text{cm}^2\). 3. Rectangle C: Half the perimeter is \(18\div 2=9\,\text{cm}\), so the width is \(9-5=4\,\text{cm}\). Its area is \(5\times 4=20\,\text{cm}^2\). 4. Since \(30>20>16\), Rectangle A has the greatest area.

Rectangle A: width \(5\,\text{cm}\), area \(30\,\text{cm}^2\) Rectangle B: area \(16\,\text{cm}^2\) Rectangle C: width \(4\,\text{cm}\), area \(20\,\text{cm}^2\) Rectangle A has the greatest area.

- Treat Plot 1's missing width as a factor that must make the given area. - Use half of Plot 2's perimeter to find its missing length. - Compare both perimeters and both areas after the missing dimensions are known.

1. For Plot 1, \(20 \times 24=480\), so the width is \(20\,\text{ft}\). Its perimeter is \(2 \times (24+20)=88\,\text{ft}\). 2. For Plot 2, half the perimeter is \(46\,\text{ft}\). Its length is \(46-20=26\,\text{ft}\), so its area is \(26 \times 20=520\,\text{ft}^2\). 3. Plot 2 needs more fencing because \(92\,\text{ft}>88\,\text{ft}\), and it has the greater area because \(520\,\text{ft}^2>480\,\text{ft}^2\).

Plot 2 needs more fencing and has the greater area. Its perimeter is \(92\,\text{ft}\), and its area is \(520\,\text{ft}^2\).

- Half the perimeter is the sum of the length and width. - Try making one side short and the other side close to \(10\,\text{ft}\). - Check both the perimeter and area of your example.

1. A perimeter of \(20\,\text{ft}\) means that one length plus one width equals \(10\,\text{ft}\). 2. Choose side lengths \(1\,\text{ft}\) and \(9\,\text{ft}\). Their sum is \(10\,\text{ft}\), so the perimeter is \(20\,\text{ft}\). 3. The area is \(1\times 9=9\,\text{ft}^2\). 4. Since \(9\,\text{ft}^2<10\,\text{ft}^2\), such a rectangle is possible.

Yes. For example, a \(1\,\text{ft}\) by \(9\,\text{ft}\) rectangle has perimeter \(20\,\text{ft}\) and area \(9\,\text{ft}^2\).

- Find the square's side length from its area. - Divide Rectangle B's area by its width to find its length. - Compare the two distances around the rectangles.

1. Rectangle A has side length \(10\,\text{in.}\), because \(10\times 10=100\). Its perimeter is \(4\times 10=40\,\text{in.}\). 2. Rectangle B has length \(100\div 1=100\,\text{in.}\). Its perimeter is \(2\times(100+1)=202\,\text{in.}\). 3. The rectangles have the same area, but the long, narrow rectangle has a much greater perimeter.

a) \(40\,\text{in.}\) b) \(202\,\text{in.}\) c) Rectangles with the same area can have very different perimeters.

- Half the perimeter is one length plus one width. - Represent the length as five equal width parts. - Find the dimensions before calculating area.

1. Half the perimeter is \(96\div 2=48\,\text{ft}\). This equals one width plus five widths, or six equal width parts. 2. The width is \(48\div 6=8\,\text{ft}\). 3. The length is \(5\times 8=40\,\text{ft}\). 4. The area is \(40\times 8=320\,\text{ft}^2\).

The width is \(8\,\text{ft}\), the length is \(40\,\text{ft}\), and the area is \(320\,\text{ft}^2\).

- Picture the rectangle as three equal squares in a row. - Divide the total area by \(3\) to find the area of one square. - Use the square's side length as the rectangle's width.

1. Think of the rectangle as three equal squares placed in a row. Each square has area \(108\div 3=36\,\text{cm}^2\). 2. Each square has side length \(6\,\text{cm}\), because \(6\times 6=36\). This is the rectangle's width. 3. The length is \(3\times 6=18\,\text{cm}\). 4. The perimeter is \(2\times(18+6)=48\,\text{cm}\).

The width is \(6\,\text{cm}\), the length is \(18\,\text{cm}\), and the perimeter is \(48\,\text{cm}\).

- Find the square's area first. - Picture the rectangle as four equal squares in a row. - Find both perimeters and subtract.

1. The square's area is \(6\times 6=36\,\text{cm}^2\). 2. Think of the rectangle as four equal squares in a row. Each small square has area \(36\div 4=9\,\text{cm}^2\), so its side length is \(3\,\text{cm}\). 3. The rectangle's width is \(3\,\text{cm}\), and its length is \(4\times 3=12\,\text{cm}\). 4. The square's perimeter is \(4\times 6=24\,\text{cm}\). The rectangle's perimeter is \(2\times(12+3)=30\,\text{cm}\). 5. The difference is \(30-24=6\,\text{cm}\).

The rectangle is \(3\,\text{cm}\) by \(12\,\text{cm}\). Its perimeter is \(6\,\text{cm}\) greater than the square's perimeter.

- Use the pen's area and length to find its width. - The walking path extends beyond both sides of each pen dimension. - Find the perimeter of the larger rectangle.

1. The pen's width is \(24 \div 6 = 4\,\text{ft}\). 2. The path is \(2\,\text{ft}\) beyond each side, so its dimensions are \(6 + 2 + 2 = 10\,\text{ft}\) by \(4 + 2 + 2 = 8\,\text{ft}\). 3. The path's perimeter is \(2 \times (10 + 8) = 36\,\text{ft}\).

a) \(4\,\text{ft}\) b) \(36\,\text{ft}\)

- Find the square's side length from its perimeter. - Use the square's area as the rectangle's area. - Find the rectangle's missing length before comparing perimeters.

1. The square's side length is \(48 \div 4 = 12\,\text{ft}\). 2. Its area is \(12 \times 12 = 144\,\text{ft}^2\). 3. The rectangle's length is \(144 \div 9 = 16\,\text{ft}\). 4. The rectangle's perimeter is \(2 \times (16 + 9) = 50\,\text{ft}\). 5. The difference is \(50\,\text{ft} - 48\,\text{ft} = 2\,\text{ft}\).

The rectangle's perimeter is \(2\,\text{ft}\) greater.

- Find the patio area and the area of one tile in the same unit. - Divide to find the number of tiles. - Convert the tile count to packages before finding the cost.

1. The patio area is \(12\times 8=96\,\text{ft}^2\). 2. One tile covers \(2\times 2=4\,\text{ft}^2\). 3. The number of tiles needed is \(96\div 4=24\). 4. The number of packages is \(24\div 6=4\). 5. The total cost is \(4\times\$24.50=\$98.00\).

The total cost is \(\$98.00\).

- Use half the rectangle's perimeter to find the missing side. - Use place value to break the rectangle's area multiplication into easier steps. - A square divides its perimeter equally among four sides; then compare the two areas.

1. Half the rectangle's perimeter is \(200\,\text{ft}\), so the other side is \(200-120=80\,\text{ft}\). 2. For the rectangle, \(120 \times 8=960\), then multiplying by \(10\) gives \(120 \times 80=9600\,\text{ft}^2\). 3. If the path is a square, each side is \(100\,\text{ft}\). One hundred groups of \(100\) make \(10{,}000\), so the square's area is \(10{,}000\,\text{ft}^2\). 4. The difference is \(10{,}000-9600=400\,\text{ft}^2\).

a) \(9600\,\text{ft}^2\) b) \(10{,}000\,\text{ft}^2\) c) \(400\,\text{ft}^2\)

- List whole-number pairs that add to half the string length. - Compare the products of very unequal and nearly equal pairs. - Include the square in your list.

1. Half the perimeter is \(36\div 2=18\,\text{in.}\), so the length and width must have a sum of \(18\). 2. The most unequal positive whole-number pair is \(1\) and \(17\). Its area is \(1\times 17=17\,\text{in.}^2\). 3. The most equal pair is \(9\) and \(9\). Its area is \(9\times 9=81\,\text{in.}^2\).

a) \(1\,\text{in.}\) by \(17\,\text{in.}\), with area \(17\,\text{in.}^2\) b) \(9\,\text{in.}\) by \(9\,\text{in.}\), with area \(81\,\text{in.}^2\)

- Find the square's side length from its area. - The inside strip reduces both dimensions of the open square. - Find the habitat area before applying the stated cost per square foot.

1. The original side length is \(50\,\text{ft}\), because \(50 \times 50 = 2500\). 2. The inner side length is \(50 - 2 - 2 = 46\,\text{ft}\). 3. The remaining meadow area is \(46 \times 46 = 2116\,\text{ft}^2\). 4. The habitat area is \(2500 - 2116 = 384\,\text{ft}^2\). 5. The estimated loss is \(384 \times \$0.20 = \$76.80\).

a) \(50\,\text{ft}\) b) \(2116\,\text{ft}^2\) c) \(\$76.80\)

- Find the total rectangular area. Use the trailing zeros to simplify the multiplication. - Subtract the border width from both sides of each dimension. - Find the inner area, then subtract it from the total area.

1. Use place value to find the total area: \(220\times 90=(22\times 9)\times 100=19{,}800\,\text{ft}^2\). 2. The inner dimensions are \(220-10=210\,\text{ft}\) and \(90-10=80\,\text{ft}\). 3. Use place value again: \(210\times 80=(21\times 8)\times 100=16{,}800\,\text{ft}^2\). 4. Subtract the inner area from the total area. The difference between \(19{,}800\,\text{ft}^2\) and \(16{,}800\,\text{ft}^2\) is \(3000\,\text{ft}^2\).

a) \(19{,}800\,\text{ft}^2\) b) \(3000\,\text{ft}^2\)

- Use multiplication to find the missing width of Field B. - Subtract the border width twice from each dimension. - Find each border area by subtracting the inner area from the total area. - Break apart a large product using a nearby multiple of \(10\) when useful.

1. Since \(250\times 40=10{,}000\), Field B's width is \(40\,\text{ft}\). 2. Field A is \(100\,\text{ft}\) by \(100\,\text{ft}\). Its inner dimensions are \(98\,\text{ft}\) by \(98\,\text{ft}\), and \(98\times 98=9604\,\text{ft}^2\). The difference between the total area of \(10{,}000\,\text{ft}^2\) and the inner area of \(9604\,\text{ft}^2\) is \(396\,\text{ft}^2\). 3. Field B's inner dimensions are \(248\,\text{ft}\) by \(38\,\text{ft}\). Use the distributive property: \(248\times 38=248\times 40-248\times 2=9920-496=9424\,\text{ft}^2\). The difference between the total area of \(10{,}000\,\text{ft}^2\) and the inner area of \(9424\,\text{ft}^2\) is \(576\,\text{ft}^2\). 4. Field B loses more usable area.

a) \(40\,\text{ft}\) b) Field A: \(396\,\text{ft}^2\); Field B: \(576\,\text{ft}^2\). Field B loses more usable area.

- Add the path width on both sides of each patio dimension. - Subtract the patio area from the outer area. - Multiply the path area by the cost per square foot.

1. The outer dimensions are \(8+2+2=12\,\text{ft}\) and \(5+2+2=9\,\text{ft}\). 2. The outer area is \(12\times 9=108\,\text{ft}^2\). 3. The patio area is \(8\times 5=40\,\text{ft}^2\). 4. The path area is \(108-40=68\,\text{ft}^2\). 5. The total cost is \(68\times\$15=\$1020\).

The pavers cost \(\$1020\).

- Add the deck width on both sides of each pool dimension. - Subtract the pool area from the outer area. - Multiply the deck area by the weight per square foot. - Compare the result with the truck's capacity.

1. The outer dimensions are \(10+1.5+1.5=13\,\text{ft}\) and \(5+1.5+1.5=8\,\text{ft}\). 2. The outer area is \(13\times 8=104\,\text{ft}^2\). 3. The pool area is \(10\times 5=50\,\text{ft}^2\), so the deck area is \(104-50=54\,\text{ft}^2\). 4. The tile weight is \(54\times 25=1350\,\text{lb}\). 5. Since \(1350\,\text{lb}>1000\,\text{lb}\), the truck cannot carry all the tiles in one trip.

The tiles weigh \(1350\,\text{lb}\). No, the truck cannot carry them all in one trip.

- List every whole-number factor pair of \(24\). - Find the perimeter for each rectangle. - Compare the results systematically.

1. The whole-number factor pairs of \(24\) are \(1\times 24\), \(2\times 12\), \(3\times 8\), and \(4\times 6\). 2. Their perimeters are \(2\times(1+24)=50\,\text{ft}\), \(2\times(2+12)=28\,\text{ft}\), \(2\times(3+8)=22\,\text{ft}\), and \(2\times(4+6)=20\,\text{ft}\). 3. The \(4\,\text{ft}\) by \(6\,\text{ft}\) patio has the least perimeter.

a) \(1\,\text{ft}\times 24\,\text{ft}\), \(2\,\text{ft}\times 12\,\text{ft}\), \(3\,\text{ft}\times 8\,\text{ft}\), and \(4\,\text{ft}\times 6\,\text{ft}\) b) \(50\,\text{ft}\), \(28\,\text{ft}\), \(22\,\text{ft}\), and \(20\,\text{ft}\), respectively c) \(4\,\text{ft}\) by \(6\,\text{ft}\)

- Use place value to find the deck area without treating \(120 \times 200\) as one large algorithm step. - Treat each space count as a missing factor in an area multiplication equation. - Use \(144\,\text{in.}^2=1\,\text{ft}^2\) to interpret the motorcycle-space area.

1. First, \(120 \times 2=240\). Since \(200=2 \times 100\), the deck area is \(24{,}000\,\text{ft}^2\). 2. Since \(80 \times 300\,\text{ft}^2=24{,}000\,\text{ft}^2\), the deck could theoretically fit \(80\) car spaces. 3. Since \(100 \times 144\,\text{in.}^2=14{,}400\,\text{in.}^2\), one motorcycle space is \(100\,\text{ft}^2\). 4. Since \(240 \times 100\,\text{ft}^2=24{,}000\,\text{ft}^2\), the deck could theoretically fit \(240\) motorcycle spaces.

Under the simplified assumption, the deck could fit \(80\) cars or \(240\) motorcycles.

- Treat each missing side as a factor that must produce the given area. - After shortening the \(60\)-centimeter side, use the same area to determine the new other side. - Find both perimeters before comparing them.

1. Since \(40 \times 60=2400\), the original missing side is \(40\,\text{cm}\). 2. The shortened side is \(60-10=50\,\text{cm}\). 3. Since \(48 \times 50=2400\), the new other side is \(48\,\text{cm}\). 4. The original perimeter is \(2 \times (60+40)=200\,\text{cm}\). The new perimeter is \(2 \times (50+48)=196\,\text{cm}\). 5. The new perimeter is \(200-196=4\,\text{cm}\) less.

a) \(40\,\text{cm}\) b) \(48\,\text{cm}\) c) The perimeter decreases by \(4\,\text{cm}\), from \(200\,\text{cm}\) to \(196\,\text{cm}\).

- Find whole-number factor pairs of \(24\). - Test the change using one pair from part a. - Think about what happens to a product when one factor doubles and the other is halved.

1. Whole-number factor pairs of \(24\) include \(24\) and \(1\), \(12\) and \(2\), \(8\) and \(3\), and \(6\) and \(4\). Any three pairs work. 2. Doubling the length multiplies the area by \(2\). Halving the width multiplies the area by \(\frac{1}{2}\). 3. The combined factor is \(2\times\frac{1}{2}=1\), so the area remains \(24\,\text{ft}^2\).

a) Possible pairs include \(8\,\text{ft}\times 3\,\text{ft}\), \(6\,\text{ft}\times 4\,\text{ft}\), and \(12\,\text{ft}\times 2\,\text{ft}\). b) The area stays \(24\,\text{ft}^2\) because doubling one factor and halving the other cancel each other.

- For subtraction, use the enclosing rectangle and the square cutout. - For addition, find the height below the cutout and split the U into three nonoverlapping rectangles. - Include the inner edges of the cutout when finding perimeter.

1. Subtraction method: The enclosing rectangle has area \(10\times 6=60\,\text{cm}^2\). The cutout has area \(4\times 4=16\,\text{cm}^2\). The template area is \(60-16=44\,\text{cm}^2\). 2. Addition method: The two side rectangles each have area \(3\times 6=18\,\text{cm}^2\). The bottom middle height is \(6-4=2\,\text{cm}\), so the bottom middle rectangle has area \(4\times 2=8\,\text{cm}^2\). The total is \(18+18+8=44\,\text{cm}^2\). 3. Add all eight boundary lengths: \(10+6+3+4+4+4+3+6=40\,\text{cm}\).

a) \(44\,\text{cm}^2\) by either method b) \(40\,\text{cm}\)

- Read Rectangle A's dimensions from the diagram and find its area. - Find a square side length whose product with itself gives that same area. - Calculate and compare the two perimeters.

1. From the diagram, Rectangle A measures \(18\,\text{ft}\times8\,\text{ft}\), so its area is \(18 \times 8=144\,\text{ft}^2\). 2. Square B has the same area. Since \(12 \times 12=144\), \(s=12\,\text{ft}\). 3. Rectangle A has perimeter \(2 \times (18+8)=52\,\text{ft}\). 4. Square B has perimeter \(4 \times 12=48\,\text{ft}\). 5. Square B needs \(52-48=4\,\text{ft}\) less fencing.

a) \(144\,\text{ft}^2\) b) \(s=12\,\text{ft}\) c) Rectangle A: \(52\,\text{ft}\); Square B: \(48\,\text{ft}\). Square B saves \(4\,\text{ft}\) of fencing.

- Split Figure 1 into three rectangles. - Count the entire boundary of each figure, including the inside edges of the U-shape. - Compare how compact the two shapes are.

1. Figure 1 can be split into two \(1\times 4\) rectangles and one \(2\times 1\) rectangle. Its area is \(1\times 4+1\times 4+2\times 1=10\) square units. Figure 2 is a \(5\times 2\) rectangle, so its area is also \(10\) square units. 2. Adding all outer and inner boundary segments of Figure 1 gives \(4+4+1+3+2+3+1+4=22\) length units. 3. Figure 2 has perimeter \(2\times(5+2)=14\) length units. 4. Figure 1 has an indentation, which adds boundary segments without adding area. The compact rectangle has a shorter boundary.

a) \(10\) square units b) Figure 1: \(22\) length units Figure 2: \(14\) length units c) Figure 1's indentation creates additional boundary segments, so it has a greater perimeter.

- Find the full rectangle's area, then subtract the notch. - The notch adds boundary segments to the perimeter. - Count and add all eight sides.

1. The original rectangle measures \(20\,\text{cm}\times 12\,\text{cm}\), so its area is \(240\,\text{cm}^2\). The notch measures \(6\,\text{cm}\times 4\,\text{cm}\), so its area is \(24\,\text{cm}^2\). The remaining area is \(240-24=216\,\text{cm}^2\). 2. Adding all boundary segments, including the three sides of the notch, gives \(20+4+6+4+6+4+20+12=76\,\text{cm}\).

Area: \(216\,\text{cm}^2\) Perimeter: \(76\,\text{cm}\)

- Try one horizontal dividing line and one vertical dividing line. - Make sure the rectangles in each decomposition cover the shaded cells without overlap.

1. One decomposition gives a \(6 \times 3\) rectangle and a \(2 \times 7\) rectangle: \(18 + 14 = 32\) square units. 2. Another decomposition gives an \(8 \times 3\) rectangle and a \(2 \times 4\) rectangle: \(24 + 8 = 32\) square units.

The area is \(32\) square units. Two decompositions are \(6 \times 3 + 2 \times 7\) and \(8 \times 3 + 2 \times 4\).